Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

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5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 2

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 3

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 4

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 5

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 6

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 7

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 8

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 9

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 10

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 11

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 12

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 13

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 14

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 15

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 16

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 17

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 18

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 19

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 20

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 21

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 22

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 23

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 24

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 25

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 26

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 27

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 28

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 29

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 30

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 31

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 32

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 33

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 34

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 35

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 36

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 37

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 38

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 39

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 40

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 41

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 42

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 43

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 44

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 45

5-6 Graphing Inequalities in Two Variables

Graph each inequality.

1. y > x + 3

SOLUTION:  

y > x + 3   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 3, shade the half–plane that does not contain (0, 0).  

2. y ≥ −8

SOLUTION:  

y ≥ −8   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  0 ≥ −8   Since 0 is greater than or equal to −8, shade the half–plane that contains (0, 0).  

3. x + y > 1

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 1, shade the half–plane that does not contain (0, 0).  

4. y ≤ x − 6

SOLUTION:  

y ≤ x − 6   Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is not less than or equal to −6, shade the half–plane that does not contain (0, 0).  

5. y < 2x − 4

SOLUTION:  

y < 2x − 4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −4, shade the half–plane that does not contain (0, 0).  

6. x − y ≤ 4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 4, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

7. 7x + 1 < 15

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

  Because the inequality involves <, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since 1 is less than 15, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 2.  

8. −3x − 2 ≥ 11

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and get 0 on a side by itself.  

 

Because the inequality involves ≥, graph x = with a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –2 is not greater than or equal to 11, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

9. 3y − 5 ≤ 34

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y = 13 using a solid line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –5 is less than or equal 34, shade the half–plane that contains the point (0, 0).

Notice that the y–intercept of the graph is at 13. Since the half–plane below the y–intercept is shaded, the solution is y ≤ 13.  

 

10. 4y − 21 > 1

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves >, graph using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since –21 not is greater than 1, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 5 . Since the half–plane above the y–intercept is shaded, the solution

is .

 

11. FINANCIAL LITERACY  The surf shop has a weekly overhead of $2300. a.  Write an inequality to represent the number of skimboards and longboards the shop sells each week to make a profit. b.  How many skimboards and longboards must the shop sell each week to make a profit?  

SOLUTION:  a. Let x represent the skimboards and y represent the longboards. 115x + 685y > 2300 b. To graph the inequality, first solve for x.  

  Graph. then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0 > 2300 which is false. Shade the side that does not include the origin.   

  One possible solution would be to sell 1 skimboard. Calculate how many longboards must be sold to make a profit when just 1 skimboard is sold.   

  So, in order to make a profit, if the shop sells 1 skimboard, then it must sell at least 4 longboards. Thus, one possible solution is 1 skimboard and 4 longboards.

Graph each inequality.

12. y < x − 3

SOLUTION:  

y < x − 3   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

13. y > x + 12

SOLUTION:  

y > x + 12   Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than 12, shade the half–plane that does not contain (0, 0).  

14. y ≥ 3x − 1

SOLUTION:  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −1, shade the half–plane that contains (0, 0).  

 

15. y ≤ −4x + 12

SOLUTION:  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 12, shade the half–plane that contains (0, 0).  

16. 6x + 3y > 12

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 4, shade the half–plane that does not contain (0, 0).  

17. 2x + 2y < 18

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 9, shade the half–plane that contains (0, 0).  

18. 5x + y > 10

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not greater than 10, shade the half–plane that does not contain (0, 0).  

19. 2x + y < −3

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is not less than −3, shade the half–plane that does not contain (0, 0).  

20. −2x + y ≥ −4

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −4, shade the half–plane that contains (0, 0).  

21. 8x + y ≤ 6

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 6, shade the half–plane that contains (0, 0).  

22. 10x + 2y ≤ 14

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is less than or equal to 7, shade the half–plane that contains (0, 0).  

23. −24x + 8y ≥ −48

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point. 

  Since 0 is greater than or equal to −6, shade the half–plane that contains (0, 0).  

Use a graph to solve each inequality.

24. 10x − 8 < 22

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves <, graph x = 3 using a dashed line. Choose (0, 0) as a test point in the original inequality.  

.

  Since –8 is less than 22, shade the half–plane that contains the point (0, 0).

Notice that the x–intercept of the graph is at 3. Since the half–plane to the left of the x–intercept is shaded, the solution is x < 3.  

25. 20x − 5 > 35

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for x.  

  Because the inequality involves >, graph x = 2 using a dashed line. Choose (0, 0) as a test point in the original equation.  

  Since –5 is not greater than 35, shade the half–plane that does not contain the point (0, 0).

Notice that the x–intercept of the graph is at 2. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 2.  

26. 4y − 77 ≥ 23

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign, and solve for y .  

  Because the inequality involves ≥, graph y = 25 using a solid line. Choose (0, 0) as a test point in the original equation.  

  Since –77 is not greater than or equal 23, shade the half–plane that does not contain the point (0, 0).

Notice that the y–intercept of the graph is at 25. Since the half–plane above the y–intercept is shaded, the solution is y ≥ 25.  

 

27. 5y + 8 ≤ 33

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for y .  

  Because the inequality involves ≤, graph y   = 5 using a solid line. Choose (0, 0) as a test point in the original equation.  

.

  Since 8 is less than or equal 33, shade the half–plane that contains the point (0, 0).  

  Notice that the y–intercept of the graph is at 5. Since the half–plane under the y–intercept is shaded, the solution is y≤ 5.  

28. 35x + 25 < 6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since 25 is not less than 6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

29. 14x − 12 > −31

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves >, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original equation.  

.

  Since –12 is greater than –31, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

30. DECORATING  Sybrina is decorating her bedroom. She has $300 to spend on paint and bed linens. A gallon of paint costs $14, while a set of bed linens costs $60. a.  Write an inequality for this situation. b.  How many gallons of paint and bed linen sets can Sybrina buy and stay within her budget?

SOLUTION:  a. Let x represent the number of gallons of paint and y represent the number of sets of bed linens. 14x + 60y  ≤ 300 b. To graph the inequality, solve for x first.   

  Graph. Then use the test point (0, 0) to determine the shading. The point (0, 0) results in 0≤ 5 which is true. Shadethe side that includes the origin.   

  One possible solution would be to buy 5 gallons of paint Calculate how many bed linen sets must be purchased when 5 gallons of paint are bought.   

  Thus when 5 gallons of paint are purchased, 3 bed linen sets can be purchased. 

Use a graph to solve each inequality.

31. 3x + 2 < 0

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

 

Because the inequality involves <, graph  with a dashed line.

  Choose (0, 0) as a test point in the original inequality.  

.

  Since 2 is not less than 0, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

32. 4x − 1 > 3

SOLUTION:  First, graph the boundary, which is the related equation. Replace the inequality sign with an equals sign and solve for x.  

  Because the inequality involves >, graph x = 1 with a dashed line.   Choose (0, 0) as a test point in the original inequality.  

.

  Since –1 is not greater than 3, shade the half–plane that does not contain the point (0, 0).  

  Notice that the x–intercept of the graph is at 1. Since the half–plane to the right of the x–intercept is shaded, the solution is x > 1.  

33. −6x − 8 ≥ −4

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≥, graph x = – using a solid line.

Choose (0, 0) as a test point in the original inequality.  

  Since –8 is not greater than or equal to –4, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

 

34. −5x + 1 < 3

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves <, graph x = – using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 1 is less than 3, shade the half–plane that contains the point (0, 0).  

 

Notice that the x–intercept of the graph is at – . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

35. −7x + 13 < 10

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

Because the inequality involves <, graph x = using a dashed line.

Choose (0, 0) as a test point in the original inequality.  

  Since 13 is not less than 10, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the right of the x–intercept is shaded, the

solution is .

 

36. −4x − 4 ≤ −6

SOLUTION:  Graph the boundary, which is the related function. Replace the inequality sign with an equal sign and solve for x.  

 

Because the inequality involves ≤, graph x = using a solid line.

Choose (0, 0) as a test point in the original inequality.  

.

  Since –4 is not less than or equal to –6, shade the half–plane that does not contain the point (0, 0).  

 

Notice that the x–intercept of the graph is at . Since the half–plane to the left of the x–intercept is shaded, the

solution is .

     

37. SOCCER  The girls’ soccer team wants to raise $2000 to buy new goals. How many of each item must they sell to buy the goals?

a.  Write an inequality that represents this situation. b.  Graph this inequality. c.  Make a table of values that shows at least five possible solutions. d.  Plot the solutions from part c.

SOLUTION:  

a. Let x represent the number of hot dogs and let y represent the number of sodas. x + 1.25y ≥ 2000

b. c. Sample answer:

d. Sample points should be in the shaded region of the graph in part b.

x y 400 1600 200 1500 300 1400 400 1300 1000 1000

Graph each inequality. Determine which of the ordered pairs are part of the solution set for each inequality.

38. y ≥ 6; {(0, 4), (−2, 7), (4, 8), (−4, −8), (1, 6)}

SOLUTION:  

y ≥ 6   Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.   0 6   Since 0 is not greater than or equal to 6, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (–2, 7), (4, 8) and (1, 6) are in the shaded half–plane or on the boundary line, they are part of the solution set.

39. x < −4; {(2, 1), (−3, 0), (0, −3), (−5, −5), (−4, 2)}

SOLUTION:  

y < –4   Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point.  0 –4   Since 0 is not less than –4, shade the half–plane that does not contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (–5, –5) is in the shaded half–plane, it is part of the solution set.

40. 2x − 3y ≤ 1; {(2, 3), (3, 1), (0, 0), (0, −1), (5, 3)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is greater than or equal to , shade the half–plane that contains (0, 0).

 

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (2, 3), (0, 0), and (5, 3) are in the shaded half–plane or on the boundary line, they are part of the solution set.

41. 5x + 7y ≥ 10; {(−2, −2), (1, −1), (1, 1), (2, 5), (6, 0)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

 

Since 0 is not greater than or equal to , shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

    Because (1, 1), (2, 5), and (6, 0) are in the shaded half–plane or on the boundary line, they are part of the solution set.

42. −3x + 5y < 10; {(3, −1), (1, 1), (0, 8), (−2, 0), (0, 2)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves <, graph the boundary using a dashed line. Choose (0, 0) as a test point. 

  Since 0 is less than 10, shade the half–plane that contains (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (3, –1), (1, 1), and (2, 0) are in the shaded half–plane, they are part of the solution set.

43. 2x − 2y ≥ 4; {(0, 0), (0, 7), (7, 5), (5, 3), (2, −5)}

SOLUTION:  Solve for y in terms of x.  

  Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.  

  Since 0 is not greater than or equal to 4, shade the half–plane that does not contain (0, 0).  

  To determine which of the ordered pairs are part of the solution set, plot them on the graph.   

  Because (7, 5), (5, 3), and (2, –5) are in the shaded half–plane or on the boundary line, they are part of the solution set.

44. RECYCLING  Mr. Jones would like to spend no more than $37.50 per week on recycling. A curbside recycling service will remove up to 50 pounds of plastic bottles and paper products per week. They charge $0.25 per pound of plastic and $0.75 per pound of paper products. a.  Write an inequality that describes the number of pounds of each product that can be included in the curbside service. b.  Write an inequality that describes Mr. Jones’ weekly cost for the service if he stays within his budget. c.  Graph an inequality for the weekly costs for the service.  

SOLUTION:  a. Let x represent the pounds of plastic and y represent the pounds of paper products. They will remove up to 50 pounds, so use ≤.   x + y  ≤ 50 b.  The sum of the costs for paper and plastic products must be less than or equal to $37.50 per week.  0.25x + 0.75y  ≤ 37.50 c.

 

45. MULTIPLE REPRESENTATIONS  Use inequalities A and B to investigate graphing compound inequalities on acoordinate plane. A.  7(y + 6) ≤ 21x + 14 B.  −3y ≤ 3x −12 a.  NUMERICAL  Solve each inequality for y . b.  GRAPHICAL  Graph both inequalities on one graph. Shade the half–plane that makes A true in red. Shade the half–plane that makes B true in blue. c.  VERBAL  What does the overlapping region represent?

SOLUTION:  a. For equation A:

For equation B:

b.

c. The overlapping region represents the solutions that make both A and B true.

46. ERROR ANALYSIS  Reiko and Kristin are solving  by graphing. Is either of them correct? Explain your

reasoning.

SOLUTION:  

Kristin used a test point located on the line and shaded the incorrect half-plane. Reiko is correct.

47. CHALLENGE Write a linear inequality for which (–1, 2), (1, 1), and (3, –4) are solutions but (0, 1) is not.

SOLUTION:  We want to find a linear inequality for which three points are solutions and one point is not a solution. The best way to find this is to plot the points on the coordinate plane and visualize the line that would be satisfactory.   Plot all of the points and label (1,1) with an X to separate it from the others.  

From the graph, it looks like the three points that are included in the solution almost form a straight line. It looks like agood line to use would be the one that includes (–1, 2) and (0, 1). It also looks like point X would be above this line and the other point would be below it. Therefore, the linear inequality should be close to this line and the shading should be below it.  

Using points (–1, 2) and (0, 1), you can determine the equation for the line to be . The points on the lineare a part of the solution and we are shading below the line, so the inequality is .

48. REASONING  Explain why a point on the boundary should not be used as a test point.

SOLUTION:  

The test point is used to determine which half–plane is the solution set for the inequality. A test point on the boundary does not show which half–plane contains the points that make the inequality true.

49. OPEN ENDED  Write a two-variable inequality with a restricted domain and range to represent a real-world situation. Give the domain and range, and explain why they are restricted.

SOLUTION:  

Sample answer: The inequality y > 10x + 45 represents the cost of a monthly smartphone data plan with a flat rate of $45 for the first 2 GB of data used, plus $10 per each additional GB of data used. Both the domain and range are nonnegative real numbers because the GB used, and the total cost cannot be negative.  

50. WRITING IN MATH  Summarize the steps to graph an inequality in two variables.

SOLUTION:  

Sample answer: First solve the inequality for y . Then change the inequality sign to an equal sign and graph the boundary. If < or > is used, the boundary is not included in the graph and the line is dashed. Otherwise, the boundary is included and the line is solid. Then choose a test point not on the boundary. Substitute the coordinates of the test point into the original inequality. If the result makes the inequality true, then shade the half-plane that includes the test point. If the result makes the inequality false, shade the half-plane that does not include the test point. Lastly, check your solution by choosing a test point that is in the half-plane that is not shaded. This second test point should make the inequality false if the solution is correct.

51. What is the domain of this function?

A  {x | 0 ≤ x ≤ 3} B  {x | 0 ≤ x ≤ 9} C  {y | 0 ≤ y ≤ 9} D  {y | 0 ≤ y ≤ 3}

SOLUTION:  The domain of a function is the x–values, which means that you can delete choices C and D. This graph shows values from 0 to 9, so the correct choice is B.

52. EXTENDED RESPONSE  An arboretum will close for the winter when all of the trees have lost their leaves. The table shows the number of trees each day that still have leaves.  

  a.  Write an equation that represents the number of trees with leaves y after d days. b.  Find the y-intercept. What does it mean in the context of this problem? c.  After how many days will the arboretum close? Explain how you got your answer.

SOLUTION:  a. Choose two points (5, 325) and (20, 130). Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in slope–intercept form.  

  b. The y–intercept is 390. This represents the number of trees in the arboretum before any have lost their leaves.c. To determine when the arboretum will close, let y = 0.  

 

  The arboretum will close after 30 days.

53. Which inequality best represents the statement below? A jar contains 832 gumballs. Ebony’s guess was within 46 pieces. F  |g − 832| ≤ 46 G  |g + 832| ≤ 46 H  |g − 832| ≥ 46 J  |g + 832| ≥ 46

SOLUTION:  The guess was within 46, means less than or equal to 46, which eliminates choices H and J. For |g + 832 | ≤ 46, g +

832 ≤ 46 is one solution, which makes g ≤ -786. The number of gumballs cannot be negative, so G is eliminated.   The correct choice is F.

54. GEOMETRY  If the rectangular prism has a volume of 10,080 cm3, what is the value of x?

A  12 B  14 C  16 D  18

SOLUTION:  Volume of a rectangular prism is equal to length times width times height. 

  So, the correct choice is D.

Solve each open sentence.

55. |y − 2| > 4

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {y | y > 6 or y < −2}.

56. |t − 6| ≤ 5

SOLUTION:  Case 1:

  Case 2:

  So the solution set is {t | 1 ≤ t ≤ 11}.

57. |3 + d| < −4

SOLUTION:  This problem has an absolute value equal to a negative number. That means that the distance between 3 and d is equal to –4, but distances cannot be negative, so there is no solution. So the solution set is an empty set, Ø.

Solve each compound inequality.

58. 4c − 4 < 8c − 16 < 6c − 6

SOLUTION:  SOLUTION:  

  and   

              

  So, the solution set is {c | 3 < c < 5}.

59. 

SOLUTION:  

and             

  So, the solution set is {p | 4 < p < 10}.

60. 0.5n ≥ −7 or 2.5n + 2 ≤ 9

SOLUTION:  

or             

  Notice the two inequalities overlap so that every point is a solution. So, the solution set is {n | n is any real number}.

Write an equation of the line that passes through each pair of points.

61. (1, −3) and (2, 5)

SOLUTION:  Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept. 

  Write the equation in slope–intercept form.  

62. (−2, −4) and (−7, 3)

SOLUTION:  Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept. 

  Write the equation in slope–intercept form.  

63. (−6, −8) and (−8, −5)

SOLUTION:  Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept. 

  Write the equation in slope–intercept form.  

64. FITNESS  The table shows the maximum heart rate to maintain during aerobic activities. Write an equation in function notation for the relation. Determine what would be the maximum heart rate to maintain in aerobic training for an 80-year-old.  

SOLUTION:  Choose two points: (20, 175) and (70, 130). Find the slope of the line containing the given points. 

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in function notation.  

  To determine what the maximum heart rate is for an 80 year–old, substitute 80 in for x.  

  The maximum heart rate for an 80-year-old is 121 beats/min.

65. WORK  The formula  is used to find keyboarding speeds. In the formula, s represents the speed in

words per minute, w the number of words typed, r  the number of errors, and m the number of minutes typed. Solve for r.

SOLUTION:  

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5-6 Graphing Inequalities in Two Variables

SOLUTION:  

  and   

              

  So, the solution set is {c | 3 < c < 5}.

59. 

SOLUTION:  

and             

  So, the solution set is {p | 4 < p < 10}.

60. 0.5n ≥ −7 or 2.5n + 2 ≤ 9

SOLUTION:  

or             

  Notice the two inequalities overlap so that every point is a solution. So, the solution set is {n | n is any real number}.

Write an equation of the line that passes through each pair of points.

61. (1, −3) and (2, 5)

SOLUTION:  Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept. 

  Write the equation in slope–intercept form.  

62. (−2, −4) and (−7, 3)

SOLUTION:  Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept. 

  Write the equation in slope–intercept form.  

63. (−6, −8) and (−8, −5)

SOLUTION:  Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept. 

  Write the equation in slope–intercept form.  

64. FITNESS  The table shows the maximum heart rate to maintain during aerobic activities. Write an equation in function notation for the relation. Determine what would be the maximum heart rate to maintain in aerobic training for an 80-year-old.  

SOLUTION:  Choose two points: (20, 175) and (70, 130). Find the slope of the line containing the given points. 

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in function notation.  

  To determine what the maximum heart rate is for an 80 year–old, substitute 80 in for x.  

  The maximum heart rate for an 80-year-old is 121 beats/min.

65. WORK  The formula  is used to find keyboarding speeds. In the formula, s represents the speed in

words per minute, w the number of words typed, r  the number of errors, and m the number of minutes typed. Solve for r.

SOLUTION:  

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5-6 Graphing Inequalities in Two Variables

SOLUTION:  

  and   

              

  So, the solution set is {c | 3 < c < 5}.

59. 

SOLUTION:  

and             

  So, the solution set is {p | 4 < p < 10}.

60. 0.5n ≥ −7 or 2.5n + 2 ≤ 9

SOLUTION:  

or             

  Notice the two inequalities overlap so that every point is a solution. So, the solution set is {n | n is any real number}.

Write an equation of the line that passes through each pair of points.

61. (1, −3) and (2, 5)

SOLUTION:  Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept. 

  Write the equation in slope–intercept form.  

62. (−2, −4) and (−7, 3)

SOLUTION:  Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept. 

  Write the equation in slope–intercept form.  

63. (−6, −8) and (−8, −5)

SOLUTION:  Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept. 

  Write the equation in slope–intercept form.  

64. FITNESS  The table shows the maximum heart rate to maintain during aerobic activities. Write an equation in function notation for the relation. Determine what would be the maximum heart rate to maintain in aerobic training for an 80-year-old.  

SOLUTION:  Choose two points: (20, 175) and (70, 130). Find the slope of the line containing the given points. 

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in function notation.  

  To determine what the maximum heart rate is for an 80 year–old, substitute 80 in for x.  

  The maximum heart rate for an 80-year-old is 121 beats/min.

65. WORK  The formula  is used to find keyboarding speeds. In the formula, s represents the speed in

words per minute, w the number of words typed, r  the number of errors, and m the number of minutes typed. Solve for r.

SOLUTION:  

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5-6 Graphing Inequalities in Two Variables

SOLUTION:  

  and   

              

  So, the solution set is {c | 3 < c < 5}.

59. 

SOLUTION:  

and             

  So, the solution set is {p | 4 < p < 10}.

60. 0.5n ≥ −7 or 2.5n + 2 ≤ 9

SOLUTION:  

or             

  Notice the two inequalities overlap so that every point is a solution. So, the solution set is {n | n is any real number}.

Write an equation of the line that passes through each pair of points.

61. (1, −3) and (2, 5)

SOLUTION:  Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept. 

  Write the equation in slope–intercept form.  

62. (−2, −4) and (−7, 3)

SOLUTION:  Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept. 

  Write the equation in slope–intercept form.  

63. (−6, −8) and (−8, −5)

SOLUTION:  Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept. 

  Write the equation in slope–intercept form.  

64. FITNESS  The table shows the maximum heart rate to maintain during aerobic activities. Write an equation in function notation for the relation. Determine what would be the maximum heart rate to maintain in aerobic training for an 80-year-old.  

SOLUTION:  Choose two points: (20, 175) and (70, 130). Find the slope of the line containing the given points. 

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in function notation.  

  To determine what the maximum heart rate is for an 80 year–old, substitute 80 in for x.  

  The maximum heart rate for an 80-year-old is 121 beats/min.

65. WORK  The formula  is used to find keyboarding speeds. In the formula, s represents the speed in

words per minute, w the number of words typed, r  the number of errors, and m the number of minutes typed. Solve for r.

SOLUTION:  

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5-6 Graphing Inequalities in Two Variables

SOLUTION:  

  and   

              

  So, the solution set is {c | 3 < c < 5}.

59. 

SOLUTION:  

and             

  So, the solution set is {p | 4 < p < 10}.

60. 0.5n ≥ −7 or 2.5n + 2 ≤ 9

SOLUTION:  

or             

  Notice the two inequalities overlap so that every point is a solution. So, the solution set is {n | n is any real number}.

Write an equation of the line that passes through each pair of points.

61. (1, −3) and (2, 5)

SOLUTION:  Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept. 

  Write the equation in slope–intercept form.  

62. (−2, −4) and (−7, 3)

SOLUTION:  Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept. 

  Write the equation in slope–intercept form.  

63. (−6, −8) and (−8, −5)

SOLUTION:  Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept. 

  Write the equation in slope–intercept form.  

64. FITNESS  The table shows the maximum heart rate to maintain during aerobic activities. Write an equation in function notation for the relation. Determine what would be the maximum heart rate to maintain in aerobic training for an 80-year-old.  

SOLUTION:  Choose two points: (20, 175) and (70, 130). Find the slope of the line containing the given points. 

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in function notation.  

  To determine what the maximum heart rate is for an 80 year–old, substitute 80 in for x.  

  The maximum heart rate for an 80-year-old is 121 beats/min.

65. WORK  The formula  is used to find keyboarding speeds. In the formula, s represents the speed in

words per minute, w the number of words typed, r  the number of errors, and m the number of minutes typed. Solve for r.

SOLUTION:  

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5-6 Graphing Inequalities in Two Variables

SOLUTION:  

  and   

              

  So, the solution set is {c | 3 < c < 5}.

59. 

SOLUTION:  

and             

  So, the solution set is {p | 4 < p < 10}.

60. 0.5n ≥ −7 or 2.5n + 2 ≤ 9

SOLUTION:  

or             

  Notice the two inequalities overlap so that every point is a solution. So, the solution set is {n | n is any real number}.

Write an equation of the line that passes through each pair of points.

61. (1, −3) and (2, 5)

SOLUTION:  Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept. 

  Write the equation in slope–intercept form.  

62. (−2, −4) and (−7, 3)

SOLUTION:  Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept. 

  Write the equation in slope–intercept form.  

63. (−6, −8) and (−8, −5)

SOLUTION:  Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept. 

  Write the equation in slope–intercept form.  

64. FITNESS  The table shows the maximum heart rate to maintain during aerobic activities. Write an equation in function notation for the relation. Determine what would be the maximum heart rate to maintain in aerobic training for an 80-year-old.  

SOLUTION:  Choose two points: (20, 175) and (70, 130). Find the slope of the line containing the given points. 

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in function notation.  

  To determine what the maximum heart rate is for an 80 year–old, substitute 80 in for x.  

  The maximum heart rate for an 80-year-old is 121 beats/min.

65. WORK  The formula  is used to find keyboarding speeds. In the formula, s represents the speed in

words per minute, w the number of words typed, r  the number of errors, and m the number of minutes typed. Solve for r.

SOLUTION:  eSolutions Manual - Powered by Cognero Page 51

5-6 Graphing Inequalities in Two Variables

SOLUTION:  

  and   

              

  So, the solution set is {c | 3 < c < 5}.

59. 

SOLUTION:  

and             

  So, the solution set is {p | 4 < p < 10}.

60. 0.5n ≥ −7 or 2.5n + 2 ≤ 9

SOLUTION:  

or             

  Notice the two inequalities overlap so that every point is a solution. So, the solution set is {n | n is any real number}.

Write an equation of the line that passes through each pair of points.

61. (1, −3) and (2, 5)

SOLUTION:  Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept. 

  Write the equation in slope–intercept form.  

62. (−2, −4) and (−7, 3)

SOLUTION:  Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept. 

  Write the equation in slope–intercept form.  

63. (−6, −8) and (−8, −5)

SOLUTION:  Find the slope of the line containing the given points.  

  Use the slope and either of the two points to find the y–intercept. 

  Write the equation in slope–intercept form.  

64. FITNESS  The table shows the maximum heart rate to maintain during aerobic activities. Write an equation in function notation for the relation. Determine what would be the maximum heart rate to maintain in aerobic training for an 80-year-old.  

SOLUTION:  Choose two points: (20, 175) and (70, 130). Find the slope of the line containing the given points. 

  Use the slope and either of the two points to find the y–intercept.  

  Write the equation in function notation.  

  To determine what the maximum heart rate is for an 80 year–old, substitute 80 in for x.  

  The maximum heart rate for an 80-year-old is 121 beats/min.

65. WORK  The formula  is used to find keyboarding speeds. In the formula, s represents the speed in

words per minute, w the number of words typed, r  the number of errors, and m the number of minutes typed. Solve for r.

SOLUTION:  

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5-6 Graphing Inequalities in Two Variables