Lesson 1: Graphing Points on a Coordinate Plane Equations/Graphin… · ... Graph the following points on the coordinate plane. (0,0) Y axis Example: Z. (-4, 6) x axis x y A. (2,4)

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Unit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing EquationsUnit 2: Graphing Equations

Lesson 1: Graphing Points on a Coordinate Plane




Directions: Graph the following points on the coordinate plane. (0,0) Y axis

Example: Z. (-4, 6) x axis

x y

A. (2,4) F. (-2,-8)

B. (-4, 1) G. (1,8)

C. (-5, -2) H. (-2,4)

D. (0, -7) I. (5, -2)

E. (8, 0) J. (4, -4)



Part 1: Identify the coordinates of the points graphed on the grid. (1 point each)

1. Point A

2. Point B

3. Point C

Part 2: Graph the following points on the graph. Label your points. (1 point each)

1. Point D: (-5, 8)

2. Point E: (2, -4)

3. Point F: (10, 0)

4. Point G: (-9, -1)

A

B C




Answer Key

Directions: Graph the following points on the coordinate plane. (0,0) Y axis

Example: Z. (-4, 6) x axis

x y

A. (2,4) F. (-2,-8)

B. (-4, 1) G. (1,8)

C. (-5, -2) H. (-2,4)

D. (0, -7) I. (5, -2)

E. (8, 0) J. (4, -4)



Part 1: Identify the coordinates of the points graphed on the grid. (1 point each)

1. Point A: (6,3)

2. Point B: (-7, -5)

3. Point C: (0, -4)

Part 2: Graph the following points on the graph. Label your points. (1 point each)

1. Point D: (-5, 8)

2. Point E: (2, -4)

3. Point F: (10, 0)

4. Point G: (-9, -1)

A

B C

D

E

F

G



Lesson 2: Using a Table of Values

Example 1

y = -4x – 8

x -4x – 8 y

-4( ) – 8

-4( ) – 8

-4( ) – 8

Example 2

y = -2/3x + 1

x -2/3x + 1 y



Lesson 2: Graphing Linear Equations Using a Table of Values

Directions: Use a table of values to plot points and graph the line of the following linear equations.

1. y = 2x -4

x 2x - 4 y

Ordered Pairs:

2. y = -x + 3

x -x + 3 y

Ordered Pairs:



3. y = -3x – 5

x -3x - 5 y

Ordered Pairs:

4. y = ½x + 4

x ½x + 4 y

Ordered Pairs:



5. y = -¾x - 2

x -¾x - 2 y

Ordered Pairs:

6. y = ⅕x - 5

x ⅕x - 5 y

Ordered Pairs:



Part 2: Which points are solutions to the equations? Justify your answer algebraically

and graphically.

7. y = 4x – 1 (3,12) (2,7) (-1,-5)

Solutions:

8. y = -1/2x + 2 (-2,1) (2,1) (-4,0)

Solutions:



Part 1: Complete the table of values. (3 points) Then

graph the line on the grid. (1 point) (4 total)

1. y = -1/2x + 2

x y

Ordered Pairs:

Part 2: Which points are solutions to the

given equation? Justify your answer

algebraically and by graphing. (3 points)

2. y = 3x - 2 (3,7) (1,-1) (-2,-4)

Solutions:



Lesson 2: Graphing Using a Table of Values

Answer Key

1. y = 2x -4

x 2x - 4 y

0 2(0) -4 -4

1 2(1) -4 -2

2 2(2)-4 0

Ordered Pairs: (0,-4) (1,-2) (2,0)

2. y = -x + 3

x -x + 3 y

0 0+3 3

1 -1 +3 2

2 -2 +3 1

Ordered Pairs: (0,3) (1,2) (2,1)



3. y = -3x – 5

x -3x - 5 y

0 -3(0) -5 -5

1 -3(1)-5 -8

-1 -3(-1) -5 -2

Ordered Pairs: (0,-5) (1,-8) (-1,-2)

4. y = ½x + 4

x ½x + 4 y

0 ½(0) + 4 4

2 ½(2) + 4 5

4 ½(4) + 4 6

Ordered Pairs: (0,4) (2,5) (4,6)



5. y = -¾x - 2

x -¾x - 2 y

0 y = -¾(0) - 2

-2

4 y = -¾(4) - 2

-5

8 y = -¾(8) - 2

-8

Ordered Pairs: (0,-2) (4,-5) (8,-8)

6. y = ⅕x - 5

x ⅕x - 5 y

0 ⅕(0) - 5 -5

5 ⅕(5) - 5 -4

10 ⅕(10) - 5 -3

Ordered Pairs: (0,-5) (5,-4) (10,-3)



Part 2: Which points are solutions to the equations? Justify your answer algebraically and

graphically.

7. y = 4x -1 (3,12) (2,7) (-1,-5)

4x -1

3 4(3) -1 11

2 4(2) -1 7

-1 4(-1) -1 -5

Solutions:

(2,7) (-1,-5)

8. y = -½x + 2 (-2, 1) (2,1) (-4,0)

-½x + 2

-2 -½(-2) + 2 3

2 -½(2)+ 2 1

-4 -½(-4) + 2 4

Solutions:

(2,1)



(Answers in the table of values may

vary depending on what x values you

choose. )

1. y = -1/2x + 2

x -1/2x + 2 y

0 -1/2(0) + 2 2

2 -1/2(2) +2 1

-2 -1/2(-2) +2 3

Ordered Pairs:

(0,2) (2, 1) (-2, 3)

Part 2: Which points are solutions to the

given equation? Justify your answer

algebraically and by graphing. (3 points)

2. y = 3x - 2 (3,7) (1,-1) (-2,-4)

X Y = 3x-2 Y

3 Y = 3(3) – 2 7

1 Y = 3(1) – 2 1

-2 Y = 3(-2) – 2 -8

The point (3,7) is the only point that is a

solution to this equation.



Lesson 3: Calculating Slope

Example 1

Rise = ______________ Run = ______________ Slope = rise run __________

Formula for Slope



Example 2

Example 3

Rise = _________ Run = _________ Slope = rise run _________

Rise = _________ Run = _________ Slope = rise run _________



Example 4

Special Notes:

A horizontal line always has a

slope of _____________because

_____________________________

A vertical line has an ___________

slope because ________________

_____________________________



Lesson 3: Calculating Slope

Directions: Calculate the slope for each graph.

1.

2.

3.

4.

Rise = ________ Slope =

Run = _________

Slope = rise

run

Rise = ________ Slope =

Run = _________

Rise = ________ Slope =

Run = _________

Rise = ________ Slope =

Run = _________



5.

6.

7.

8. John calculated that the slope of the

following line was 3/2. Is he correct?

Explain why or why not!

Rise = ________ Slope =

Run = _________

Rise = ________ Slope =

Run = _________

Rise = ________ Slope =

Run = _________

Explanation:



Part 1: Calculate the slope of each line. (1 point each)

1. 2.

Part 2: Graph the points. Then calculate the slope of the line. (2 points each)

1. (2, 5) & (-2, 0) 2. (-3, 8) & (2, -3)



Lesson 3: Calculating Slope – Answer Key

Directions: Calculate the slope for each graph.

1.

2.

3.

4.

1 = run

3 = rise

2 = run

1 = run

1 = run

1 = rise

Slope = Rise = 1 SLOPE= 1

Run 1


Run 1

- 4 = rise (down)

Slope = Rise = -4 SLOPE = -4

Run 1

1 = rise

Slope = Rise = 1 SLOPE= 1/2

Run 2



5.

6.

7. 8. John calculated that the slope of the

following line was 3/2. Is he correct?

Explain why or why not!

3 = run

-2 = rise (down)

Slope = Rise = -2 SLOPE= -2/3

Run 3

4 = rise

3 = run

Slope = Rise = 4 SLOPE= 4/3

Run 3


Run 1

For all horizontal lines, the slope is 0.

The slope is not 3/2, it is -3/2. The line is

falling from left to right, so the slope is

negative. There is a rise (fall) of -3 and a run

of 2. Therefore, the slope is -3/2.



1. 2.

Rise: 5 The slope is 5/6. Rise = -1 The slope is -1/4 Run 6 Run = 4

Part 2: Graph the points. Then calculate the slope of the line. (2 points each)

1. (2, 5) & (-2, 0) 2. (-3, 8) & (2, -3)

Rise = 5 The slope is 5/4. Rise = -11 The slope is -11/5

Run = 4 Run = 5



**BONUS LESSON** - Predicting Slope

There are a couple of rules that you can use to predict the slope of a line. These rules will

also help you to check your answers as you calculate slope!

For each graph below, determine whether the slope of the line will be positive or negative.

You read a graph from left to right just as you do

a book. If the line is falling (from left to right),

then the slope is negative.

The slope of this line is negative!

If the line is rising (from left to right), then the

slope of the line is positive.

The slope of this line is positive!



Predicting Slope – Answer Key

For each graph below, determine whether the slope of the line will be positive or negative.

Negative Positive Negative

Notice that the sign (negative or positive) only dictates whether the line is rising or falling. It

has no impact on the steepness of the line. In order to identify the steepness of the slope,

we will look at the absolute value of the slope. (By absolute value, I am referring to the

number without regards to it’s’ sign). Take a look at the following examples.

Notice that in the first graph, the slopes of the lines are less than 1. The slopes of the lines in

the second graph are 1 or more. Notice how the slope of 1 and 2 are greater than the slopes

of ½ or ¼. The larger the absolute value of the slope, the greater (steeper) the slope of the

line.



Lesson 4: Graphing Slope

Example 1

Example 2

1. Start with a point on (0,-4).

2. Draw a line with a slope of

¾.

1. Start with a point on (0,3).


-2.



Example 3

Example 4



3/5.

Part 1:

1. Start with a point on (0,-2).

2. Draw a line with a slope of 0.

Part 2:


4. Draw a line with an undefined

slope through this point.



Lesson 4: Graphing Slope

Directions: Start at the point given to you on the graph. Use the slope given to plot the next 2

points and then draw a line to create your graph.

1. Slope = 3 2. Slope = -2

3. Slope = 2/3 4. Slope = -1/2



5. Slope = 4/3 6. Slope = 0

7. Slope = -1/3 8. Slope = Undefined

Quiz #1 is coming up next! You will need to be able to:

• graph an equation using a table of values

• calculate slope

• graph slope.

**Hint: Create yourself a study guide to help you prepare for the quiz.



Directions: Start at the point given to you on the graph.

Use the slope given to plot the next 2 points

and then draw a line to create your graph. (2 points each)

1. Slope = 1/3 2. Slope = -4

3. Slope = -3/4 4. Slope = 0



Lesson 4: Graphing Slope – Answer Key

1. Slope = 3 2. Slope = -2

Rise = 3 (up)

Run 1 (right)

Rise = -2 (down)

Run 1 (right)

Rise = 2 (up)

Run 3 (right)

3. Slope = 2/3 4. Slope = -1/2

Rise = -1 (down)

Run 2 (right)



5. Slope = 4/3 6. Slope = 0

7. Slope = -1/3 8. Slope = Undefined

Rise = 4 (up)

Run 3 (right)

Rise = 0 **A slope of 0 always results

Run 1 in a horizontal line.

Rise = -1 (down)

Run 3 (right)

A slope that’s undefined always

results in a vertical line.



3. Slope = -3/4 4. Slope = 0

Rise 1

Run 3

Rise -4 OR Rise 4

Run 1 Run -1

Rise -3 Or Rise 3

Run 4 Run -4

A line with a slope of 0, is always a

horizontal line.



Quiz # 1

1. Which line on the graph has a slope of -2/3?

2. What is the slope of the line that passes 3. Start with a point on (-4,8). Draw a

through (-2, 4) (5, 8)? line with a slope of -2.

A. Line A

B. Line B

C. Line C

D. Line D



4. Start with a point on the origin. Draw a line with a slope of 3/5.

5. Complete the table of values for the following equation. Then find the slope of the line.

y = 1/4x - 2

x y



Quiz # 1 – Answer Key

1. Which line on the graph has a slope of -2/3? (1 point)

2. What is the slope of the line that passes 3. Start with a point on (-4,8). Draw a

through (-2, 4) (5, 8)? (2 points) line with a slope of -2. (2 points)

A. Line A

B. Line B

C. Line C

D. Line D

B

You can eliminate answers C and

A because they are positive slopes.

(When read from left to right, they

rise).

Letters D and B are negative

slopes, but only letter B rises -2

and runs 3.

Rise = 4

Run = 7

Slope = rise/run

Slope = 4/7

(-4, 8)

A slope of -2 means that you count

down 2 and right 1.

Slope = rise/run

Slope = -2/1



4. Start with a point on the origin. Draw a line with a slope of 3/5. (2 points)

5. Complete the table of values for the following equation. Then find the slope of the line.

y = 1/4x – 2 (3 points)

x y = 1/4x - 2

y

0 y = 1/4x - 2

y = ¼(0) – 2

-2

4 y = 1/4x - 2

y = ¼(4) – 2

-1

-4 y = 1/4x - 2

y = ¼(-4) -2

-3

Origin (0,0)

A slope of 3/5 means that you count up 3 and

right 5.

Slope = rise/run

Slope = 3/5

Ordered Pairs

(0, -2)

(4, -1)

(-4, -3)

This quiz is worth 10 points.

The slope of the line is

¼. You rise 1 and run

4 to get from one

point to the next

point.



Lesson 5: Graphing Using Slope Intercept Form

Example 1 Example 2

Slope = ________ Y-intercept = __________

Slope = ________ Y-intercept = __________

Notes



Example 3

Example 4



Lesson 5: Using Slope Intercept Form to Graph Equations

Directions: Graph each equation using the following directions:

Example:

1. Circle the y-intercept in the equation.

Then plot this point on your graph.

y = 1/2x +2

2. Put a square around the slope in the

equation.

Then use the slope to plot your next point.

y = ½ x +2

3. Draw a line through your two points.

(0,2)

Rise = 1 (up 1)

Run 2 (right 2)



1. y = 2x +2 2. y = -3x +1

3. y = 1/3x - 4 4. y = -2/5x - 2



5. y = -3/4x 6. y = x +5

7. y = -2 8. y = -4x



9. Which graph represents the equation: y = -2/3x – 7?

A. Line A

B. Line B

C. Line C

D. Line D

10. Graph the equation: y = 3x -8

on the grid. Label this line A.

Identify three solutions for this

equation.

11. Graph the equation: y = -x +10

on the grid. Label this line B.


equation.



1. Which line on the graph represents the equation: y = -3/4x – 2? (1 point)

A. Line A

B. Line B

C. Line C

D. Line D

Part 2: Identify the slope and y-intercept in each equation. Then graph the equation on the grid.

(3 points each)

1. y = 3x -6 2. y = -1/4x

Slope: _______ Slope: __________

Y-intercept:_________ Y-intercept:___________

A

B

C

D



Lesson 5: Using Slope Intercept Form - Answer Key

1. y = 2x + 2

3. y = 1/3x - 4 4. y = -2/5x - 2

2. y = -3x +1



5. y = -3/4x 6. y= x +5

7. y = -2 8. y = -4x

**The y-intercept is (0,0). When no

number is written in the y-intercept’s

place, it is assumed to be 0.

**The slope is 1. When there is no

coefficient (slope), the slope is assumed

to be 1. (1 times x is still x)

**There is no variable term in this

equation; therefore the slope is 0.

Anytime the slope is 0, the line is

horizontal. Therefore, we have a

horizontal line through the y-intercept.

**The y-intercept is (0,0). When no

number is written in the y-intercept’s

place, it is assumed to be 0.



9. Which graph represents the equation: y = -2/3x – 7?

A. Line A

B. Line B

C. Line C

D. Line D

10. Graph the equation: y = 3x -8

on the grid. Label this line A.


equation.

11. Graph the equation: y = -x +10

On the grid. Label this line B.


equation.

Since the equation has a slope of

-2/3, you can eliminate C and D

because both lines have a

positive slope (rising from left to

right.)

The equation y = -2/3x – 7 has a

y-intercept of -7. Therefore, letter

B, must be the correct answer

because the line passes through

the point (0, -7).

Letter B has a slope of -2/3. y-intercept = -7

Possible solutions:

(4,4) (3,1) (2,-2) (1,-5) (0,-8)

(Or any point on the line)

Possible solutions:

(0,10) (1,9) (2,8) (3,7) (4,6)

(Or any point on the line)



1. Which line on the graph represents the equation: y = -3/4x – 2? (1 point)

A. Line A

B. Line B

C. Line C

D. Line D

Part 2: Identify the slope and y-intercept in each equation. Then graph the equation on the grid.

(3 points each)

1. y = 3x -6 2. y = -1/4x +0

Slope: 3 Slope: -1/4

Y-intercept: -6 Y-intercept: 0 (this is the origin)

A

B

C

D

Line B has a y-intercept of -2. The

slope is -3/4, so if you count down

three and right 4, you will end up on

the next point on the line. The slope is

negative, so you can immediately

eliminate letter D since its slope is

positive.



Lesson 6: Finding Slope Given Two Points

Example 1 Example 2

Find the slope of the line that passes through

(-3,0) and (3,6)

Find the slope of the line that passes through

(4, -4) and (6,7)

Formula for Finding Slope Given Two Points

Using two points: (x1, y1) (x2, y2)



Lesson 6: Finding Slope Given Two Points

Part 1: Find the slope of a line given two points.

1. (2,7) (-5,-3)

2. (10, 3) (9,2) 3. Find the slope of a line that passes

through the origin and the point (-6,2) .

4. Which number represents the slope of the line that contains the following points:

(-2, 8) & (7, -10)

A. -2/3 B. 2 C. -2 D. 2/3

Graph the line for the two points above (-2, 8) & (7, -10) to check your answer. Was your answer

correct? Explain.

Formula For Finding Slope:

Using Two Points: (x1, y1) & (x2, y2)

y2 – y1

x2 - x1

Explain Your Answer:



5. Line A passes through the points (2, -7) & ( -3,5). Line B passes through the points

(10, -8) & (22, 9). Which line has a greater slope? Explain your answer.

6. Find the slope of the line that passes through the points (2,3) & (2,-5). Describe what this line

would look like when graphed. Make a sketch of the line.

7. Find the slope of the line that passes through the points (8,4) & (-7,4). Describe what this line




1. Using the slope formula, find the slope of the line that contains the points: (-3,6) & (5,10) (2 points) 2. The three points are vertices of a triangle. Find the slope of each side of the triangle. (6 points) A (-5,1) B (-2, 5) C (1, -4) AB_______ AC_______ BC_______

3. Find the missing value using the given slope and points. (1 point)

1. m = 3/2 and (x, 4) (5,7)



Lesson 6: Finding Slope Given Two Points – Answer Key

Part 1: Find the slope of a line given two points.

1. (2, 7) (-5,-3)

x1 y1 x2 y2

y2 – y1 = -3 – 7 = -10 = 10

x2 – x1 -5 -2 -7 7

Slope = 10/7

2. (10, 3) (9, 2)

x1 y1 x2 y2

y2 – y1 = 2 - 3 = -1= 1

x2 – x1 9 – 10 -1

Slope = 1

4. Which number represents the slope of the line that contains the following points:

(-2, 8) & (7, -10) y2 – y1 = -10 – 8 = -18 = -2

x2 – x1 7 – (-2) 9

A. -2/3 B. 2 C. -2 D. 2/3

b. Graph the line for the two points above (-2, 8) & (7, -10) to check your answer. Was your

answer correct? Explain.

Formula For Finding Slope:

Using Two Points: (x1, y1) & (x2, y2)

y2 – y1

x2 - x1

Explain Your Answer: (Your answer may vary based on

your answer above.)

My answer was correct because the slope is -2. From each

point on the line, if I count a rise of 2 and a run of -1 (to the

left), I will find the next point on the line. Therefore, the

slope is -2.

3. Find the slope of the line that passes

through the origin and the point (-6,2)

(0,0) (-6, 2)

x1 y1 x2 y2

y2 – y1 = 2 – 0 = 2 = -1

x2 – x1 -6 -0 -6 3

Slope = -1/3



5. Line A passes through the points (2, -7) & ( -3,5). Line B passes through the points

(10, -8) & (22, 9). Which line has a greater slope? Explain your answer.

Line A: y2 – y1 = 5 – (-7) = 12 = -2 &2/5

x2 – x1 -3 – 2 -5

Line B: y2 – y1 = 9 – (-8) = 17 = 1 & 5/12

x2 – x1 22 – 10 12

Line A has the greater slope. Line A’s slope is – 12/5 and line B’s slope is 17/12. 12/5 is greater

than 17/12. When you are determining the steepness of the slope, disregard the negative signs.

Take the absolute value of the number and the larger the number the greater the slope. The

positive and negative signs only determine whether the line rises or falls.

6. Find the slope of the line that passes through the points (2,3) & (2,-5). Describe what this line


y2 – y1 = -5 – 3 = -8

x2 – x1 2 -2 0

Since the slope has a 0 in the denominator, the slope of the

line is undefined. That means that this is a vertical line.

7. Find the slope of the line that passes through the points (8,4) & (-7,4). Describe what this line


y2 – y1 = 4-4 = 0

x2 – x1 -7 -8 -15

Since this slope has a 0 in the numerator, the slope

of the line is 0. This is a horizontal line through (0,4).



1. Using the slope formula, find the slope of the line that contains the points: (-3,6) & (5,10) (2 points) y2 – y1 10 – 6 = 4 = 1 x2 - x1 5-(-3) 8 2 The slope of the line is 1/2 2. The three points are vertices of a triangle. Find the slope of each side of the triangle. (6 points) A (-5,1) B (-2, 5) C (1, -4) AB 4/3 AC -5/6 BC -3

3. Find the missing value using the given slope and points. (1 point)

1. m = 3/2 and (x, 4) (5,7)

y2 – y1 7 – 4 = 3 x2 - x1 5 – x 2

x = 3 because 5 – 3 = 2. Since the slope is 3/2, we know that the denominator must equal 3 once

the coordinates are substituted into the slope formula.

Line segment AB

A (-5,1) B (-2,5)

y2 – y1 5 – 1 = 4 x2 - x1 -2 – (-5) 3

Slope of AB = 4/3

Line segment AC

A (-5,1) C (1, -4)

y2 – y1 -4 - 1 = -5 x2 - x1 1 – (-5) 6

Slope of AC= 5/6

Line segment BC

B (-2,5) C (1, -4)

y2 – y1 -4 –5 = -9 = -3 x2 - x1 1-(-2) 3

Slope of BC = -3



Lesson 7: Rate of Change

Example 1

Formula for Finding Slope Given Two Points

Using two points: (x1, y1) (x2, y2)

In real world problems, slope is related to how something changes (usually over time). In this

case, slope is referred to as rate of change. Therefore, in real world problems when you are

asked to find the rate of change, you are actually finding the slope. The same formulas that

you use for slope will also be used for rate of change.

1. What is the average rate of change

between hours 4 and 9?

2. What is the average rate of change

between hours 9 and 11? What most likely

occurred during this time?



Example 2

Jerry purchased a house in 1994 for $230,000. In 2008, he sold the house for $378,000. What is

the average rate of change in the value of the house per year?



Lesson 7: Rate of Change Practice

The following graph shows the weekly price of a stock over a ten week period. Use the graph to

answer the following questions:

1a. How much was the stock worth at the very beginning (week 0)? ___________

Write this as an ordered pair. (# of week, Price of stock) (0, ____)

b. How much was the stock worth during week 3? ___________

Write this as an ordered pair. (# of week, Price of stock) (3, ____)

c. What is the average rate of change during this time frame (week 0 to week 3)?

Write your two ordered pairs: (___ , ____) (___ , ____). Find the slope. (Slope and rate of

Slope = y2 – y1 change are synonymous)

x2 - x1

The rate of change from week 0 to week 3 is ________________. This means

that_________________________________________________________________________



2. What is the average rate of change from week 3 to week 5?

Week 3 ordered pair: ___________ Week 5 ordered pair _____________

Rate of Change = slope = y2 – y1

x2 - x1

The rate of change from week 3 to week 5 is _____________.

6b. Explain what the rate of change means during this time frame.

3. Find the rate of change for the company’s final downfall, between weeks 7 and 10.



4. As part of a New Year’s Resolution plan, Brenda started a weight loss program on January 1,

2009. On January 1, 2009 she weighed 185 pounds. Six months later she recorded a weight of

140 pounds. Find the average monthly rate of change of her weight in pounds per month.

Ordered Pairs: ___________________ ___________________________

January weight June weight

5. Jonathan was driving from Maryland to Virginia for vacation. He started his trip at 5:00am. By

7:00 am he had travelled 120 miles. By 9:00 am he had travelled 210 miles. He stopped for

breakfast between 9 and 10:00 am. By 12:00 pm he had reached his destination and had travelled

480 miles.

a. Create a graph to illustrate this situation. Don’t forget to label your axis and title your graph.

b. Find the average rate of change in miles per hour between 5:00 am and 9:00 am.

c. Find the average rate of change in miles per hour for the entire trip.



6. Michael bought 573 shares of stock for $730 in 1995. When he sold the stock in 2007, the stock

was valued at $1735. What is the average rate of change of the price of the stock per year?

7. Claudia bought her first brand new car in 2004 for $23,458. She sold that car in 2010 for

$12,010. What is the average rate of change of the price of the car per year?

8. In the first year that Patricia started her career as a teacher, she made $26,800 per year. She is

now in her 17th year of teaching and her annual salary is $62,000 per year. What is the average

rate of change in dollars per year?



1. Josie tracks the balance in her savings account

very carefully. She opened an account and in

January she deposited $480. In February she was

able to deposit $130 dollars more into her account.

In March, she did not work much, so she made no

deposits or withdrawals. In April she deposited

another $440. In May she bought a computer, so

she withdrew $500.

A. Create a graph to represent the balance in Josie’s savings account. (3 points)

B. What is the average rate of change between January and April? (2 points)

C. What is the average rate of change from the time she opened the account to the end of

May? (2 points)

D. What is the average rate of change between February and March? How do you know?

(2 points)



Lesson 7: Rate of Change – Answer Key The following graph shows the weekly price of a stock over a ten week period. Use the graph to

answer the following

questions:

1a. How much was the stock worth at the very beginning (week 0)? $12

Write this as an ordered pair. (# of week, Price of stock) (0, 12)

b. How much was the stock worth during week 3? $19

Write this as an ordered pair. (# of week, Price of stock) (3, 19)

c. What is the average rate of change during this time frame (week 0 to week 3)?

Write your two ordered pairs: (0, 12) (3, 19) Find the slope. (Slope and rate of

Slope = y2 – y1 = 19 - 12 = 7 change are synonymous)

x2 - x1 3 -0 3

The rate of change from week 0 to week 3 is 7/3. This means that between weeks 0 and 3, the

average price of the stock rose $2.33 cents per week. (Since we are talking about money, I

converted the fraction 7/3 to a decimal. 7/3 = 2 – 1/3 which is 2.33333…)



2. What is the average rate of change from week 3 to week 5?

Week 3 ordered pair: (3, 19) Week 5 ordered pair (5, 8)

Rate of Change = slope = y2 – y1 = 8 – 19 = -11

x2 - x1 5 – 3 2

The rate of change from week 3 to week 5 is -11/2.

2b. Explain what the rate of change means during this time frame.

During the time period between weeks 3 and 5, the average price of the stock fell $5.50 per week. I

changed 11/2 to a decimal since we were talking about money. 11/2 = 5.50. The price of the stock

fell because the slope is negative.

3. Find the rate of change for the company’s final downfall, between weeks 7 and 10.

Week 7 ordered pair (7, 13) Week 10 ordered pair (10, 2) x1 y1 x2 y2

y2 – y1 = 2 – 13 = -11

x2 - x1 10 - 7 3

The rate of change between weeks 7 and 10 is -11/3

or -3.67. The average price of the stock fell $3.67 per

week between the weeks of 7 and 10.



4. As part of a New Year’s Resolution plan, Brenda started a weight loss program on January

1, 2009. On January 1, 2009 she weighed 185 pounds. Six months later she recorded a

weight of 140 pounds. Find the average monthly rate of change of her weight in pounds per

month.

Ordered Pairs: (1, 185) (6, 140)

January weight June weight

y2 – y1 = 140- 185 = -45 = -9

x2 - x1 6-1 5

5. Jonathan was driving from Maryland to Virginia for vacation. He started his trip at

5:00am. By 7:00 am he had travelled 120 miles. By 9:00 am he had travelled 210 miles. He

stopped for breakfast between 9 and 10:00 am. By 12:00 pm he had reached his destination

and had travelled 480 miles.

a. Create a graph to illustrate this situation. Don’t forget to label your axis and title your

graph.

b. Find the average rate of change in miles per hour between 5:00 am and 9:00 am.

(5, 0) (9, 210)

y2 – y1 = 210-0 = 210 = 52.5

x2 - x1 9-5 4

Brenda’s average rate of change is -9. This means

that she lost an average of 9 pounds per month

between January and June.

The average rate of change between 5 and 9 am is 52.5

miles per hour.



c. Find the average rate of change in miles per hour for the entire trip.

(5, 0) (12, 480)

y2 – y1 = 480-0 = 480 = 68.6

x2 - x1 12-5 7

6. Michael bought 573 shares of stock for $730 in 1995. When he sold the stock in 2007, the stock

was valued at $1735. What is the average rate of change of the price of the stock per year?

7. Claudia bought her first brand new car in 2004 for $23,458. She sold that car in 2010 for

$12,010. What is the average rate of change of the price of the car per year?

8. In the first year that Patricia started her career as a teacher, she made $26,800 per year. She is

now in her 17th year of teaching and her annual salary is $62,000 per year. What is the average

rate of change in dollars per year?

The average rate of change over the entire trip

is 68.6 miles per hour.

(year, price of stock)

(1995, 730) (2007, 1735)

y2 – y1 = 1735 – 730 = 1005 = 83.75

x2 - x1 2007-1995 12

The stock rose an average of $83.75 per year.

(year, price of car)

(2004, 23,458) (2010, 12,010)

y2 – y1 = 12010-23458 = -11448 = -1908

x2 - x1 2010-2004 6

The car’s value decreased in value by $1908 a year.

(year, salary)

(1, 26800) (17, 62000)

y2 – y1 = 62000-26800 = 35200 = 2200

x2 - x1 17-1 16

Patricia’s salary has an average rate of change

of $2200 per year. Therefore, her salary

increases by about $2200 per year.

**Note: The number of shares (573) is

irrelevant information in this problem. It is not

needed to find the rate of change.



1. Josie tracks the balance in her savings account

very carefully. She opened an account on

December 31st and in January she deposited $480.

In February she was able to deposit $130 dollars

more into her account. In March, she did not work

much, so she made no deposits or withdrawals. In

April she deposited another $440. In May she

bought a computer, so she withdrew $500.

A. Create a graph to represent the balance in Josie’s savings account. (3 points)

B. What is the average rate of change between January and April? (2 points)

January (1, 480) April (4, 1050)

y2 – y1 1050 – 480 = 570 = $190 x2 - x1 4-1 3

The rate of change between January and April is $190 per month.

C. What is the average rate of change from the time she opened the account to the end of

May? (2 points)

Open account (0,0) May (5,550)

y2 – y1 550-0 = 550 = $110 x2 - x1 5-0 5

The rate of change from the time she opened the account to the end of May is $110 per

month.



D. What is the average rate of change between February and March? How do you know?

(2 points)

Between February and March, the rate of change is 0 because she did not make any

deposits or withdrawals. Her account balance remained the same. The graph shows a

horizontal line which also indicates a slope of 0.



Lesson 8: Graphing Standard Form Equations – Lesson 1 of 2

A Close Look at Intercepts

The y-intercept is on the ________________. The y-intercepts all have _______________

___________________________________.

The x-intercept is on the ________________. The x-intercepts all have _______________

___________________________________.

Finding Intercepts

To find the y-intercept: ________________________

To find the x-intercept: ________________________



Example 1

Example 2

Graph the line:

4x – 3y = -24

X-intercept: (Let y = 0)

Y-intercept: (Let x = 0)

Graph the line:

-2x - 9y = 18

X-intercept: (Let y = 0)

Y-intercept: (Let x = 0)



Example 3

Example 4

Mary is preparing for a Super Bowl Party. She has $200 to spend on pizza and wings. One pizza

costs $10. One order of wings costs $5. The equation representing x pizzas and y orders of

wings is: 10x + 5y = 200

• Graph the equation on the grid. Let x = the number of pizzas and y = the number of orders

of wings.

• Using your graph, list one realistic option for buying pizza and wings.

Find the x-intercept of the line: y = -3x – 2.



Lesson 8: Graphing Standard Form Equations

1. Which line has an x intercept of 8 and a y intercept of 5?

2.. Find the x intercept of the line: y = -3x -1

A. -1 C. -1/3

B. -3 D. 1

3. Find the x intercept of the line: y = 2x +4

HINT:

Let y = 0 to find the x-intercept.

A. Line A

B. Line B

C. Line C

D. None



4. The slope of a line is ½. The y- intercept is -3. Find the x-intercept.

5. Given the equation: 3x +6y = 9, find the x and y intercepts.

6. Graph the following equations on the grid provided.

A. -2x +8y = 16 B. x – 4y = 8

X intercept = __________ X intercept = __________

Y intercept = __________ Y intercept = __________



7. John has started a summer business mowing grass. He charges $10 per hour, but had to spend

$50 on equipment. The equation, y = 10x – 50 is used to determine the total profit after mowing x

lawns.

• Graph the equation on the grid. Let x = the number of hours worked. Let y = the profit.

• Use your graph to determine the amount of profit for 10 hours. Justify your answer.

• Explain why the y-intercept is a negative number in this problem.

Hint: Since the numbers in the equation are so

large, it will be easier to find the x and y

intercepts.



8. Jerry spent $30 on ice cream and cookies for his children’s party. Ice cream costs $2 a cone and

cookies cost $1.50 a piece. The equation representing x ice cream cones and y cookies is

2x +1.50y = $30.

• Graph the equation on the grid. Let x = number of ice cream cones and y = number of

cookies.

• Suppose Jerry bought 6 ice cream cones, how many cookies can he buy for $30? Explain

your answer.

9. John is supplying hot dogs and sodas for the high school football game. He has $450 to spend.

The hot dogs cost $3 per pack of 6. The sodas cost $0.50 a piece. The equation representing x

hot dogs and y sodas is: 3x +.50y = 450.

• Graph the equation on the grid.

Let x = the number of hot dogs and

y = the number of sodas.

• Using your graph, list three different realistic options for buying hot dogs and sodas.

(For example, John could buy ____ packages of hot dogs and ____ sodas for $450).

Quiz #2 is next! You will need to be able to graph equations written in slope intercept form, find the

slope of a line given 2 points, and calculate rate of change. Don’t forget to create a study guide.



1. The slope of a line is 5 and the y- intercept is -4. Find the x-intercept.

(2 points)

For problems 2-3, identify the x and y-intercepts. Then graph the equations on the grid (3 points ea)

2. 3x – 2y = 18 3. x + 3y = 9

x-intercept: __________ x-intercept: ________

y-intercept: __________ y-intercept: ________

4. Kerry is purchasing food and drinks for an annual Halloween party. She has a budget of $250

and she must provide appetizers and drinks for 40 people. Appetizers average about $9 per box

and sodas average $.90 a 2 liter bottle. Let x represent the number of boxes of appetizers and let y

= the number of 2 liter bottles of soda. The equation that represents this situation is: 9x + .90y =

250. (3 points)


Let x = the number of appetizers and

y = the number of 2 liters of soda.

• Using your graph, list two different realistic options for buying appetizers and sodas.

(For example, Kerry could buy ____ packages of appetizers and ____ sodas for $250).



Lesson 8: Graphing Standard Form Equations– Answer Key

1. Which line has an x intercept of 8 and a y intercept of 5?

2.. Find the x intercept of the line: y = -3x -1

X intercept: Let y = 0

y = -3x - 1

0 = -3x -1

0+1 = -3x -1 + 1

1 = -3x

1/-3 = -3x/-3

-1/3 = x

A. -1 C. -1/3

B. -3 D. 1

A. Line A

B. Line B

C. Line C

D. None

B

HINT:

You can use the same method to find

the x intercept even if the equation is

written in slope intercept form.

Let y = 0 to find the x-intercept.



3. Find the x intercept of the line: y = 2x +4

Let y = 0

Y = 2x + 4

0 = 2x + 4 Substitute 0 for y.

0 – 4 = 2x + 4 -4 Subtract 4 from both sides.

-4 = 2x

-4/2 = 2x /2 Divide by 2 on both sides.

-2 = x The X intercept is -2

4. The slope of a line is ½. The y- intercept is -3. Find the x-intercept.

Since I know the slope and y intercept, I can write an equation in slope intercept form:

y = mx +b m = slope (1/2) b = y intercept (-3)

y = 1/2x -3 Write the equation in slope intercept form.

0 = 1/2x – 3 Let y = 0 to find the x-intercept

0 +3 = 1/2x – 3 + 3 Add 3 to both sides.

3 = 1/2x

(2)3 = (2)1/2x Multiply by 2 on both sides.

6 = x The x intercept is 6

5. Given the equation: 3x +6y = 9, find the x and y intercepts.

X intercept: Let y = 0 Y Intercept: Let x = 0

3x +6y = 9 3x +6y = 9

3x +6(0) = 9 3(0) +6y = 9

3x = 9 6y = 9

3 3 6 6

x = 3 X intercept = 3 y = 3/2 Y intercept = 3/2



6. Graph the following equations on the grid provided.

A. -2x +8y = 16 B. x – 4y = 8

X Intercept: y=0 Y Intercept: x=0 X intercept: y=0 y intercept: x=0

(Eliminates y term) (Eliminates x term) x = 8

-2x = 16 8y = 16 -4y = 8 -2 -2 8 8 -4 -4 X = -8 y = 2 y = -2

X intercept = ___-8_______ X intercept = ____8______

Y intercept = ____2______ Y intercept = _____-2_____



7. John has started a summer business mowing grass. He charges $10 per hour, but had to spend

$50 on equipment. The equation, y = 10x – 50 is used to determine the total profit after mowing x

lawns.

• Graph the equation on the grid. Let x = the number of hours worked. Let y = the profit.

• Use your graph to determine the amount of profit for 10 hours. Justify your answer.

• Explain why the y-intercept is a negative number in this problem.

Since the equation is written in slope intercept form, we

know the y intercept is -50. Since this is a large number

and we know that we cannot use a scale of 1 on the

graph, it would be easiest to find the x intercept.

Y = 10x-50

0 = 10x – 50 Let y = 0

0 + 50 = 10x – 50 + 50

50 = 10x

50/10 = 10x/10

5 = x

X intercept = 5 Y intercept = -50

**You need to establish your scale for the x and y axis.

Since the x intercept is 5, and you only need to

determine a profit for 10 hours, you can use a scale of 1.

Since the y intercept is -50, we can use a scale of 10.

• According to the graph, the amount of profit for 10 hours would be $50.

• Justify:

y = 10x-50 Original Equation

y = 10(10) – 50 Substitute 50 hours for x

y = 50 y is the profit, so our profit is $50.

• The y-intercept is a negative number in this problem because before John starts

working (0 hours), he has lost money. Since he had to buy equipment for $50, he is in

the negative. He would have to work 5 hours before breaking even at $0 and then he

will start making a profit.



8. Jerry spent $30 on ice cream and cookies for his children’s party. Ice cream costs $2 a cone and

cookies cost $1.50 a piece. The equation representing x ice cream cones and y cookies is

2x +1.50y = $30.

• Graph the equation on the grid. Let x = number of ice cream cones and y = number of

cookies.

• Suppose Jerry bought 6 ice cream cones, how many cookies can he buy for $30? Explain

your answer.

Since this equation is already written in standard form,

it is easiest to find the x and y intercepts in order to

graph this equation on the grid.

2x + 1.50y = 30

X Intercept: y = 0 Y Intercept: x =

0

2x + 1.50y = 30 2x + 1.50y = 30

2x + 1.50(0) = 30 2(0) +1.50y = 30

2x = 30 1.50y = 30

2 2 1.50 1.50

x = 15 y = 20

X intercept = 15 Y Intercept = 20

**I must choose a scale that will fit on my graph. I

chose to count by 2’s because then I can maximize the

space in my graph. This graph has endpoints at the x

and y intercept since it represents Jerry’s spending to

$30. It doesn’t go on until infinity like other graphs.

According to the graph, if Jerry bought 6 ice cream cones, he could buy 12 cookies for $30. The point on the

line with an x coordinate of 6 is (6, 12). That means 6 ice cream cones and 12 cookies can be purchased for

$30. Let’s prove it by substituting into the original equation.

2x + 1.50y = 30

2(6) +1.50(12) =

12 + 18 = 30



9. John is supplying hot dogs and sodas for the high school football game. He has $450 to spend.

The hot dogs cost $3 per pack of 6. The sodas cost $0.50 a piece. The equation representing x

hot dogs and y sodas is: 3x +.50y = 450.


Let x = the number of hot dogs and

y = the number of sodas.

• Using your graph, list three different realistic options for buying hot dogs and sodas.

(For example, John could buy ____ packages of hot dogs and ____ sodas for $450).

Since this equation is already written in

standard form, it would be easiest to graph

using the x and y intercepts. We also know

that our endpoints on the graph are the x and

y intercept since John’s spending is limited to

$450.

3x + .50y = 450

X intercept: y = 0 Y intercept: x = 0

3x + .50y = 450 3x +.50y = 450

3x +.50(0) = 450 3(0) +.50y = 450

3x = 450 .50y = 450

3 3 .50 .50

x = 150 y = 900

I need to choose a scale for my graph based

on the x and y intercept. I have 10 spaces on

the horizontal axis and it must reach 150. I

am going to use a scale of 20 on the x axis.

The y axis must reach 900. So, I will need to

use a scale of 100.

With $450, John could buy:

50 packages of hot dogs and 600 sodas.

3x +.50y = 450

3(50) +.50(600) = 450


3x +.50y = 450

3(60) +.50(540) = 450


3x +.50y = 450

3(80) +.50(420) = 450



1. The slope of a line is 5 and the y- intercept is -4. Find the x-intercept. (2 points)

Slope = 5 Y-int = -4. Equation in slope intercept form: y = 5x – 4

If we are finding the x-intercept, then we will let y = 0.

Y = 5x – 4

0 = 5x – 4 let y = 0

0+4 = 5x – 4+ 4 Add 4 to both sides

4 = 5x Simplify

4/5 = 5x/5 Divide by 5 on both sides

4/5 = x Simplify

The x-intercept is equal to 4/5.

For problems 2-3, identify the x and y-intercepts. Then graph the equations on the grid. (3 points

ea)

2. 3x – 2y = 18 3. x + 3y = 9

x-intercept: 6 x-intercept: 9

y-intercept: -9 y-intercept: 3

X-intercept:

Let y = 0

3x – 2(0) = 18

3x = 18

3x/3 = 18/3

x= 6

Y-intercept:

Let x = 0

3(0)– 2y = 18

-2y = 18

-2y/-2 = 18/-2

x= -9

X-intercept:

Let y = 0

x +3(0) = 9

x = 9

Y-intercept:

Let x = 0

(0)+ 3y = 9

3y = 9

3y/3 = 9/3

x= 3



4. Kerry is purchasing food and drinks for an annual Halloween party. She has a budget of $250

and she must provide appetizers and drinks for 40 people. Appetizers average about $9 per box

and sodas average $.90 a 2 liter bottle. Let x represent the number of boxes of appetizers and let y

= the number of 2 liter bottles of soda. The equation that represents this situation is: 9x + .90y =

250.

(3 points)


Let x = the number of appetizers and

y = the number of 2 liters of soda.

• Using your graph, list two different realistic options for buying appetizers and sodas.

(For example, Kerry could buy ____ packages of appetizers and ____ sodas for $250).

The x-intercept is 27.8 and the y-intercept is 277.8. Kerry could buy 21 boxes of appetizers

and 60 2-liters of soda. She could also buy 24 boxes of appetizers and 30 2-liters of soda.

(Answers may vary)

Find the x and y intercepts in order to graph the

equation on the grid.

X-intercept:

Let y = 0

9x+.9(0) = 250

9x = 250

9x/9 = 250/9

x= 27.8

Y-intercept:

Let x = 0

9(0)+.9y = 250

.9y = 250

.9y/.9 = 250/.9

y= 277.8



Graphing Equations - Quiz #2

1. Graph the following 3 equations on the graph.

Label each line on the graph.

y = 4x -6

y = 2/3x + 2

y = -5x + 8

2. Use the slope formula to find the slope of the line that passes through the

points (4, 6) & (-2, -1)

3. Lindsey has been tracking her savings account in order to keep track of her spending habits.

The first month account balance was $672. Eight months later her account balance was

$389. What is the average rate of change in dollars per month?



Graphing Equations - Quiz #2 (Answer Key)

1. Graph the following 3 equations on the graph.

Label each line on the graph.

y = 4x -6 (2 points)

y = 2/3x + 2 (2 points)

y = -5x + 8 (2 points)

2. Use the slope formula to find the slope of the line that passes through the

points (4, 6) & (-2, -1) (2 points)

3. Lindsey has been tracking her savings account in order to keep track of her spending habits.

The first month account balance was $672. Eight months later her account balance was

$389. What is the average rate of change in dollars per month? (2 points)

y2 – y1 = -1 - 6 = -7 = 7/6

x2 - x1 -2 - 4 -6

The slope of the line is 7/6

(month, balance)

(1, 672) (8, 389)

y2 – y1 = 389 - 672 = -283 = -40.43

x2 - x1 8-1 7

Lindsey’s rate of change is -40.43. Therefore, on average, she spent about $40.43 per month.

This quiz is worth 10 points.





Lesson 9: Graphing Standard Form Equations – Lesson 2 of 2

Example 1

Method 2: Rewriting the equation in slope intercept form

Use the same strategies that were used for solving equations:

1. ______________________________________________________________

2. ______________________________________________________________

Your goal is to solve for _______________.

Graph the following equations:

6x – 8y = -16

x – y = -9



Lesson 9: Graphing an Equation in Standard Form (2 of 2)

Directions: First rewrite each equation in slope intercept form. Then identify the slope and y-

intercept. Last, graph your equation on the grid.

1. -4x – y = -2

Slope = ____ y-intercept = ____

2. -2x – 3y = 6




3. -4x – y = -2


4. -3x +9y = -9




5. Which one of the following equations shows 12x – 3y = 6 in slope intercept form?

A. 3y = 12x -6

B. y = 4x -2

C. –y = -4x + 2

D. y = 4x +2

6. Write an equation (in slope intercept form) that is equivalent to: 8x -2y = 12

7. Which one of the following equations shows -3y = 6 – 12x in slope intercept form?

A. -3y = 6 – 12x

B. y = 6 – 12x

C. y = 4x – 2

D. y = -4x + 2

8. Given the following equation: 3x – 5y = 10, identify the slope of the line.



9. Given the following equation: 3x – 8y = 16, identify the slope and y-intercept of the line.

10. Are the following equations equivalent? Justify your answer.

2x – 4y = 8 & y = -1/2x -2

1. Write an equation in slope intercept form that is equivalent to: 2x – 5y = 12

(2 points)

2. Given the equation: 4x + 3y = 8. Identify the slope and y-intercept of the line. (2 points)

3. Are the following equations equivalent? Explain your answer, then justify by graphing each line

on the grid below. (4 points)

2x – 3y = 15

4x = 6y + 36



Lesson 9: Graphing an Equation in Standard Form – Answer Key

Directions: For each problem below, identify the slope and the y-intercept.

1. 3x +2y = -4

3x – 3x +2y = -4 – 3x Subtract 3x

2y = -4 – 3x Divide by 2

2 2 2

y = -2 - 3x Simplify

2

y = -3/2x -2 Switch terms around

Slope = -3/2 y-intercept = -2

2. -2x – 3y = 6

-2x +2x – 3y = 6 +2x Add 2x

-3y = 6 +2x Divide by -3

-3 -3 -3

y = -2 – 2/3x Simplify

y = -2/3x – 2 Switch terms around

Slope = -2/3 y-intercept = -2



3. -4x – y = -2

-4x +4x –y = -2 +4x Add 4x

-y = -2 +4x Divide by -1

-1 -1 -1

y = 2 – 4x Simplify

y = -4x +2 Switch terms around

Slope = -4 y-intercept = 2

4. -3x +9y = -9

-3x +3x +9y = -9 +3x Add 3x

9y = -9 + 3x Divide by 9

9 9 9

y = -1 + 1/3x Simplify

y = 1/3 x – 1 Switch terms around

Slope = 1/3 y-intercept = -1



5. Which one of the following equations shows 12x – 3y = 6 in slope intercept form?

A. 3y = 12x -6

B. y = 4x -2

C. –y = -4x + 2

D. y = 4x +2

6. Write an equation (in slope intercept form) that is equivalent to: 8x -2y = 12

7. Which one of the following equations shows -3y = 6 – 12x in slope intercept form?

A. -3y = 6 – 12x

B. y = 6 – 12x

C. y = 4x – 2

D. y = -4x + 2

8. Given the following equation: 3x – 5y = 10, identify the slope of the line.

You can eliminate A and C immediately because slope

intercept form must by y = . The y cannot have a coefficient,

nor can it be negative. Eliminating answers that do not

make sense is a great test taking strategy!

12x – 3y = 6

12x -12x -3y = 6 – 12x

-3y = -12x +6

-3 -3 -3

y = 4x -2 - The answer is B

8x -8x – 2y = 12 -8x Subtract 8x from both sides

-2y = -8x + 12 Simplify & reverse the terms on the right hand side.

-2y = -8x + 12 Divide all terms by -2 -2 -2 -2

y = 4x – 6 The equation written in slope intercept form.

You can eliminate letter A, because slope intercept form must be solved for y.

This equation is almost in slope intercept form. We must get y by itself on the left

hand side; therefore, we need to get rid of the coefficient of -3.

-3y = 6 – 12x Divide all terms by -3

-3 -3 -3

y = -2 + 4x OR y = 4x – 2 Equation written in slope intercept form.

In order to find the slope, we must rewrite the equation in slope intercept form.

3x -3x -5y = 10 -3x Subtract 3x from both sides


-5y = -3x + 10 Divide all terms by -5

-5 -5 -5

y = 3/5x - 2 Equation written in slope intercept form.

The slope of the line is 3/5. (3/5 is the coefficient of x, therefore, it is the slope)



9. Given the following equation: 3x – 8y = 16, identify the slope and y-intercept of the line.

10. Are the following equations equivalent? Justify your answer.

2x – 4y = 8 & y = -1/2x -2

3x -3x – 8y = 16 – 3x Subtract 3x from both sides.



-8 -8 -8

y = 3/8x – 2 Equation written in slope intercept form.

The slope of the line is 3/8 and the y-intercept is -2.

In order to determine if the equations are equivalent, you must rewrite the standard form in slope intercept

form. If the equations are equivalent, then they will be the exact same equation when written in slope

intercept form.

Let’s rewrite the standard form equation in slope intercept form.

2x – 4y = 8

2x -2x – 4y = 8 -2x Subtract 2x from both sides.



-4 -4 -4

y = 1/2x - 2 Equation written in slope intercept form.

The equations are not equivalent.

y = 1/2x -2 & y = -1/2x – 2 differ because equation #1 has a positive slope and equation #2 has a

negative slope.



1. Write an equation in slope intercept form that is equivalent to: 2x – 5y = 12 (2 points)

In order to write the equation in slope intercept form, I must solve for y.

2x – 5y = 12 Original equation

2x – 2x – 5y = 12 – 2x Subtract 2x from both sides

-5y = -2x + 12 Simplify and rewrite in correct format

-5y/-5 = -2x/-5 + 12/-5 Divide ALL terms by -5

y = 2/5x – 12/5 Simplify

The equation in slope intercept form is: y = 2/5x – 12/5

2. Given the equation: 4x + 3y = 8. Identify the slope and y-intercept of the line. (2 points)

In order to identify the slope and y-intercept of the line, we must rewrite the equation in slope

intercept form. Therefore, we must solve for y.

4x + 3y = 8 Original equation

4x – 4x + 3y = 8 – 4x Subtract 4x from both sides

3y = -4x + 8 Simplify and rewrite in correct format

3y/3 = -4x/3 + 8/3 Divide ALL terms by 3

y = -4/3x + 8/3 Simplify

The equation in slope intercept form is: y = -4/3x + 8/3. Therefore, the slope is -4/3 and the y-

intercept is 8/3.



3. Are the following equations equivalent? Explain your answer, then justify by graphing each line

on the grid below. (4 points)

2x – 3y = 15

4x = 6y + 36

We can tell if the equations are equivalent by rewriting both

in slope intercept form:

2x – 3y = 15 Original equation

2x-2x – 3y = 15 – 2x Subtract 2x from both sides

-3y = -2x + 15 Simplify

-3y/-3 = -2x/-3 + 15/-3 Divide All terms by -3

y = 2/3x – 5 Equation in slope intercept form

4x = 6y +36 Original equation

4x – 36= 6y + 36 -36 Subtract 36 from both sides

4x – 36 = 6y Simplify

4x/6 – 36/6 = 6y/6 Divide All terms by 6

2/3x – 6 = y Equation in slope intercept form

y = 2/3x - 6

These equations are not equivalent because when written in slope intercept form, they are

not the same exact equation. The graph shows parallel lines, which means that the

equations are not equivalent. They have the same slope but different y-intercepts and this is

why they are parallel.

Copyright © 2009-2012 Karin Hutchinson – Algebra-class.com

Algebra 1Algebra 1Algebra 1Algebra 1

Unit 2: Unit 2: Unit 2: Unit 2: Graphing EquationsGraphing EquationsGraphing EquationsGraphing Equations

Lesson 10: Graphing Absolute Value Equations

Example 1

Example 2

Graph: y = |x|

Graph: y = |x| +2

Give two solutions to this

equation.




Example 3

Example 4

Graph: y = -|x-2| +1

Analyze the following graphs.

Y = |1/2x| (red graph)

Y = |2x| (green graph)

Y = |4x| (yellow graph)

Y = -|4x| (blue graph)

What do you notice about the

size of the graphs in regards

to the coefficient of x?




Lesson 10: Graphing Absolute Value Equations and Inequalities – Practice Problems

Part 1: Graph the following absolute value equations and inequalities on the grid.

1 y= |x+2|

2. y = |x-1| +2 Identify two solutions.

3. y = |3x|+1

4. y = |2x+2|




5. y = |x-1| -3

6. y = - |x| +4

7. y = -|x-1| Identify two solutions.

8. y = |1/2x| -3




1. Graph the following inequality on

the grid. (3 points)

Y = |2x-4| -2

2. Identify two solutions. (2 points)




Lesson 10: Graphing Absolute Value Equations and Inequalities –Answer Key

Part 1: Graph the following absolute value equations and inequalities on the grid.

1. y= |x+2|

First create a table of values with positive and

negative x values. Then plot the points and

draw your V shaped graph.

2. y = |x-1| +2 Identify two solutions.

First create a table of values with positive and negative x

values. Then plot the points and draw your V shaped

graph.

Two Solutions are: (-3, 6) and (0,3). These are points

on the graph. You can choose any two points on the

graph: (Hint: any point in the table of values is a

solution.)




3. y = |3x|+1


negative x values. Then plot the points and draw

your V shaped graph.

4. y = |2x+2|







5. y= |x-1| -3




6. y = - |x| +4



your upside down V shaped graph. (This graph is

upside down since we have a negative absolute

value.




7. y = -|x-1| Identify two solutions.



your upside down V shaped graph.

Two solutions are: (-3,-4) and (-1,-2) (Any

point in the table of values is a solution).

8. y = |1/2x| -3







1. Graph the following inequality on

the grid. (3 points)

Y= |2x-4| -2

2. Identify two solutions. (2 points)

First create a table of values with positive and negative

x values. Then plot the points and draw your V shaped

graph.

Two solutions: (0,2) & (3,0) (Any point in the table of

values is a solution)

Graphs and table of values were created using a free calculator on rentcalculators.org

Unit 2: Graphing Equations

Copyright © 2011 Karin Hutchinson – Algebra-class.com

Graphing Equations Chapter Test Review

Part 1: Calculate the slope of the following lines: (Lesson 3)

2. Find the slope of a line that has a 3. Find the slope of the line that

y-intercept of -4 and contains the point (3, -2) contains the following points: (-5, 4) (2, -4)



Part 2: Graphing Slope and Slope Intercept Form (Lessons 4 and 5)

4. Write the equation for the line represented on the graph:

5. Graph the following equations on the grid:

a. y = -3x+ 2 b. y = 1/3 x – 8



Part 3: Finding Slope Given Two Points and Rate of Change (Lessons 6 & 7)

6. Find the slope of the line that contains the following points: (-6, 9)(-4, -4)

7. A line has a y-intercept of 8 and passes through the point (-2, -9). Without graphing the line,

determine the slope of the line.

8. In Jessica’s first year in her career, she made $39000. In her sixth year, she makes $52500.

What is Jessica’s average rate of change over her six years in her career.

9. Jesse is investing money in the stock market. His initial investment when he opened the account

was $50. By the end of January he had gained $250. By the end of February he had lost $150. At

the end of March he remained steady with no gains or losses. By the end of April he gained $440.

a. Create a graph to show the total amounts in Jesse’s account each month.



b. What was Jesse’s total amount at the end of April?

c. What was the rate of change between his initial deposit and his amount at the end of April?

d. What was the rate of change between February and march? Explain how you determined

your answer.

Part 4: Graphing Standard Form Equations (Lessons 8 & 9)

10. Identify the x and y intercepts for the following equation: 4x – 3y = 24

11. Write an equation that is equivalent to each of the following equations:

a. 6x + 2y = -14 2. 3x – 2y = -12

12. Graph the following equations:

a. 2x + 4y = -16 b. -6x – 4y = -24



13. Mel is ordering t-shirts for the lacrosse team. Short sleeve t-shirts cost $10 and long sleeve t-

shirts cost $17. The total bill came to $293.00. The equation that represents x short sleeve t-shirts

and y long sleeve t-shirts is:

10x+17y = 293

• Graph the equation on the grid. Let x = the number of short sleeve t-shirts

y = the number of long sleeve t-shirts

• If 14 people ordered short sleeve t-shirts, how many long sleeve t-shirts were ordered? Justify

your answer mathematically.

Part 5: Graphing Absolute Value Equations

1. y= |x+2 |

2. y= - |x+2| -1 Identify two solutions.



Graphing Equations Chapter Test Review – Answer Key

Part 1: Calculate the slope of the following lines: (Lesson 3)

2. Find the slope of a line that has a 3. Find the slope of the line that

y-intercept of -4 and contains the point (3, -2) contains the following points: (-5, 4) (2, -4)

Slope = rise Slope = 10 = 5 Run 4 2 The slope is 5/2.

Slope = rise Slope = -6 = - 3 Run 2 The slope is -3.

Slope = rise Slope = 2 Run 3 The slope of this line is 2/3.

Slope = rise Slope = 8 Run -7 The slope of this line is -8/7.



Part 2: Graphing Slope and Slope Intercept Form (Lessons 4 and 5)

4. Write the equation for the line represented on the graph:

5. Graph the following equations on the grid:

a. y = -3x+ 2 b. y = 1/3 x – 8

To write the equation in slope

intercept form, we need to know the

slope and y-intercept. Y = mx +b

Slope (m) = -1 Y-int (b) = -2

Y = -x - 2

y-int

To write the equation in slope

intercept form, we need to know the

slope and y-intercept. Y = mx +b

Slope (m) = 1/2 Y-int (b) = 4

Y = 1/2x + 4

We can start by plotting the y-intercept

which is y = 2. From that point, graph

the slope. Slope is a rise of -3 and run

of 1.

We can start by plotting the y-intercept

which is y = -8. From that point, graph

the slope. Slope is a rise of 1 and run

of 3.



Part 3: Finding Slope Given Two Points and Rate of Change (Lessons 6 & 7)

6. Find the slope of the line that contains the following points: (-6, 9)(-4, -4)

7. A line has a y-intercept of 8 and passes through the point (-2, -9). Without graphing the line,

determine the slope of the line.

8. In Jessica’s first year in her career, she made $39000. In her sixth year, she makes $52500.

What is Jessica’s average rate of change over her six years in her career.

Use the slope formula to calculate the slope when given two points.

y2 – y1 -4 – 9 = -13 or -6.5

x2 - x1 -4 –(-6) 2

The slope of the line is: -13/2 or -6.5

In order to determine the slope of the line, we must have two points. We know one point is (-2,-9)

and we know the y-intercept is 8. The y-intercept can also be written as (0,8) because the y-

intercept is the point on the y-axis, so we know the x-coordinate is 0. Now we have two points.

(0,8) (-2,-9) Now we can use the slope formula to find the slope.

y2 – y1 = -9-8 = -17 or 8.5

x2 - x1 -2 – 0 -2

The slope of the line is 17/2 or 8.5

If we write two ordered pairs for this problem, then we can determine the rate of change, which is

also the slope. Therefore, we can use the slope formula.

Let x = the number of years & y = her salary

First year - $39000 can be written as: (1, 39000)

Sixth year - $52500 can be written as: (6, 52500)

Now we can use the slope formula to determine the rate of change.

y2 – y1 = 52500 – 39000 = 13500 = 2700

x2 - x1 6-1 5

Jessica’s rate of change is $2700 per year.



9. Jesse is investing money in the stock market. His initial investment when he opened the account

was $50. By the end of January he had gained $250. By the end of February he had lost $150. At

the end of March he remained steady with no gains or losses. By the end of April he gained $440.

a. Create a graph to show the total amounts in Jesse’s account each month.

b. What was Jesse’s total amount at the end of April?

c. What was the rate of change between his initial deposit and his amount at the end of April?

d. What was the rate of change between February and March? Explain how you determined your

answer.

At the end of April Jesse’s total was $590. (50+250-150+0+440 = 590)

Let’s write two ordered pairs for this information.

Initial Deposit(month 0) - $50 is (0, 50) April (4th month) – 590 is (4, 590)

Let x = the month Let y = the total amount)

Now we can use our slope formula to find the rate of change.

y2 – y1 = 590 – 50 = 540 = 135

x2 - x1 4-0 4

The rate of change between the initial deposit and the end of April is $135.

The rate of change between February and March is 0. The problem states that at the end of March is

account remained steady with no gains or losses. The ordered pairs would be: (2, 150) (3, 150)



Part 4: Graphing Standard Form Equations (Lessons 8 & 9)

10. Identify the x and y intercepts for the following equation: 4x – 3y = 24

11. Write an equation that is equivalent to each of the following equations:

a. 6x + 2y = -14 2. 3x – 2y = -12

To identify the x intercept, we will let y = 0

4x – 3(0) = 24 Substitute 0 for y

4x = 24 Simplify

4x/4 = 24/4 Divide by 4

X = 6

The x-intercept is 6.

To identify the y-intercept, we will let x = 0

4(0) – 3y = 24 Substitute 0 for x

-3y = 24 Simplify

-3y/-3 = 24/-3 Divide by -3

Y = -8

The y-intercept is -8

I need to write an equation in slope intercept

form in order for the equation to be

equivalent. Therefore, we will solve for y.

6x – 6x + 2y = -14 – 6x Subtract 6x

2y = -6x – 14 Simplify

2y/2 = -6x/2- 14/2 Divide by 2

Y = -3x – 7

Y = -3x – 7 is an equivalent equation.

I need to write an equation in slope intercept form in

order for the equation to be equivalent. Therefore, we

will solve for y.

3x – 3x – 2y = - 12 – 3x Subtract 3x

-2y = -3x – 12 Simplify

-2y/-2 = -3x/-2 – 12/-2 Divide by -2

Y = 3/2x + 6

Y = 3/2x + 6 is an equivalent equation.



12. Graph the following equations:

a. 2x + 4y = -16 b. -6x – 4y = -24

Since this equation is written in standard

form, you can find the x and y intercepts.

(You may also rewrite it in slope intercept

form). We will find the intercepts.

X-intercept (Let y = 0) Y-int (Let x=0)

2x +4(0) = -16 2(0) +4y = -16

2x = -16 4y = -16

2x/2 = -16/2 4y/4=-16/4

X = -8 Y = -4

**If you rewrote the equation in slope

intercept form, it would be:

Y = -1/2x 4

Since this equation is written in standard form,

you can find the x and y intercepts. (You may

also rewrite it in slope intercept form). We will

find the intercepts.

X-intercept (Let y = 0) Y-int (Let x=0)

-6x – 4(0) = -24 -6(0) – 4y = -24

-6x = -24 -4y = -24

-6x/-6 = -24/-6 -4y/-4 = -24/-4

X = 4 y = 6

**If you rewrote the equation in slope intercept

form, it would be:

Y = -3/2 + 6



13. Mel is ordering t-shirts for the lacrosse team. Short sleeve t-shirts cost $10 and long sleeve t-

shirts cost $17. The total bill came to $293.00. The equation that represents x short sleeve t-shirts

and y long sleeve t-shirts is:

10x+17y = 293

• Graph the equation on the grid. Let x = the number of short sleeve t-shirts

y = the number of long sleeve t-shirts

• If 14 people ordered short sleeve t-shirts, how many long sleeve t-shirts were ordered? Justify

your answer mathematically.

Since this equation is written in standard form, I can find the x and y intercepts in order to graph. Since

we are dealing with larger numbers, this is definitely the easiest way to graph this equation.

10x + 17y = 293

X int: (Let y = 0) Y int: (Let x = 0)

10x + 17(0) = 293 10(0) + 17y = 293

10x = 293 17y = 293

10x/10 = 293/10 17y/17 = 293/17

X = 29.3 y = 17.2

According to the graph, if 14 people ordered short sleeve t-shirts, then 9 long sleeve t-shirts were ordered.

Justify: 10(14) + 17(9) = 293

140 + 153 = 293

293= 293



Part 5: Graphing Absolute Value Equations and Inequalities

**Graphs and table of values were created using a free calculator on rentcalculators.org**

1. y= |x+2 |

2. y = -|x+2| -1 Identify two solutions.

Two solutions are: (-5, -4) and (-4, -3)



Graphing Equations Chapter Test

1. Which line on the graph has a slope of 2/3?

2. Which equation is represented on the graph?

A. Line A

B. Line B

C. Line C

D. Line D

A. y = 4x – 6

B. y = -4x – 6

C. y = 4x +6

D. y = 6x +4



3. Find the slope of the line that has a y-intercept of 3 and contains the point (6,-3).

Patti is tracking the price of gas over a 10 week period. See the graph below to answer questions 4

– 6.

4. What is the y-intercept in this problem? What does it mean in the context of this problem?



5. What is the rate of change between week 3 and week 6?

6. What is the rate of change between week 0 and Week 3? Explain what the rate of change means

in the context of this problem.

7. What is the x intercept for the equation: 3x + 5y = 27?

8. Judy bought 10 shares of a stock on February 1, 2008 for $350. On October 1, 2008 she sold the

stock for $210. Find Judy’s average monthly rate of change on the stock.

9. Graph the following equation on the grid. Identify the slope and the y intercept.

2x +5y = -15

Slope = ___________

Y intercept = ___________



10. Which equation is equivalent to: 4x +2y = -12

A. y = -2x – 6 C. y = -4x - 12

B. y = 2x – 6 D. y = -2x + 6

11. Identify the graph that represents the function: y = |x+3|

A. B. C. D.

12. Joe is traveling to his sister’s house for Thanksgiving. He begins his drive at 3am. He drives

210 miles in 3 hours. He then stops for a half hour break and drives another 90 miles in the

next 2 hours. His trip ends at 10 am after driving a total of 377.5 miles.

• Create a graph to illustrate this problem.

• What is Joe’s average rate of change between 3 am and 6 am?

• If Joe’s rate of change for the first three hours had remained constant throughout the entire

trip, what time would he have arrived at his sister’s house?



13. A final exam contains multiple choice questions that are worth 2 points and free response

questions that are worth 3 points. The test is worth 60 points. The equation that represents x

multiple choice questions and y free response questions is:

2x +3y = 60.

• Graph the equation on the grid. Let x = the number of multiple choice questions and

y = the number of free response questions.

• If there are 15 multiple choice questions on the test, how many free response questions are

on the test? Justify your answer.

• Identify one other possibility for the number of each question on the test: Justify your

answer.

There could be _______ multiple choice and _______ free response.



Graphing Equations Chapter Test – Answer Key

1. Which line on the graph has a slope of 2/3? (1 Point)

2. Which equation is represented on the graph? (1 Point)

A. Line A

B. Line B

C. Line C

D. Line D

A. y = 4x – 6

B. y = -4x – 6

C. y = 4x +6

D. y = 6x +4

A

**Note: You can eliminate C & D

immediately because they have negative

slopes.

Then you only need to count the slope for A

& B.

A

.

**This is another problem where you can

eliminate answers that don’t make sense. You

know the y – intercept is -6, so C & D can be

eliminated!

The line is rising so the slope is positive.

Therefore, B can also be eliminated!



3. Find the slope of the line that has a y-intercept of 3 and contains the point (6,-3). (2 points)

Patti is tracking the price of gas over a 10 week period. See the graph below to answer questions 4

– 6.

4. What is the y-intercept in this problem? What does it mean in the context of this problem?

(2 points)

The y-intercept is 2.40. In this problem the y-intercept indicates that when Patti first began tracking

gas prices (week 0), the price was $2.40.

5. What is the rate of change between week 3 and week 6? (2 Points)

Ordered pairs: (3, 3.20) (6, 2.20)

y2 – y1 = 2.20 – 3.20 = -1

x2 – x1 6 – 3 3

Slope = rise from one point to the next

run

Slope = 1 = -1

-1

The rate of change between week 3 and 6 is -1/3. This means

that the price of gas dropped about $0.33 each week.



6. What is the rate of change between week 0 and Week 3? Explain what the rate of change means

in the context of this problem. (3 Points)

Ordered pairs: (0, 2.40) (3, 3.20)

y2 – y1 = 3.20 – 2.40= .8 = .27

x2 – x1 3 – 0 3

7. What is the x intercept for the equation: 3x + 5y = 27? (2 points)

X intercept: Let y = 0

3x +5(0) = 27 3x = 27 x = 9 The x intercept is 9.

3x = 27 3 3

8. Judy bought 10 shares of a stock on February 1, 2008 for $350. On October 1, 2008 she sold the

stock for $210. Find Judy’s average monthly rate of change on the stock. (2 points)

Ordered pairs: (2, 350) (10, 210) y2 – y1 = 210 – 350 = -140 = -17.5

x2 – x1 10-2 8

Feb is month 2 October is month 10

9. Graph the following equation on the grid. Identify the slope and the y intercept. (3 points)

2x +5y = -15

10. Which equation is equivalent to: 4x +2y = -12 (1 point)

A. y = -2x – 6 C. y = -4x - 12

B. y = 2x – 6 D. y = -2x + 6

Slope = -2/5

Y intercept = -3

The rate of change between week 0 and 3 is .27. This means

that the price of gas rose about $0.27 each week between

weeks 0 and 3.

Judy’s average monthly rate of change is -17.5.

She lost 17.5 dollars per month while holding

that stock.

I am going to rewrite it in slope

intercept form:

2x -2x +5y = -15 – 2x

5y = -2x – 15

5y = -2x – 15

5 5 5

y = -2/5x -3

4x – 4x +2y = -12 – 4x

2y = -4x – 12

2y = -4x – 12

2 2 2

y = -2x -6



11. Identify the graph that represents the function: y = |x+3| (1 point)

A. B. C. D.

You can eliminate answer B because it represents a negative absolute value, since it is upside

down. We also know that the graph shifts up if there is a constant after the absolute value.

Therefore, we’ve narrowed it down to A or D. Since y = 0, which x value would make sense, -3 or

3? -3 + 3 = 0, so the correct answer is A.

You can also make a table of value and see which graph has the same points as the table of

values.



12. Joe is traveling to his sister’s house for Thanksgiving. He begins his drive at 3am. He drives

210 miles in 3 hours. He then stops for a half hour break and drives another 90 miles in the

next 2 hours. His trip ends at 10 am after driving a total of 377.5 miles.

(4 points)

• Create a graph to illustrate this problem. (2/4 points)

• What is Joe’s average rate of change between 3 am and 6 am?

• If Joe’s rate of change for the first three hours had remained constant throughout the entire

trip, what time would he have arrived at his sister’s house?

Rate of change = slope Two points: 3am (3, 0) 6am (6, 210)

Slope = y2 – y1 = 210-0 = 210 = 70

X2 – x1 6 – 3 3

The rate of change between 3am and 6 am is 70. Therefore, Joe maintained a constant rate

of 70 miles per hour during this three hour time period.

If Joe had been able to maintain a constant rate of 70 miles per hour throughout his entire trip he would have arrived

at his sister’s house at 8:24 am.

D= rt Distance = 377.5 miles Rate = 70 miles per hour Time = ?

377.5 = 70t 377.5 = 70t

70 70

T = 5.4 (5 hours and 24 minutes) ** .4 = 4/10 = 2/5. 2/5 of 60 minutes = 2/5 •60 = 24

3 am plus 5 hours and 24 minutes = 8:24 am.



12. A final exam contains multiple choice questions that are worth 2 points and free response

questions that are worth 3 points. The test is worth 60 points. The equation that represents x

multiple choice questions and y free response questions is:

2x +3y = 60. (4 points)

• Graph the equation on the grid. Let x = the number of multiple choice questions and

y = the number of free response questions.

• If there are 15 multiple choice questions on the test, how many free response questions are

on the test? Justify your answer.

•

• Identify one other possibility for the number of each question on the test: Justify your

answer.

There could be _______ multiple choice and _______ free response.

In order to graph the equation, we must find

the x and y intercepts:

2x +3y = 60

X Intercept Y Intercept

2x +3(0) = 60 2(0) +3y = 60

2x = 60 3y = 60

2 2 3 3

x = 30 y = 20

X intercept = 30 Y intercept = 20

Since we know there are 15 multiple choice questions, we can substitute 15 for x and solve for y.

2(15) + 3y = 60

30+3y = 60

30 – 30 +3y = 60 – 30

3y = 30

3y/3 = 30/3

Y= 10

If there are 15 multiple choice questions, then there are 10 free

response questions.

Justify: 2(15) + 3(10) = 60

30+30 = 60

There could be 21 multiple choice and 6 free response.

2(21) + 3(6) = 60

42 + 18 = 60

60 = 60

This test is worth 28 points

Documents

Lesson 1: Graphing Points on a Coordinate Plane Equations/Graphin… · ... Graph the following points on the coordinate plane. (0,0) Y axis Example: Z. (-4, 6) x axis x y A. (2,4)