Solving Systems by Graphing ALGEBRA 1 LESSON 7-1 Solve each equation. 1. 2n + 3 = 5n – 22. 8 – 4z = 2z – 133. 8q – 12 = 3q + 23 Graph each pair of equations

Solving Systems by GraphingSolving Systems by GraphingALGEBRA 1 LESSON 7-1ALGEBRA 1 LESSON 7-1

Solve each equation.

1. 2n + 3 = 5n – 2 2. 8 – 4z = 2z – 13 3. 8q – 12 = 3q + 23

Graph each pair of equations on the same coordinate plane.

4. y = 3x – 6 5. y = 6x + 1 y = –x + 2 y = 6x – 4

(For help, go to Lessons 2-4 and 6-2.)

6. y = 2x – 5 7. y = x + 5 6x – 3y = 15 y = –3x + 5

7-1

Solving Systems by GraphingSolving Systems by Graphing

1. 2n + 3 = 5n – 2 2. 8 – 4z = 2z – 13

2n – 2n + 3 = 5n – 2n – 2 8 – 4z + 4z = 2z + 4z – 13

3 = 3n – 2 8 = 6z – 13

5 = 3n 21 = 6z

1 = n 3 = z

3. 8q – 12 = 3q + 23

8q – 3q – 12 = 3q – 3q + 23

5q – 12 = 23

5q = 35

q = 7

12

23

ALGEBRA 1 LESSON 7-1ALGEBRA 1 LESSON 7-1

Solutions

7-1


4. y = 3x – 6 5. y = 6x + 1

y = –x + 2 y = 6x – 4

6. y = 2x – 5 7. y = x + 5

6x – 3y = 15 y = –3x + 5

–3y = –6x – 15

y =

y = 2x – 5

–6x 15 –3


Solutions (continued)

7-1

Solve by graphing. Check your solutions.


y = 2x + 1y = 3x – 1

Graph both equations on the same coordinate plane.

y = 2x + 1 The slope is 2. The y-intercept is 1.y = 3x – 1 The slope is 3. The y-intercept is –1.


7-1

(continued)


Find the point of intersection.

The lines intersect at (2, 5), so (2, 5) is the solution of the system.

y = 2x + 1 y = 3x – 1

5 2(2) + 1 Substitute (2, 5) for (x, y). 5 3(2) – 1

5 4 + 1 5 6 – 1

5 = 5 5 = 5

Check: See if (2, 5) makes both equations true.


7-1

Suppose you plan to start taking an aerobics class. Non-members pay $4 per class while members pay $10 a month plus an additional $2 per class. After how many classes will the cost be the same? What is that cost?


Define: Let = number of classes.

Let = total cost of the classes.

c

T(c)

Relate: cost is membership plus cost of classes fee attended

Write: member = 10 + 2

non-member = 0 + 4T(c)

c

c

T(c)


7-1

(continued)


Method 1: Using paper and pencil.

T(c) = 2c + 10 The slope is 2. The intercept on the vertical axis is 10.

T(c) = 4c The slope is 4. The intercept on the vertical axis is 0.

The lines intersect at (5, 20).

After 5 classes, both will cost

$20.

Graph the equations.

T(c) = 2c + 10

T(c) = 4c


7-1

(continued)


Method 2: Using a graphing calculator.First rewrite the equations using x and y.T(c) = 2c + 10 y = 2x + 10T(c) = 4c y = 4x

Then graph the equations using a graphing calculator.

Set an appropriate range.

The lines intersect at (5, 20). After 5 classes, both will cost $20.

Then graph the equations. Use the key to find the coordinates of the intersection point.


7-1


Solve by graphing. y = 3x + 2

y = 3x – 2


The lines are parallel. There is no solution.

y = 3x + 2 The slope is 3. The y-intercept is 2.

y = 3x – 2 The slope is 3. The y-intercept is –2.


7-1


Solve by graphing. 3x + 4y = 12

y = – x + 3


The graphs are the same line. The solutions are an

infinite number of ordered pairs (x, y), such that

y = – x + 3.34

34

3x + 4y = 12 The y-intercept is 3.The x-intercept is 4.

y = – x + 3 The slope is – . The y-intercept is 3.34

34


7-1

6. (0, 0);

7. (1, 1);


pages 343–345 Exercises

1. Yes, (–1, 5) makes both equations true.

2. No, (–1, 5) makes only one equation true.



5. (0, 2);

8. (1, 5);

9. (6, –1);


7-1

10. (2, 2);

11. (4, 0);


12. (2, 3);

13. a. 3 weeks b. $35

14. 7 weeks

15. no solution;

16. no solution;

17. infinitely many solutions;


7-1


18. no solution;

19. no solution; same slope, different y-int.

20. inf. many solutions; equivalent equations

21. one solution; different slopes

22. inf. many solutions; equivalent equations

23. A

24. 5 min

25. (20, 60);

26. Answers may vary. Sample: y = –1; x = 2

27. Answers may vary. Sample: y = 2x – 1, y = 2x + 5

28. Answers may vary. Sample: x + y = 3, 3x + 3y = 9


7-1


29. (2, 20)

30. (15, 40)

31. (12, 30)

32. (–20, 0)

33. a. time on the horizontal anddistance on the vertical

b. Red represents the tortoise because it shows distance changing steadily over time. Blue represents the hare because it is steeper than the other line at the ends but shows no change in distance while the hare is napping.

c. The point of intersection shows when the tortoise passed the sleeping hare.

34. (–12, –16)

35. (–2, 10)


7-1


36. (–30, –2.5)

37. (–0.9, 1.6)

38. (2, 3)

39. a. c = 100 + 50t; c = 50 + 75t; (2, 200);

b. The cost of renting either studio for 2 h is the same, $200.

40. a. no valuesb. w = vc. w = v

/

41. a. sometimesb. never

42. (–9, –2)

43. D

44. F

45. [2] a. Answers may vary. Sample: x – 2y = 6

b.Since the lines do not intersect, the lines are parallel. Parallel lines have the same slope but different intercepts.

[1] incorrect equation OR incorrect explanation


7-1


46. [4] a. y = 150 + 0.20xy = 200 + 0.10x

b. $500c. cellular phone sales

[3] appropriate methods but one computational error

[2] incorrect system solved correctly OR correct system solved incorrectly

[1] no work shown

47. It is translated up 2 units.

48. It is translated 3 units left.

49. It is translated 5 units up and 2 units right.

50. 25% increase

51. 33 % decrease

52. 20% increase

53. 150% increase

54. 33 % decrease

55. 25% increase

56. 10% increase

57. 12.5% decrease

13

13


7-1

Solve by graphing.

1. y = –x – 2 2. y = –x + 3 3. y = 3x + 2 y = x + 3 y = 2x – 6 6x – 2y = –4

4. 2x – 3y = 9 5. –2x + 4y = 12

y = x – 5 – x + y = –3


23

12

(3, 1) (3, 0) Infinitely many solutions

(6, 1) no solution


7-1

Solve each equation.

1. m – 6 = 4m + 8 2. 4n = 9 – 2n 3. t + 5 = 10

Solving Systems Using SubstitutionSolving Systems Using Substitution

(For help, go to Lessons2-4 and 7-1.)

13

For each system, is the ordered pair a solution of both equations?

4. (5, 1) y = –x + 4 5. (2, 2.4) 4x + 5y = 20

y = x – 6 2x + 6y = 10


7-2


1. m – 6 = 4m + 8 2. 4n = 9 – 2n

m – m – 6 = 4m – m + 8 4n + 2n = 9 – 2n + 2n

–6 = 3m + 8 6n = 9

–14 = 3m n = 1

–4 = m

3. t + 5 = 10

t = 5

t = 15

122

3

13

13


Solutions

7-2


4. (5, 1) in y = –x + 4 (5, 1) in y = x – 6

1 –5 + 4 1 5 – 6

1 = –1 1 = –1

no no

No, (5, 1) is not a solution of both equations.

5. (2, 2.4) in 4x + 5y = 20 (2, 2.4) in 2x + 6y = 10

4(2) + 5(2.4) 20 2(2) + 6(2.4) 10

8 + 12 20 4 + 14.4 10

20 = 20 18.4 = 10

yes no

No, (2, 2.4) is not a solution of both equations.

/

//



7-2


Solve using substitution. y = 2x + 2

y = –3x + 4

Step 1: Write an equation containing only one variable and solve.

y = 2x + 2 Start with one equation. –3x + 4 = 2x + 2 Substitute –3x + 4 for y in that equation.

4 = 5x + 2 Add 3x to each side. 2 = 5x Subtract 2 from each side. 0.4 = x Divide each side by 5.

Step 2: Solve for the other variable.

y = 2(0.4) + 2 Substitute 0.4 for x in either equation.y = 0.8 + 2 Simplify. y = 2.8


7-2


(continued)

Since x = 0.4 and y = 2.8, the solution is (0.4, 2.8).

Check: See if (0.4, 2.8) satisfies y = –3x + 4 since y = 2x + 2 was used in Step 2.

2.8 –3(0.4) + 4 Substitute (0.4, 2.8) for (x, y) in the equation.2.8 –1.2 + 42.8 = 2.8


7-2


Solve using substitution.

–2x + y = –1

4x + 2y = 12

Step 1: Solve the first equation for y because it has a coefficient of 1.

–2x + y = –1 y = 2x –1 Add 2x to each side.

Step 2: Write an equation containing only one variable and solve.

4x + 2y = 12 Start with the other equation. 4x + 2(2x –1) = 12 Substitute 2x –1 for y in that equation.

4x + 4x –2 = 12 Use the Distributive Property. 8x = 14 Combine like terms and add 2 to each

side. x = 1.75 Divide each side by 8.


7-2


(continued)

Step 3: Solve for y in the other equation.

Since x = 1.75 and y = 2.5, the solution is (1.75, 2.5).

–2(1.75) + y = 1 Substitute 1.75 for x. –3.5 + y = –1 Simplify.

y = 2.5 Add 3.5 to each side.


7-2

Let v = number of vans and c = number of cars.

Drivers v + c = 5

Persons 7 v + 5 c = 31


A youth group with 26 members is going to the beach. There

will also be five chaperones that will each drive a van or a car. Each

van seats 7 persons, including the driver. Each car seats 5 persons,

including the driver. How many vans and cars will be needed?

Solve using substitution.Step 1: Write an equation containing only one variable.

v + c = 5 Solve the first equation for c. c = –v + 5


7-2


(continued)

Step 2: Write and solve an equation containing the variable v.

7v + 5c = 31 7v + 5(–v + 5) = 31 Substitute –v + 5 for c in the second

equation.

7v – 5v + 25 = 31 Solve for v. 2v + 25 = 31 2v = 6 v = 3

Step 3: Solve for c in either equation.

3 + c = 5 Substitute 3 for v in the first equation. c = 2


7-2


(continued)

Three vans and two cars are needed to transport 31 persons.

Check: Is the answer reasonable? Three vans each transporting 7 persons is 3(7), of 21 persons. Two cars each transporting 5 persons is 2(5), or 10 persons. The total number of persons transported by vans and cars is 21 + 10, or 31. The answer is correct.


7-2

10. ( , 9 )

11. (2, 0)

12. (7 , 11 )

13. (6, –2)

14. (3, –2)

15. (8, –7)

16. (–3, 9.4)

17. 4 cm by 13 cm

18. 4 wk

19. (15, 15)

20. (9, 126)

21. (–4, 4)



1. D

2. C

3. B

4. A

5–16. Coordinates given in alphabetical order.

5. (9, 28)

6. (– , –4 )

7. (6 , – )

8. (2, 4 )

9. (4, 20)

12

1213

13

12

34

38 22. 15 video rentals

23. 80 acres flax, 160 acres sunflowers

24. 9 yr

25. estimate: ( , 1);

; ( , 1)

7 17

8 17

12

12


7-2


26. estimate: (–2, 3);

; (–2, 3)

27. estimate: (–1, 1);

(–1, 1)

28. estimate: (–3.5, –3.5);

(– , – )

29. estimate: (– , 4 );

(– , )

34

34

103

113

143

23


7-2


30. estimate: (–4, 0);

(– , – )

31. a. Let n = number of nickels, let d = number of dimes.

n + d = 280.05n + 0.10d = 2.05

b. Solve the first eq. for either var. Sub. the expression into the second eq. Solve this eq. for the other var., and then sub. its value into the first eq. and solve for the first var.

c. (15, 13)

5011

2 11

32. Answers may vary. Sample: y = x and y = –3x + 2;

( , )

33. a. (x, y) such that y = 0.5x + 4b.

c. Graphing shows only one line. Substitution results in a true equation with no variables.

12

12


7-2


34. a. no solutionb.

c. Graphing shows 2 parallel lines. Substitution results in a false equation with no variables.

35. (2, 4)

36. (– , – )

37. (2, –4)

12

12

38. (2, )

39. (– , 0)

40. (4, –2)

41. inf. many solutions

42. no solution

43. 1 solution

44. a. g = b

b. (b, g) = (572, 598)c. 26

45. a. (t, d) = (9, 79.2)b. yes

46. (r, s, t) = (7, 9, 4)

47. 29.8

48. 4803

49. 52050. [2] 7(–7) – 4(–2) 29 –49 + 8 29 –41 29

No, (–2, –7) must satisfy both equations to be asolution of the system.

[1] no explanation given

12

12


7-2

2322

=/


51. (12, 10);

52. (2, 1);

53. (4, 2);

54.

55.

56.

57.

58.

59.


7-2


Solve each system using substitution.

1. 5x + 4y = 5 2. 3x + y = 4 3. 6m – 2n = 7

y = 5x 2x – y = 6 3m + n = 4

(0.2, 1) (2, 2) (1.25, 0.25)


7-2

Solving Systems Using EliminationSolving Systems Using Elimination

(For help, go to Lesson 7-2.)

Solve each system using substitution.

1. y = 4x – 3 2. y + 5x = 4 3. y = –2x + 2

y = 2x + 13 y = 7x – 20 3x – 17 = 2y


7-3


1. y = 4x – 3y = 2x + 13Substitute 4x – 3 for y in the second equation. y = 2x + 13 4x – 3 = 2x + 13 4x – 2x – 3 = 2x – 2x + 13 2x – 3 = 13 2x = 16 x = 8 y = 4x – 3 = 4(8) – 3 = 32 – 3 = 29Since x = 8 and y = 29, the solution is (8, 29).


Solutions

7-3


2. y + 5x = 4y = 7x – 20Substitute 7x – 20 for y in the first equation. y + 5x = 4 7x – 20 + 5x = 4 12x – 20 = 4 12x = 24 x = 2 y = 7x – 20 = 7(2) – 20 = 14 – 20 = –6Since x = 2 and y = –6, the solution is (2, –6).



7-3


3. y = –2x + 23x – 17 = 2ySubstitute –2x + 2 for y in the second equation. 3x – 17 = 2y 3x – 17 = 2(–2x + 2) 3x – 17 = –4x + 4 7x – 17 = 4 7x = 21 x = 3 y = –2x + 2 = –2(3) + 2 = –6 + 2 –4Since x = 3 and y = –4, the solution is (3, –4).



7-3


Solve by elimination. 2x + 3y = 11

–2x + 9y = 1

Step 1: Eliminate x because the sum of the coefficients is 0.

2x + 3y = 11

–2x + 9y =1

0 + 12y = 12 Addition Property of Equality

y = 1 Solve for y.

Step 2: Solve for the eliminated variable x using either original equation.

2x + 3y = 11 Choose the first equation.

2x + 3(1) = 11 Substitute 1 for y.

2x + 3 = 11 Solve for x.

2x = 8

x = 4


7-3


(continued)

Since x = 4 and y = 1, the solution is (4, 1).

Check: See if (4, 1) makes true the equation not used in Step 2.

–2(4) + 9(1) 1 Substitute 4 for x and 1 for y into the

second equation.

–8 + 9 1

1 = 1


7-3

Define: Let a = number of adults

Let s = number of students

Relate: total number at the game total amount collected

Write: a + s = 1139 5 a + s = 3067


On a special day, tickets for a minor league baseball game

were $5 for adults and $1 for students. The attendance that day was

1139, and $3067 was collected. Write and solve a system of

equations to find the number of adults and the number of students

that attended the game.

Solve by elimination.


7-3


(continued)

Step 1: Eliminate one variable. a + s = 1139 5a + s = 3067 –4a + 0 = –1928 Subtraction Property of Equality a = 482 Solve for a.

Step 2: Solve for the eliminated variable using either of the original equations.

a + s = 1139 Choose the first equation. 482 + s = 1139 Substitute 482 for a. s = 657 Solve for s.

There were 482 adults and 657 students at the game.

Check: Is the solution reasonable? The total number at the game was 482 + 657, or 1139. The money

collected was $5(482), or $2410, plus $1(657), or $657, which is $3067. The solution is correct.


7-3


Solve by elimination. 3x + 6y = –6

–5x – 2y = –14

Step 1: Eliminate one variable.

Start with the givensystem.

3x + 6y = –6 –5x – 2y = –14

Add the equations to eliminate y.

3x + 6y = –6 –15x – 6y = –42 –12x – 0 = –48

To prepare to eliminate y, multiply the second equation by 3. 3x + 6y = –6 3(–5x – 2y = –14)

Step 2: Solve for x. –12x = 48 x = 4


7-3


(continued)


3x + 6y = –6 Choose the first equation.3(4) + 6y = –6 Substitute 4 for x. 12 + 6y = –6 Solve for y. 6y = –18 y = –3

The solution is (4, –3).


7-3

Define: Let p = number of cans of popcorn sold.

Let n = number of cans of nuts sold.

Relate: total number of cans total amount of sales

Write: p + n = 240 5 p + 8 n = 1614


Suppose the band sells cans of popcorn for $5 per can and

cans of mixed nuts for $8 per can. The band sells a total of 240 cans

and receives a total of $1614. Find the number of cans of popcorn

and the number of cans of mixed nuts sold.


7-3


(continued)


Subtract the equations to eliminate p.

5p + 5n = 1200 5p + 8n = 1614 0 – 3n = –414

Step 2: Solve for n. –3n = –414 n = 138


p + n = 240 5p + 8n = 1614

To prepare to eliminate p, multiply the first equation by 5. 5(p + n = 240) 5p + 8n = 1614


7-3


(continued)


p + n = 240 Choose the first equation. p + 138 = 240 Substitute 138 for n. p = 102 Solve for p.

The band sold 102 cans of popcorn and 138 cans of mixed nuts.


7-3


Solve by elimination. 3x + 5y = 10

5x + 7y = 10


Subtract the equations to eliminate x.

15x + 25y = 50 15x + 21y = 30 0 + 4y = 20


3x + 5y = 10 5x + 7y = 10

To prepare to eliminate x, multiply one equation by 5 and the other equation by 3. 5(3x + 5y = 10) 3(5x + 7y = 10)

Step 2: Solve for y. 4y = 20 y = 5


7-3


(continued)

Step 3: Solve for the eliminated variable x using either of the original equations.

3x + 5y = 10 Use the first equation. 3x + 5(5) = 10 Substitute 5 for y. 3x + 25 = 10 3x = –15 x = –5

The solution is (–5, 5).


7-3



1. (1, 3)

2. (2, –2)

3. (5, –17)

4. (–3, 4)

5. (–9, )

6. (– , 10)

7. a. x + y = 20, x – y = 4b. 12 and 8

8. a. a + s = 456, 3.5a + s = 1131b. 270 adult, 186 student

9. (–5, 1)

10. (11, –3)

11. (–2, – )

12. (1, 14)

13. ( , 1)

14. (–2, 3)

15. a. 30w + = 17.65, 20w + 3 = 25.65

b. $.39 for a wallet size,

$5.95 for an 8 1016. a. x = burritos, y = tacos;

3x + 4y = 11.33, 9x + 5y = 23.56b. $1.79 for a burrito, $1.49 for a taco

17. (–1, –3)

18. (2.5, 1)

19. (2, –2)

12

12

52

12


7-3

27. (5, 1); substitution; one eq. solved for x

28. ( , 2 ); substitution;

eqs. solved for y

29. one night: $81.25; one meal: $8.13

30. a. brass: $6; steel: $3b. $99

31. She forgot to multiply –8 by 6.

32. Answers may vary. Sample: 2x – 3y = 6, x + 3y = 9; (5, )

33. (10, –6)

34. (8, 12)


20. (10, 8)

21. (–1, )

22. (1, 5)

23–28. Methods may vary. Samples are given.

23. (–1, –2); substitution; both solved for y

24. (15, –10); elimination; equations not solved for y

25. (10, 2); substitution; one eq. solved for x

26. (–3, 11); elimination; eqs. not solved for a variable

32 1

313

43


7-3


35. (–15, –1)

36. (20, 12)

37. (–2, 16)

38. (33, –48)

39. 9

40. Answers may vary. Sample: You solve a system using the elimination method by adding or subtracting the eqs. to eliminate one of the variables. This sum or difference is one eq. with one variable that can be solved.

Use addition: Use subtraction: Use multiplication:3x + 2y = 6 5x + 3y = 15 4x + 5y = 20–x – 2y = 4 5x – 2y = 10 2x – y = 10

41. B1 = 3 volts; B2 = 1.5 volts

42. (–3, 2)

43. ( , 0) (a = 0, b = 1)

44. (8, 13, 20)

45. CD: $3.40, cassette: $1.80

46. a. 2.8 g of goldb. about 2.7%

47. D

48. H

ca

/ /


7-3


49. [2] y – x = 13 7y + x = 11

8y = 24 y = 3

3 – x = 13 x = –10; (–10, 3)

[1] no work shown50. [4]

A = h(b1 + b2)

= (4)(4 + 2)

= 2(6)

= 12The area is 12 square units.

[3] graph and formula with one mathematical error[2] graph but error in formula[1] graph only

1212

51–53. Coordinates given in alphabetical order.

51. (6, 26)

52. (–1, 4)

53. (9, –5)

54.

55.

56.

57. 71

58. –18

59. –44

60. 22

61. 83

62. –6

1929 1 15


7-3


Solve using elimination.

1. 3x – 4y = 7 2. 5m + 3n = 222x + 4y = 8 5m + 6n = 34

3. –6x + 5y = 4 4. 7p + 5q = 2 3x + 4y = 11 8p – 9q = 17

(3, 0.5) (2, 4)

(1, 2) (1, 1)


7-3

Applications of Linear SystemsApplications of Linear Systems

(For help, go to Lesson 2-5.)

1. Two trains run on parallel tracks. The first train leaves a city hour

before the second train. The first train travels at 55 mi/h. The

second train travels at 65 mi/h. Find how long it takes for the second

train to pass the first train.

2. Luis and Carl drive to the beach at an average speed of 50 mi/h.

They return home on the same road at an average speed of 55

mi/h. The trip home takes 30 min. less. What is the distance from

their home to the beach?

12


7-4


1. 55t = 65(t – 0.5)

55t = 65t – 32.5

32.5 = 10t

3.25 = t

It takes 3.25 hours for the second train to pass the first train.

2. 50t = 55(t – 0.5)

50t = 55t – 27.5

27.5 = 5t

5.5 = t

50t = 50(5.5) = 275

It is 275 miles to the beach from their home.


Solutions

7-4

Define: Let a = volume of the Let b = volume of the

50% solution. 25% solution.

Relate: volume of solution amount of acid

Write: a + b = 10 0.5 a + 0.25 b =

0.4(10)


A chemist has one solution that is 50% acid. She has

another solution that is 25% acid. How many liters of each type of acid

solution should she combine to get 10 liters of a 40% acid solution?

Step 1: Choose one of the equations and solve for a variable. a + b = 10 Solve for a. a = 10 – b Subtract b from each side.


7-4


(continued)

Step 2: Find b.

0.5a + 0.25b = 0.4(10) 0.5(10 – b) + 0.25b = 0.4(10) Substitute 10 – b for a.

Use parentheses. 5 – 0.5b + 0.25b = 0.4(10) Use the Distributive Property.

5 – 0.25b = 4 Simplify. –0.25b = –1 Subtract 5 from each side. b = 4 Divide each side by –0.25.

Step 3: Find a. Substitute 4 for b in either equation. a + 4 = 10 a = 10 – 4 a = 6

To make 10 L of 40% acid solution, you need 6 L of 50% solution and 4 L of 25% solution.


7-4

Define: Let p = the number of pages.

Let d = the amount of dollars of expenses or income.

Relate: Expenses are per-page Income is priceexpenses plus times pages typed.computer purchase.

Write: d = 0.5 p + 1750 d = 5.5 p


Suppose you have a typing service. You buy a personal

computer for $1750 on which to do your typing. You charge $5.50 per

page for typing. Expenses are $.50 per page for ink, paper, electricity,

and other expenses. How many pages must you type to break even?


7-4


(continued)

Choose a method to solve this system. Use substitution since it is easy to substitute for d with these equations.

d = 0.5p + 1750 Start with one equation. 5.5p = 0.5p + 1750 Substitute 5.5p for d. 5p = 1750 Solve for p.

p = 350

To break even, you must type 350 pages.


7-4

Define: Let A = the airspeed. Let W = the wind speed.

Relate: with tail wind with head wind

(rate)(time) = distance (rate)(time) = distance

(A + W) (time) = distance (A + W) (time) =

distance

Write: (A + W)5.6 = 2800 (A + W)6.8 = 2800


Suppose it takes you 6.8 hours to fly about 2800 miles from

Miami, Florida to Seattle, Washington. At the same time, your friend

flies from Seattle to Miami. His plane travels with the same average

airspeed, but this flight only takes 5.6 hours. Find the average

airspeed of the planes. Find the average wind speed.


7-4


(continued)Solve by elimination. First divide to get the variables on the left side of each equation with coefficients of 1 or –1.(A + W)5.6 = 2800 A + W = 500 Divide each side by 5.6.(A – W)6.8 = 2800 A – W 412 Divide each side by 6.8.

Step 1: Eliminate W. A + W = 500

A – W = 412 Add the equations to eliminate W. 2A + 0 = 912

Step 2: Solve for A. A = 456 Divide each side by 2.

Step 3: Solve for W using either of the original equations. A + W = 500 Use the first equation. 456 + W = 500 Substitute 456 for A. W = 44 Solve for W.

The average airspeed of the planes is 456 mi/h. The average wind speed is 44 mi/h.


7-4



1. a. 4a + 5b = 6.71b. 5a + 3b = 7.12c. pen: $1.19, pencil: $.39

2. D

3. a.

b. a + b = 24; 0.04a + 0.08b = 1.2

c. 18 kg A, 6 kg B

4. a. at 8 wkb. $160; $62

5. 600 games

6. 40 shirts

7. a. s + c = 2.75b. s – c = 1.5c. 2.125 mi/hd. 0.625 mi/h

8. a. (A + W )4.8 = 2100(A – W )5.6 = 2100

b. 406.25 mi/hc. 31.25 mi/h

9–14. Answers may vary. Samples are given.

9. Substitution; one eq. is solved for t.

10. Substitution; both eqs. are solved for y.

11. Elimination; subtract to eliminate m.

12. Substitution; both eqs. are solved for y.


7-4


13. Elimination; mult. first eq. by 3 and add to elim. y.

14. Substitution; one eq. is solved for u.

15. a. t = 99 – 3.5m; t = 0 + 2.5m; t = 41.25°, m = 16.5 min

b. After 16.5 min, the temp. of either piece will be 41.25°C.

16. 5 cm; 12 cm

17. Answers may vary. Sample: You have 10 coins, all dimes and quarters. The value of the coins is $1.75. How many dimes do you have? How many quarters do you have?

q + d = 100.25q + 0.10d = 1.75You have 5 dimes and 5 quarters.

18. 19 small mowers, 11 large mowers

19. a. 42 mi/hb. 12 mi/h

20. a. g = b

b. g + b= 1908

g = b

901 boys, 1007 girls

21. a. 16 daysb. Answers may vary. Sample:

If you plan to ski for many years, you should buy the equipment, since you will break even at 16 days.

22. x = 2y = 4

1917

1917


7-4


24. a. 2.50s + 4.00 = 10,000

= s

800 small, 2000 largeb. 560 hc. $17.86/h

25. C

26. H

27. B

28. [2] 3V + C = 22 12V + 4C = 882V + 4C = 28 2V + 4C = 28

10V = 60V = 60

6 people/van

[1] correct answer, no work shown

52

29. (4, 1)

30. (–6, 7)

31. (3, )

32.

33. 1

34.

35. –2

36. –

37. undefined

32

32

13

4 11


7-4


38. 6 < y < 10

39. –8 < n 3

40. 1 < k < 6

41. 1 p 4

42. –5 < c 5

43. 5 > w > 1


7-4

<–

<– <–

<–


1. One antifreeze solution is 10% alcohol. Another antifreeze solution is 18% alcohol. How many liters of each antifreeze solution should be combined to create 20 liters of antifreeze solution that is 15% alcohol?

2. A local band is planning to make a compact disk. It will cost $12,500 to record and produce a master copy, and an additional $2.50 to make each sale copy. If they plan to sell the final product for $7.50, how many disks must they sell to break even?

3. Suppose it takes you and a friend 3.2 hours to canoe 12 miles downstream (with the current). During the return trip, it takes you and your friend 4.8 hours to paddle upstream (against the current) to the original starting point. Find the average paddling speed in still water of you and your friend and the average speed of the current of the river. Round answers to the nearest tenth.

7.5 L of 10% solution; 12.5 L of 18% solution

2500 disks

still water: 3.1 mi/h; current: 0.6 mi/h


7-4

Linear InequalitiesLinear Inequalities

(For help, go to Lesson 3-1 and 6-3.)

Describe each statement as always, sometimes, or never true.

1. –3 > –2 2. 8 8 3. 4n n

Write each equation is slope-intercept form.

4. 2x – 3y = 9 5. y + 3x = 6 6. 4y – 3x = 1


7-5

>–<–

1. –3 < –2 is true, so –3 > –2 is never true.2. 8 = 8 is true, so 8 8 is always true.3. 4(1) 1 is true and 4(–1) –1 is false, so 4n n is sometimes true.

4. 2x – 3y = 9 5. y + 3x = 6 –3y = –2x + 9 y = –3x + 6

y =

y = x – 3

6. 4y – 3x = 1 4y = 3x + 1

y =

y = x +

–2x + 9 –323

3x 1 434

14

Linear InequalitiesLinear InequalitiesALGEBRA 1 LESSON 7-5ALGEBRA 1 LESSON 7-5

Solutions

7-5

>–

<–>– >–


Graph y > –2x + 1.

First, graph the boundary line y = –2x + 1.

The coordinates of the points on the boundary line do not make the inequality true. So, use a dashed line.

Shade above the boundary line.

Check The point (0, 2) is in the region of the graph of the inequality.See if (0, 2) satisfies the inequality.y > –2x + 12 > –2(0) + 12 > 1

Substitute (0, 2) for (x, y).


7-5


Graph 4x – 3y 6.

Graph y = x – 2.

The coordinates of the points on the boundary line make the inequality true. So, use a solid line.

43


7-5

>–

Solve 4x – 3y 6 for y.>–4x – 3y 6 –3y –4x + 6 Subtract 4x from each side.

y x – 2 Divide each side by –3. Reverse the inequality symbol.

43

<–

>–>–

Since y x – 2, shade below the boundary

line.

43

<–

Suppose your budget allows you to spend no more than $24 for decorations for a party. Streamers cost $2 a roll and tablecloths cost $6 each. Use intercepts to graph the inequality that represents the situation. Find three possible combinations of streamers and tablecloths you can buy.


7-5

Relate: cost of plus cost of is less than total budgetstreamers tablecloths or equal to

Define: Let s = the number of rolls of streamers.

Let t = the number of rolls of streamers.

Write: 2 s + 6 t 24<–

(continued)


7-5

Graph 2s + 6t 24 by graphing the intercepts (12, 0) and (0, 4).

The coordinates of the points on the boundary line make the inequality true. So, use a solid line.

Graph only in Quadrant I, since you cannot buy a negative amount of decorations.

<–

Test the point (0, 0). 2s + 6t 24 2(0) + 6(0) 24 Substitute (0, 0) for (s, t). 0 24 Since the inequality is true, (0, 0) is a solution.

<–<–<–

(continued)


Shade the region containing (0, 0). The graph below shows all the possible solutions of the problem.

Since the boundary line is included in the graph, the intercepts are also solutions to the inequality.

The solution (9, 1) means that if you buy 9 rolls of streamers, you can buy 1 tablecloth. Three solutions are (9, 1), (6, 2), and (3, 3).


7-5



1. no

2. yes

3. yes

4. yes

5. no

6. yes

7. A

8. B

9. B

10. A

11.

12.

13.

14.

15.

16.

17.

18.

19. y x – ;23

73


7-5

<–

20. y x – 2;

21. y x – ;

22. y – x – 2;


23

53

23. a. 3x + 5y 48

b.

c. Answers may vary. Sample: 8 blue and 4 gold, 2 blue and 8 gold, 12 blue and 2 gold

d. No; you cannot buy –2 rolls of paper.

83

32


7-5

>–

<–

<–

<–


24. a. 3n + 10c > 250b.

c. Answers may vary. Sample: 30 canvas and 10 nylon,26 canvas and 20 nylon, 35 canvas and 10 nylon

d. Domain and range values must be positive integers, since you cannot buy portions of packs or a negative number of packs.

25.

26.

27.

28.

29.

30.


7-5


7-5

34. y > 2x – 1

35. x –3

36. y x – 2

37. a. 12x + 8y 180b.

c. Yes; you can buy 8 CDs and 9 tapes.

d. 43

13

<–

<–

<–

31.

32.

33. For an inequality written in the form y < or y , shade below the boundary line. For an inequality written in the form y > or y , shade above the boundary line.

>–

<–


38. x > 0;

39. y < 0;

40. y 0;

41. x < y;

42. It should be shaded above the line, and the line should be dashed.

43. y < x + 2

44. a.

b. y > x

c.

45. a. 2w + 2 50;

b. Answers may vary. Sample: 10 ft by 10 ft, 5 ft by 5 ft

c. No; (12, 15) is not in the shaded region and is not a sol. of the inequality.

46. y < x – 3

47. y 2x – 2


7-5

>–

>–

<–

5 12


48. a. Answers may vary. Sample: 2x + y > 3

b. Answers may vary. Sample: 3x + y < 1

c. If an inequality is in standard form, where A, B, and C are all positive, you shade above the line for > or and below the line for < or .

d. yes

49. a. yesb. yesc. Answers may vary.

Sample: (2, 3)d.

50. A

51. H

52. D

53. F


7-5

>–<–


54. [2] First, graph the solid line y = 3x – 4. Then shade below the line.

[1] correct graph given, no explanation

55. about 775 games

56. 0.75 mi/h, 3.25 mi/h

57. 5

58. 7

59. 11

60. –6

61. 8, 18

62. –6.7, –2.1

63. 12

64. 28

65. 11

66. 16

67. –

68. 4

69. 13

70. 6.5

13

23


7-5


1. Determine whether (4, 1) is a solution of 3x + 2y 10.

Graph each inequality.

2. x > –2 3. 5x – 2y > 10 4. 2x + 6y 0

yes


7-5

>–

<–

Systems of Linear InequalitiesSystems of Linear Inequalities

(For help, go to Lessons 7-1 and 7-5.)

Solve each system by graphing.

1. y = 3x – 6 2. y = – x + 4 3. x + y = 4

y = –x + 2 y = – x + 3 2x – y = 8

Graph each inequality.

4. y > 5 5. y x – 1 6. 4x – 8y 4

12

12

23


7-6

>–<–


1. y = 3x – 6 2. y = – x + 4

y = –x + 2 y = – x + 3

The solution is (2, 0) There is no solution.

3. x + y = 4 4. y > 5

2x – y = 8

Solve equationsfor y:

y = –x + 4

y = 2x – 8

The solution is (4, 0).

12

12


Solutions

7-6


5. y x – 1

6. 4x – 8y 4

–8y –4x + 4

y

y x –

23

4x 4 8

12

12



7-6

>–

<–

>–

<–

<–


Solve by graphing. y < –x + 3

–2x + 4y 0

Check The point (–1, 1) is in the region graphed by both inequalities. See if (–1, 1) satisfies both inequalities.


7-6

>–

Graph y < –x + 3 and –2x + 4y 0.>–


(continued)

The coordinates of the points in the [lavender] region where the graphs of the two inequalities overlap are solutions of the system.


7-6

y < –x + 3 –2x + 4y 0

1 < –(–1) + 3 Substitute (–1, 1) for (x, y). –2(–1) + 4(1) 0

1 < 4 2 + 4 0

>–

>–

>–


Write a system of inequalities for each shaded region below.

System for the [lavender] region: y < – x + 2

y < 4

12

[red] region

boundary: y = – x + 212

The region lies below the boundary line, so the inequality is

y < – x + 2.12

[blue] region

boundary: y = 4

The region lies below the boundary line, so the inequality is

y < 4.


7-6


You need to make a fence for a dog run. The length of the

run can be no more than 60 ft, and you have 136 feet of fencing that

you can use. What are the possible dimensions of the dog run?


7-6

Define: Let = length of the dog run.

Let w = width of the dog run.

Relate: The is no 60 ft. The is no 136

ft.length more than perimeter more than

Write: 60 2 + 2 w 136

<– <–

Solve by graphing. 60

2 + 2w 136

<–<–


(continued)

The solutions are the coordinates of the points that lie in the region shaded lavender and on the lines = 60 and 2 + 2w = 136.

60m = 0: b = 60

Shade below = 60.


7-6

<–

Test (0, 0). 2(0) + 2(0) 136

0 136

So shade below

2 + 2w = 136

<–<–

2 + 2w 136

Graph the intercepts (68, 0) and (0, 68).

<–


Suppose you have two jobs, babysitting that pays $5 per

hour and sacking groceries that pays $6 per hour. You can work no

more than 20 hours each week, but you need to earn at least $90 per

week. How many hours can you work at each job?


7-6

Define: Let b = hours of babysitting.

Let s = hours of sacking groceries.

Relate: The number is less 20. The amount is at 90.of hours than or earned leastworked equal to

Write: b + s 20 5 b + 6 s 90>–<–


(continued)

The solutions are all the coordinates of the points that are nonnegative integers that lie in the region shaded lavender and on the lines b + s = 20 and 5b + 6s = 90.

Solve by graphing. b + s 205b + 6s 90


7-6

>–

<–



1. no

2. yes

3. no

4.

5.

6.

7.

8.

9.

10.

11.


7-6


12.

13.

14.

15.

16. x > 5 and y –x + 3

17. y – x – 2 and y x + 2

18. y – x + 1 and y – x + 3

19. y – x – 4 and y x – 3

20. a. 1.5f + 2.5c < 9.50, f + c > 4b.

21.

12

12

15

34

23

15

22. a. 5.99x + 9.99y 50,x 0, y 1

b.

c. 2 books and 6 CDs; no, (2, 6) is not in the shaded region.

d. Answers may vary. Sample: 3 books and 3 CDs for $47.94


7-6

>–

<–>–

>– >–

<– >–

<–>– >–


23. x + y 30, 1.25x + 3y 60

24. a. x + y 12, 6x + 4y 60

b. Answers may vary. Sample: (8, 3), (9, 1), (10, 0)

25. x 3, x –3, y 3, y –3

26. y 2, x < 5, y x

27. y x – 2, y < x + 2

28. y –x – 3, y –x + 3, y x + 3, y x – 3

29. Answers may vary. Sample: x –2, x 4, y 1, y –2

30. a. –1 b. 8

31. a. triangle b. (2, 2), (–4, –1), (–4, 2) c. 9 units2


7-6

23

23

>– <–

>– <–

<– <–>– >–

>– <–

>–

>– >–<– <–

<– <–>– >–


32. a. squareb. (1, –1), (5, –1), (1, 3), (5, 3)c. 16 units2

33. a. trapezoidb. (0, –4), (0, 2), (2, –4), (2, 0)c. 10 units2

34. a. triangleb. (2, –3), (2, 2), (7, –3)c. 12.5 units2

35. a. x 1, 10.99x + 4.99y 45

b. (3, 0), (3, 1), (3, 2), (4, 0)

36. a.

b. No; they are parallel.c. nod. no

37. a.

b. No; they are parallel.

c. It is a strip between the lines.


7-6

>– <–


38–42. Answers may vary. Samples are given.

38. x 1 and y 2

39. x < 0 and y > 0

40. y > 5 and y < 3

41. x < 2 and y < 5

42. x > 0 and y < 0

43. a. s + d > 10, s + d < 20, d 3, 10d + 0.15s < 60

b. Answers may vary. Sample: (8, 4.5); 12.5g; gold: $45.00, silver: $1.20

44.

45. Answers may vary.y > x, y < 2x

46. Answers may vary.y > x + 1, y > –x + 1

47. a. h 2 + 0.400ab.

c. Answers may vary. Sample: 5 hits, 6 at-bats


7-6

12

>–

<– <–

>–


48. a. 180x + 240y 2700x + y 17x > yy 4

b. Answers may vary. Sample: ten 14-in. drums and six 18-in. drums

49. D50. G51. [2] all points on the line 3x + 4y = 12

[1] incorrect description given

52. [4] a. Let x = number of toppings, and y = cost of pizza.Maria’s: y = 0.50x + 8Tony’s: y = 0.75x + 7

b. (4, 10) With 4 toppings, the cost is$10 at either Tony’s or Maria’s.

c. Answers may vary. Sample: Since I prefer more than 4 toppings, I will go to Maria’s, because the pizza will be less expensive.

[3] (a) and (b) only done correctly

[2] (a) done correctly, but student makes a computational error in (b)

[1] error in (a), but system solved correctly


7-6

>–<–

>–


53.

54.

55.

56.

57.

58.

59.

60. –3

61. –8

62. –

63. –

64. –

65.

66.

67.

68. ƒ(x) = 7x

69. ƒ(x) = x + 6

70. ƒ(x) = x 2

52

141583

109 6 13154


7-6



1. x 0 2. 2x + 3y > 12 3. y = x – 3 y < 3 2x + 2y < 12 2x – 3y –9

4. Write a system of inequalities for the following graph.


7-6

>–>–

23



1. x 0 2. 2x + 3y > 12 3. y x – 3

y < 3 2x + 2y < 12 2x – 3y –9


7-6

>–23

>–

>–


4. Write a system of inequalities for the following graph.

y < x + 3

y > – x – 212


7-6

Systems of Equations and InequalitiesSystems of Equations and InequalitiesALGEBRA 1 CHAPTER 7ALGEBRA 1 CHAPTER 7

1. (2, –1)

2. (0, 4)

3. one

4. one

5. none

6. one

7. one

8. inf. many

9. (8, 25)

10. (3, –7)

11. (2, –9)

12. (3.5, 1)

13. (8, –24)

14. (11, – )

15. (6, 4)

16. (–5, –2)

17. b + m = 35, b + 2m = 45; $25; $10

18. = , n + p = 24;

15 novelists, 9 poets

25

53

np

19. q + n = 15,

0.25q + 0.05n = 2.75;

10 quarters, 5 nickels

20. Answers may vary. Sample: You solve a system of linear equations by finding a single point that satisfies all the equations in the system. You solve a system of linear inequalities by graphing a region that contains points that satisfy all the inequalities in the system.

21. C

7-A

Systems of Equations and InequalitiesSystems of Equations and Inequalities

22.

23.

24.

25.

26. Answers may vary. Sample: y = x + 1, y = 3x – 5; (3, 4)

27. a. 0.10d + 0.25q < 5.00;

b. 49 itemsc. 19 items

28. a. w 30, 2 + 2w 180

b.

ALGEBRA 1 CHAPTER 7ALGEBRA 1 CHAPTER 7

7-A

<– <–

Systems of Equations and InequalitiesSystems of Equations and Inequalities

29. a.

b. x + y = 200, 0.3x + 0.5y = 84; 120 liters of 50% insecticide and 80 liters of 30% insecticide

ALGEBRA 1 CHAPTER 7ALGEBRA 1 CHAPTER 7

7-A

Documents

Solving Systems by Graphing ALGEBRA 1 LESSON 7-1 Solve each equation. 1. 2n + 3 = 5n – 22. 8 – 4z = 2z – 133. 8q – 12 = 3q + 23 Graph each pair of equations