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MATH GRADE 8 UNIT 5

LINEAR EQUATIONS

EXERCISES

Copyright © 2014 Pearson Education, Inc. 2

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Grade 8 Unit 5: Linear Equations

CONTENTS EXERCISES

LESSON 1: CATCHING UP �� 4

LESSON 2: ONE-VARIABLE EQUATIONS �� 5

LESSON 3: HOW MANY SOLUTIONS �� 9

LESSON 4: RELATE ONE-VARIABLE EQUATIONS �� 13

LESSON 5: SAME LINE–DIFFERENT EQUATIONS �� 17

LESSON 6: A SYSTEM OF LINEAR EQUATIONS �� 21

LESSON 7: SOLVING EQUATIONS: SUBSTITUTION �� 26

LESSON 8: SOLVING EQUATIONS: ELIMINATION �� 31

LESSON 9: USING A SYSTEM OF EQUATIONS �� 36

LESSON 10: SOLVING PROBLEMS �� 41

LESSON 11: PUTTING IT TOGETHER �� 51



EXERCISES

EXERCISES

• Review your end of unit assessment from the previous unit.

• Write your wonderings about linear equations.

• Write a goal stating what you plan to accomplish in this unit.

• Based on your previous work, write three things you will do differently during this unit to increase your success.

LESSON 1: CATCHING UP



EXERCISES

EXERCISES

1. Solve. x + 3 = 2x – 1

Solve each equation.

2. 5n + 4 = 3(n – 2)

3. 2n + 3n – 8 = 4n + 7

4. 3(n – 4) = 5(n – 5)

5. 2n + 6 + n = 4n – 3n + 4

Solve.

6. Six times a number is equal to ten less than the sum of the number and 3. What is the number?

Challenge Problem

7. Write an equation that has a variable on both sides and a solution of 5.6. Write a word problem that matches your equation.

LESSON 2: ONE-VARIABLE EQUATIONS



EXERCISES

ANSWERS

1. x = 4

x + 3 = 2x – 1

x + 3 – x + 1 = 2x – 1 – x + 1

4 = x

2. n = –5

5n + 4 = 3(n – 2)

5n + 4 = 3n – 6

5n + 4 – 4 – 3n = 3n – 6 – 4 – 3n

2n = –10

22

–102

=n

n = –5

3. n = 15

2n + 3n – 8 = 4n + 7

5n – 8 = 4n + 7

5n – 8 + 8 – 4n = 4n + 7 + 8 – 4n

n = 15

4. n = 612

or n = 6.5

3(n – 4) = 5(n – 5)

3n – 12 = 5n – 25

3n – 12 – 3n + 25 = 5n – 25 – 3n + 25

13 = 2n

132

22

=n

612

= n




EXERCISES

5. n = –1

2n + 6 + n = 4n – 3n + 4

3n + 6 = n + 4

3n + 6 – n – 6 = n + 4 – n – 6

2n = –2

22

–22

=n

n = –1

6. n = –125

or –1.4

6n = (n + 3) – 10

6n = n – 7

6n – n = n – 7 – n

5n = –7

55

–75

=n

n = –125

Challenge Problem

7. Answers will vary. Possible answer:

Erin and Pedra are biking the same trail and in the same direction. Erin is 3 mi. from the start of the trail and bikes 3 miles per hour. Pedra is 8.6 mi. from the start of the trail and bikes 2 miles per hour. When will the two girls be the same distance from the start of the trail?

3h + 3 = 2h + 8.6

3h + 3 – 2h – 3 = 2h + 8.6 – 2h – 3

h = 5.6

The two girls will be the same distance after 5.6 hr, or 5 hr 36 min.


ANSWERS



EXERCISESLESSON 2: ONE-VARIABLE EQUATIONS

Self Assessment

After you use the answer key to check your answers, use the chart below to self-assess your work. For each exercise, place a check mark in the column that best describes how you did on that exercise.

Exercise Number Yes! I got it� I was confused, but

now I get it� I need help!

1

2

3

4

5

6

Challenge Problem

Exercise I gave it a try, but I’m not sure I did it right�

I did it, and my answer makes sense�

7



EXERCISES

EXERCISES

1. Solve.

x + 5 = 4(x – 1)

Write whether the equation has one solution, no solution, or all numbers as solutions.

2. 3x + 4 = 6 + 3x

3. x + 9 + 3x = 4x + 9

4. 3x + 4(x – 5) = 5x + 8

5. 3(x + 2) = 3x + 2

6. 2(x – 6) = –10 + 2x – 2

7. 5x + x + 4(x – 1) = 3x – 4 + x

Challenge Problem

8. a. Write an equation that has no solution.

b. Write an equation that has only 0 as a solution.

LESSON 3: HOW MANY SOLUTIONS?



EXERCISES

ANSWERS

1. x = 3

x + 5 = 4(x – 1)

x + 5 = 4x – 4

x + 5 – x + 4 = 4x – 4 – x + 4

9 = 3x

93

33

=x

3 = x

2. No solution

3x + 4 = 6 + 3x

3x + 4 – 3x – 4 = 6 + 3x – 3x – 4

0 ≠ 2

3. All numbers as solutions

x + 9 + 3x = 4x + 9

4x + 9 = 4x + 9

4x + 9 – 9 = 4x + 9 – 9

4x = 4x

x = x

4. One solution

3x + 4(x – 5) = 5x + 8

3x + 4x – 20 = 5x + 8

7x – 20 = 5x + 8

7x – 20 – 5x + 20 = 5x + 8 – 5x + 20

2x = 28

x = 14




EXERCISES

5. No solution

3(x + 2) = 3x + 2

3x + 6 – 3x – 6 = 3x + 2 – 3x – 6

0 ≠ –4


2(x – 6) = –10 + 2x – 2

2x – 12 = –12 + 2x

2x – 12 + 12 = –12 + 2x + 12

2x = 2x

x = x

7. One solution

5x + x + 4(x – 1) = 3x – 4 + x

5x + x + 4x – 4 = 4x – 4 + x

10x – 4 – 4x + 4 = 4x – 4 – 4x + 4

6x = 0

x = 0

Challenge Problem

8. Answers will vary. Possible answers:

a. x + 4 = x + 5

b. 2x = 3x


ANSWERS



EXERCISES

Self Assessment




1

2

3

4

5

6

7

Challenge Problem



8 a�

8 b�




EXERCISESLESSON 4: RELATE ONE-VARIABLE EQUATIONS

EXERCISES

1. Write the expression 4(x + 5) + 4x + 8 – 2 in simplest form.

Use the Graphing tool to decide if the equation has one solution, no solution, or all numbers as solutions.

2. 4x + 6 = 4x – 6

3. 3(x + 4) = 3x + 7

4. 4x + 9 = 2 + 2(x + 1) + 2x + 5

5. 3x + 8 = 8(x + 1)

6. 3x – 4x – 1 = 7x – 6

7. 12

x + 6 = 3x – 1

Challenge Problem

8. Graph y = 2x + 5 and y = 3x + 1 on the same coordinate grid. Explain how to use the graph to find the solution to 2x + 5 = 3x + 1.



EXERCISES

ANSWERS

1. 8x + 26 or 26 + 8x

4(x + 5) + 4x + 8 – 2

4x + 20 + 4x + 8 – 2

8x + 26

2. No solution

4x + 6 = 4x – 6

Use graphing tool to show graph of y = 4x + 6 and y = 4x – 6; label each line with its equation; these lines will be parallel.

3. No solution

3(x + 4) = 3x + 7

Use graphing tool to show graph of y = 3(x + 4) and y = 3x + 7; label each line with its equation; these lines will be parallel.


4x + 9 = 2 + 2(x + 1) + 2x + 5

Use graphing tool to show graph of y = 4x + 9 and y = 2 + 2(x + 1) + 2x + 5; label each line with its equation; these lines are the same line.

5. One solution

3x + 8 = 8(x + 1)

Use graphing tool to show graph of y = 3x + 8 and y = 8(x + 1); label each line with its equation; these lines will intersect.

6. One solution

3x – 4x – 1 = 7x – 6

Use graphing tool to show graph of y = 3x – 4x – 1 and y = 7x – 6; label each line with its equation; these lines will intersect.

LESSON 4: RELATE ONE-VARIABLE EQUATIONS




7. One solution

12

x + 6 = 3x – 1

Use graphing tool to show graph of y = 12

x + 6 and y = 3x – 1; label each line with its equation; these lines will intersect.

Challenge Problem

8. The graphs will intersect at (4, 13). So, the solution to 2x + 5 = 3x + 1 is the value of x at that point, 4.

Use graphing tool to show graph of y = 2x + 5 and y = 3x + 1 on the same coordinate grid; label each line with its equation; label point of intersection (4, 13).

ANSWERS




Self Assessment




1

2

3

4

5

6

7

Challenge Problem



8



EXERCISES

EXERCISES

1. Which equation is equivalent to y = 4x + 5?

A 4x + y = 5

B 4x – y = –5

C 4x + y = –5

D 4x – y = 5

2. Write each equation in slope-intercept form. (y = mx + b)

a. x – y = 3

b. 2x + 2y = 10

c. –5x + 5y = 35

d. x – y = 0

3. Write each equation in standard form. (ax + by = c)

a. y = 3x – 4

b. y = –12

x + 1

c. y = –2x + 2

d. y = x

4. What are the slope and y-intercept of each equation?

a. 2x + y = 8

b. x – y = 4

LESSON 5: SAME LINE—DIFFERENT EQUATIONS



EXERCISES

5. Look at the graph of this line:

123456789

10

–5–4–3–2–1 10987654321–2–3–4–5–6

x

y

Write an equation in both slope-intercept form and standard form for the line.

Challenge Problem

6. Marshall likes to use interval training. He jogs at 200 meters per minute and runs 250 meters per minute. He runs 8 km every day. Write an equation in standard form. Let x represent the number of minutes of jogging and y represent the number of minutes of running for Marshall’s interval training.


EXERCISES



EXERCISES

ANSWERS

1. B 4x – y = –5

2. a. y = x – 3

b. y = –x + 5

c. y = x + 7

d. y = x or y = x + 0

3. a. 3x – y = 4

b. 12

x + y = 1

c. 2x + y = 2

d. –x + y = 0 or x – y = 0

4. a. Slope: –2; y-intercept: 8

2x + y = 8 written in slope-intercept form is y = –2x + 8.

b. Slope: 1; y-intercept: –4

x – y = 4 written in slope-intercept form is y = x – 4.

5. Slope-intercept form: y = –x + 4; standard form: x + y = 4

The y-intercept is (0, 4). The slope is –1.

Challenge Problem

6. 200x + 250y = 8,000




EXERCISESLESSON 5: SAME LINE—DIFFERENT EQUATIONS

Self Assessment




1

2 a.

2 b.

2 c.

2 d.

3 a.

3 b.

3 c.

3 d.

4 a.

4 b.

5

Challenge Problem



6



EXERCISESLESSON 6: A SYSTEM OF LINEAR EQUATIONS

EXERCISES

1. Write the equation 6x + 3y = 12 in slope intercept form. (y = mx + b)

Solve each system of equations. If the system has no solution, write no solution. If a system has infinitely many solutions write infinitely many solutions.

2. y = 2x + 3

y = 3x – 4

3. y = 3x – 2x

y – x = 4

4. 4 + y = 4x

y = 4(x – 1)

5. 3 = 3x + y

– 3x + y = 8

6. 9 = 5x – y

–3x + y = 9

Challenge Problem

7. Write a system of two equations that has no solution. Describe what the graph of the system would look like.



EXERCISES

ANSWERS

1. y = –2x + 4

2. (7, 17)

y = 2x + 3

y = 3x – 4

2x + 3 = 3x – 4

2x + 3 – 2x + 4 = 3x – 4 – 2x + 4

7 = x

y = 2x + 3

y = 2(7) + 3

y = 14 + 3

y = 17

3. No solution

y = 3x – 2x

y – x = 4 or y = x + 4

3x – 2x = x + 4

x = x + 4

x – x = x – x + 4

0 ≠ 4

4. Infinitely many solutions

4 + y = 4x or y = 4x – 4

y = 4(x – 1) or y = 4x – 4

4x – 4 = 4x – 4

LESSON 6: A SYSTEM OF LINEAR EQUATIONS



EXERCISES

5. (–56

, 512

)

3 = 3x + y or y = –3x + 3

– 3x + y = 8 or y = 3x + 8

–3x + 3 = 3x + 8

–3x + 3 + 3x – 8 = 3x + 8 + 3x – 8

–5 = 6x

x = –56

y = 3x + 8

y = 3(–56

) + 8

y = –52

+ 8

y = 112

or 512

6. (9, 36)

9 = 5x – y or y = 5x – 9

–3x + y = 9 or y = 3x + 9

5x – 9 = 3x + 9

5x – 9 –3x + 9 = 3x + 9 – 3x + 9

2x = 18

x = 9

y = 3x + 9

y = 3(9) + 9

y = 27 + 9

y = 36


ANSWERS



EXERCISES

Challenge Problem

7. RUBRIC

4 pointsResponse shows a sound approach, a correct answer, and good communication of the process by which the answer was obtained.

3 pointsResponse shows a sound approach, but errors along the way may result in an incorrect response. Work is clearly shown, but some of the detail of the steps taken may be incomplete.

2 pointsResponse has an approach that is flawed but carried through appropriately, with work shown to document what was done.

1 point

Response begins to take an inappropriate approach and does not follow through well, with work shown being potentially sketchy. Or the correct answer is shown, but no communication is given about how the solution was obtained.

EXAMPLE OF A 4-POINT RESPONSE:

4 points Systems will vary. Possible system:

y = 4x + 5

y = 4x – 1

4x + 5 = 4x – 1

4x + 5 – 4x – 5 = 4x – 1 – 4x – 5

0 ≠ –6

The system of the two equations has no solution.

The lines will be parallel because both lines have the same slope, 4. The line y = 4x + 5 has a y-intercept of 5. The line y = 4x – 1 has a y-intercept of – 1.


ANSWERS



EXERCISESLESSON 6: A SYSTEM OF LINEAR EQUATIONS

Self Assessment




1

2

3

4

5

6

Challenge Problem



7



EXERCISESLESSON 7: SOLVING EQUATIONS: SUBSTITUTION

EXERCISES

1. Solve this equation.

3(2y – 1) + 5y = 8

Solve each system of equations. If the system has no solution, write no solution. If a system has infinitely many solutions write infinitely many solutions.

2. x = 4y + 2

2x – 3y = 9

3. y = 2x

3x + y = 12

4. 3x + y = 5

4x + 2y = 20

5. 4x + 5 = y

4x – y = 8

6. y = 3x – 7 + x

7 = 4x – y

Challenge Problem

7. Write a system of equations that has (4, 5) as the solution.




ANSWERS

1. x = 1

3(2y – 1) + 5y = 8

6y – 3 + 5y = 8

11y – 3 = 8

11y – 3 + 3 = 8 + 3

11y = 11

y = 1

2. (6, 1)

x = 4y + 2

2x – 3y = 9

2(4y + 2) – 3y = 9

8y + 4 – 3y = 9

5y = 5

y = 1

x = 4(1) + 2

y = 6

3. (2.4, 4.8) or 225

, 445

y = 2x

3x + y = 12

3x + 2x = 12

5x = 12

x =

2

125

or 225

y =

2

125

y = 245

or 445




4. (–5, 20)

3x + y = 5 or y = –3x + 5

4x + 2y = 20

4x + 2(–3x + 5) = 20

4x – 6x + 10 = 20

– 2x + 10 = 20

– 2x + 10 + 2x – 20 = 20 + 2x – 20

– 10 = 2x

x = –5

y = –3(–5) + 5

y = 15 + 5

y = 20

5. No solution

4x + 5 = y

4x – y = 8

4x – (4x + 5) = 8

4x – 4x – 5 = 8

– 5 ≠ 8

6. Infinitely many solutions

y = 3x – 7 + x

7 = 4x – y

7 = 4x – (3x – 7 + x)

7 = 4x – 3x + 7 – x

7 = 7

ANSWERS



EXERCISES

Challenge Problem

7. Answers will vary. Ask a classmate to check your system of equations.

Possible system:

y = x + 1

y = –2x + 13

x + 1 = –2x + 13

x + 1 + 2x – 1 = –2x + 13 + 2x – 1

3x = 12

x = 4

y = 4 + 1

y = 5

LESSON 7: SOLVING EQUATIONS: SUBSTITUTION

ANSWERS




Self Assessment




1

2

3

4

5

6

Challenge Problem



7



EXERCISES

EXERCISES

1. Add the two equations.

3x – y = 7

2x + y = –6

Solve each system of equations.

2. 4x – y = 9

3x + y = 5

3. x + 3y = 8

–x + 5y = 4

4. 5x + 3y = 8

4x + 3y = 5

5. 2x + 8y = 6

3x – 4y = –1

Challenge Problem

6. a. Explain why the substitution method, rather than the elimination method, might be the better choice for solving this system of equations.

x = 4y + 1

3y + 5x = 10

b. Solve the system of equations.

LESSON 8: SOLVING EQUATIONS: ELIMINATION



EXERCISESLESSON 8: SOLVING EQUATIONS: ELIMINATION

ANSWERS

1. 5x = 1

3x – y = 7

+ 2x + y = –6

5x + 0 = 1

2. (2, –1)

4x – y = 9

+ 3x + y = 5

7x + 0 = 14

x = 2

3(2) + y = 5

6 + y = 5

y = –1

3. 312

, 112

or (3.5, 1.5)

x + 3y = 8

+ –x + 5y = 4

0 + 8y = 12

y = 128

or 32

or 112

x + 3

32

= 8

x + 92

= 8

x = 312



EXERCISES

4. 3, – 213

4x + 3y = 5 or –1(4x + 3y) = –1(5) or –4x –3y = –5

5x + 3y = 8

+ –4x – 3y = –5

x + 0 = 3

x = 3

5(3) + 3y = 8

15 + 3y = 8

3y = –7

y = –73

or – 213

5. 12

,58

3x – 4y = –1 or 2(3x – 4y) = 2(–1) or 6x – 8y = –2

2x + 8y = 6

+ 6x – 8y = –2

8x + 0 = 4

x = 12

2(12

) + 8y = 6

1 + 8y = 6

8y = 5

y = 58


ANSWERS



EXERCISES

Challenge Problem

6. a. Since the first equation is solved for x, it is easy to substitute the value for x into the second equation.

b. 12023

,523

x = 4y + 1

3y + 5x = 10

3y + 5(4y + 1) = 10

3y + 20y + 5 = 10

23y = 5

y = 523

4523

1=

+x

x = 2023

+ 1

x = 12023


ANSWERS



EXERCISES

Self Assessment




1

2

3

4

5

Challenge Problem



6 a�

6 b�




EXERCISES

EXERCISES

1. The sum of two numbers is 25. Their difference is 9. What are the two numbers?

2. A fire lookout is a person who looks for fire from the top of a structure known as a fire lookout tower. These towers are used in remote areas, often on mountain tops. From these towers, the fire lookouts have a good view of the surrounding terrain and are able to spot wildfires.

The grid shows the location of two lookout towers.

123456789

10

–5–4–3–2

10987654321

x

y

Tower 1

Tower 2

A fire breaks out. The fire lookout at Tower 1 sees the fire with a line-of-sight having

slope of –12

. The fire lookout at Tower 2 sees the fire with a line-of-sight having

slope of 1.

At the coordination center, exact coordinates of the fire are needed in order to find the optimal position for a water-bombing plane.

a. What is the y-intercept and slope of each line?

b. Find the linear equations of the two lines.

c. Find the coordinates of the fire.

LESSON 9: USING A SYSTEM OF EQUATIONS



EXERCISES

Challenge Problem

3. A third tower is located at (5, 0). The lookout sees a fire with a line-of-sight with a

slope of –32

. Can this be the same fire as the one Tower 1 and Tower 2 see?

Explain.


EXERCISES



EXERCISES

ANSWERS

1. The two numbers are 17 and 8.

x + y = 25

x – y = 9

2x + 0 = 34

x = 17

17 + y = 25

y = 18

2. RUBRIC




1 point


EXAMPLE OF A 4-POINT RESPONSE:

4 pointsa. Tower 1: slope = –

12

; y-intercept = 5

Tower 2: slope = 1; y-intercept = –1




EXERCISES

b. Tower 1: y = –12

x + 5

Tower 2: y = x – 1

c. The fire is at coordinates (4, 3).

123456789

10

–5–4–3–2

10987654321

x

y

Tower 1

Tower 2

Challenge Problem

3. No. The equation for Tower 3 is y = –32

x + 712

. The point (4, 3) is not on that line.

On the line of sight for Tower 3, x = 4 and y = 112

.


ANSWERS



EXERCISESLESSON 9: USING A SYSTEM OF EQUATIONS

Self Assessment




1

2 a.

2 b.

2 c.

Challenge Problem



3



EXERCISES

EXERCISES

1. Write the equation in slope-intercept form for a line with slope 2 that passes through the point (2, 5).

2. Each table represents a linear relationship.

Table A

x –2 –1 0 2

y –4 –2 0 4

Table B

x –4 –2 0 4

y –3 0 3 9

a. Find the slope of each line.

b. Find the equation for each line.

c. Find the coordinates of the point of intersection.

LESSON 10: SOLVING PROBLEMS



EXERCISESLESSON 10: SOLVING PROBLEMS

3. Jones and Janes are two competing landscaping companies. Jones charges $40 to come to the location and then $40 per hour. Janes has made a graph representing their costs.

34032030028026024022020018016014012010080604020

543 82 71 6

x

y

Hours

Dollars

Two points on the graph are (0, 60) and (3, 150).

The horizontal axis shows the number of hours. The vertical axis shows the total cost.

a. Which company is less expensive for a 3-hour job?

b. Draw the graph of Jones in the same coordinate system.

c. What are the coordinates of the point of intersection of the two lines?

d. What is the meaning of this point?

EXERCISES



EXERCISES

Challenge Problem

4. Points A, B, C, and D represent the locations of four airports.

1

2

3

4

5

–2

–154321–1–2–3–4

x

y

D

A

C

B

a. Find the slope of each line segment shown.

b. Find the y-intercepts of each segment.

c. Write the system of linear equations that will give you the coordinates of the point of intersection of the two segments.

d. Solve the system of equations.


EXERCISES



EXERCISES

ANSWERS

1. y = 2x + 1

y = 2x + b

5 = 2(2) + b

5 = 4 + b

1 = b

y = 2x + 1

2. RUBRIC




1 point


EXAMPLE:

4 points a. Table A: slope = 2

4 02 0

( )( )

−−

= 42

= 2

Table B: slope = 32

9 34 0

( )( )

−−

= 64

= 23




EXERCISES

b. Use slope-intercept to write the equation. (y = mx + b)

Table A: y = 2x

When x = 0, y = 0, so y-intercept is 0. The slope is 2.

Table B: y = 32

x + 3

When x = 0, y = 3, so y-intercept is 3. The slope is 32

.

c. (6, 12)

y = 2x

– y = –32

x – 3

0 = 12

x – 3

3 = 12

x

6 = x

y = 2x

y = 2(6)

y = 12


ANSWERS




3. a. Janes is less expensive for a 3-hour job.

Jones: y = $40x + $40

y = $40(3) + $40

y = $120 + $40

y = $160

Janes: y = $30x + $60

y = $30(3) + $60

y = $90 + $60

y = $150

b.

34032030028026024022020018016014012010080604020

543 82 71 6

x

y

Hours

Dollars

Jones

James

ANSWERS




c. (2, 120)

y = 40x + 40

+ –y = –30x – 60

0 = 10x – 20

20 = 10x

2 = x

y = 40x + 40

y = 40(2) + 40

y = 80 + 40

y = 120

d. The cost will be the same for either company, $120, for a 2-hour visit.

Challenge Problem

4. a. Segment AC: slope = 23

; Segment DB: slope = –12

Segment AC:

8 – –14 – –2

96

23

( )( ) = =

Segment DB:

4 –12

–112

– 512

3.5–7

–12

= =

ANSWERS



EXERCISES

b. Segment AC: y-intercept = 13

; Segment DB: y-intercept = 314

Segment AC:

–1 = 23

(–2) + b

–1 = –43

+ b

13

= b

Segment DB:

12

–12

512

=

+ b

12

= –114

+ b

134

= b

b = 314

c. y = 23

x + 13

y = –12

x + 314


ANSWERS



EXERCISES

d.

2

12

, 2

y = 23

x + 13

–y = 12

x – 314

0 = 76

x – 3512

3512

= 76

x

52

= x or x = 212

y = 23

x + 13

–23

52

13

=

+y

y = 53

+ 13

y = 63

or 2


ANSWERS




Self Assessment




1

2 a.

2 b.

2 c.

3 a.

3 b.

3 c.

3 d.

Challenge Problem



4 a�

4 b�

4 c�

4 d�



EXERCISES

• Read through your work on the Self Check task and think about your other work in this lesson.

• Write what you have learned.

• What would you do differently if you were starting the Self Check task now?

• Record your ideas. Keep track of any strategies you have learned.

• Complete any exercises that you have not finished from this unit.

LESSON 11: PUTTING IT TOGETHER

linear equations - d3jc3ahdjad7x7.cloudfront.net filemath grade 8 unit 5 linear equations exercises....

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