## Engage NY Eureka Math 8th Grade Module 4 Lesson 20 Answer Key

### Eureka Math Grade 8 Module 4 Lesson 20 Exercise Answer Key

Opening Exercise

Figure 1

Answer:

The equation for the line in Figure 1 is y = \(\frac{2}{3}\) x – 3.

Figure 2

Answer:

The equation for the line in Figure 2 is y = – \(\frac{1}{4}\) x + 2.

Exercises

Exercise 1.

Write the equation that represents the line shown.

Answer:

y = 3x + 2

Use the properties of equality to change the equation from slope – intercept form, y = mx + b, to standard form, ax + by = c, where a, b, and c are integers, and a is not negative.

Answer:

y = 3x + 2

– 3x + y = 3x – 3x + 2

– 3x + y = 2

– 1( – 3x + y = 2)

3x – y = – 2

Exercise 2.

Write the equation that represents the line shown.

Answer:

y = – \(\frac{2}{3}\) x – 1

Use the properties of equality to change the equation from slope – intercept form, y = mx + b, to standard form, ax + by = c, where a, b, and c are integers, and 𝒂 is not negative.

Answer:

y = – \(\frac{2}{3}\) x – 1

(y = – \(\frac{2}{3}\) x – 1)3

3y = – 2x – 3

2x + 3y = – 2x + 2x – 3

2x + 3y = – 3

Exercise 3.

Write the equation that represents the line shown.

Answer:

y = – \(\frac{1}{5}\) x – 4

Use the properties of equality to change the equation from slope – intercept form, y = mx + b, to standard form, ax + by = c, where a, b, and c are integers, and a is not negative.

Answer:

y = – \(\frac{1}{5}\) x – 4

(y = – \(\frac{1}{5}\) x – 4) 5

5y = – x – 20

x + 5y = – x + x – 20

x + 5y = – 20

x + 5

Exercise 4.

Write the equation that represents the line shown.

Answer:

y = x

Use the properties of equality to change the equation from slope – intercept form, y = mx + b, to standard form, ax + by = c, where a, b, and c are integers, and a is not negative.

Answer:

y = x

– x + y = x – x

– x + y = 0

– 1( – x + y = 0)

x – y = 0

Exercise 5.

Write the equation that represents the line shown.

Answer:

y = \(\frac{1}{4}\) x + 5

Use the properties of equality to change the equation from slope – intercept form, y = mx + b, to standard form, ax + by = c, where a, b, and c are integers, and a is not negative.

Answer:

y = \(\frac{1}{4}\) x + 5

(y = \(\frac{1}{4}\) x + 5)4

4y = x + 20

– x + 4y = x – x + 20

– x + 4y = 20

– 1( – x + 4y = 20)

x – 4y = – 20

Exercise 6.

Write the equation that represents the line shown.

Answer:

y = – \(\frac{8}{5}\) x – 7

Use the properties of equality to change the equation from slope – intercept form, y = mx + b, to standard form, ax + by = c, where a, b, and c are integers, and a is not negative.

Answer:

y = – \(\frac{8}{5}\) x – 7

(y = – \(\frac{8}{5}\) x – 7)5

5y = – 8x – 35

8x + 5y = – 8x + 8x – 35

8x + 5y = – 35

### Eureka Math Grade 8 Module 4 Lesson 20 Problem Set Answer Key

Question 1.

Write the equation that represents the line shown.

Answer:

y = – \(\frac{2}{3}\) x – 4

Use the properties of equality to change the equation from slope – intercept form, y = mx + b, to standard form, ax + by = c, where a, b, and c are integers, and a is not negative.

Answer:

y = – \(\frac{2}{3}\) x – 4

(y = – \(\frac{2}{3}\) x – 4)3

3y = – 2x – 12

2x + 3y = – 2x + 2x – 12

2x + 3y = – 12

Question 2.

Write the equation that represents the line shown.

Answer:

y = 8x + 1

Use the properties of equality to change the equation from slope – intercept form, y = mx + b, to standard form, ax + by = c, where a, b, and c are integers, and a is not negative.

Answer:

y = 8x + 1

– 8x + y = 8x – 8x + 1

– 8x + y = 1

– 1( – 8x + y = 1)

8x – y = – 1

Question 3.

Write the equation that represents the line shown.

Answer:

y = \(\frac{1}{2}\) x – 4

Use the properties of equality to change the equation from slope – intercept form, y = mx + b, to standard form, ax + by = c, where a, b, and c are integers, and a is not negative.

Answer:

y = \(\frac{1}{2}\) x – 4

(y = \(\frac{1}{2}\) x – 4)2

2y = x – 8

– x + 2y = x – x – 8

– x + 2y = – 8

– 1( – x + 2y = – 8)

x – 2y = 8

Question 4.

Write the equation that represents the line shown.

Answer:

y = – 9x – 8

Use the properties of equality to change the equation from slope – intercept form, y = mx + b, to standard form, ax + by = c, where a, b, and c are integers, and a is not negative.

Answer:

y = – 9x – 8

9x + y = – 9x + 9x – 8

9x + y = – 8

Question 5.

Write the equation that represents the line shown.

Answer:

y = 2x – 14

Use the properties of equality to change the equation from slope – intercept form, y = mx + b, to standard form, ax + by = c, where a, b, and c are integers, and a is not negative.

Answer:

y = 2x – 14

– 2x + y = 2x – 2x – 14

– 2x + y = – 14

– 1( – 2x + y = – 14)

2x – y = 14

Question 6.

Write the equation that represents the line shown.

Answer:

y = – 5x + 45

Use the properties of equality to change the equation from slope – intercept form, y = mx + b, to standard form, ax + by = c, where a, b, and c are integers, and a is not negative.

Answer:

y = – 5x + 45

5x + y = – 5x + 5x + 45

5x + y = 45

### Eureka Math Grade 8 Module 4 Lesson 20 Exit Ticket Answer Key

Question 1.

Write an equation in slope – intercept form that represents the line shown.

Answer:

y = – \(\frac{1}{3}\) x + 1

Question 2.

Use the properties of equality to change the equation you wrote for Problem 1 from slope – intercept form, y = mx + b, to standard form, ax + by = c, where a, b, and c are integers, and a is not negative.

Answer:

y = – \(\frac{1}{3}\) x + 1

(y = – \(\frac{1}{3}\) x + 1)3

3y = – x + 3

x + 3y = – x + x + 3

x + 3y = 3

Question 3.

Write an equation in slope – intercept form that represents the line shown.

Answer:

y = \(\frac{3}{2}\) x + 2

Question 4.

Use the properties of equality to change the equation you wrote for Problem 3 from slope – intercept form, y = mx + b, to standard form, ax + by = c, where a, b, and c are integers, and a is not negative.

Answer:

y = \(\frac{3}{2}\) x + 2

(y = \(\frac{3}{2}\) x + 2)2

2y = 3x + 4

– 3x + 2y = 3x – 3x + 4

– 3x + 2y = 4

– 1( – 3x + 2y = 4)

3x – 2y = – 4