## Engage NY Eureka Math 8th Grade Module 4 Lesson 25 Answer Key

### Eureka Math Grade 8 Module 4 Lesson 25 Exercise Answer Key

Exploratory Challenge/Exercises 1–5

Exercise 1.

Sketch the graphs of the linear system on a coordinate plane:

Answer:

For the equation 2y + x = 12:

2y + 0 = 12

2y = 12

y = 6

The y – intercept point is (0, 6).

2(0) + x = 12

x = 12

The x – intercept point is (12, 0).

For the equation y = \(\frac{5}{6}\) x – 2:

The slope is \(\frac{5}{6}\), and the y – intercept point is (0, – 2).

a. Name the ordered pair where the graphs of the two linear equations intersect.

Answer:

(6, 3)

b. Verify that the ordered pair named in part (a) is a solution to 2y + x = 12.

Answer:

2(3) + 6 = 12

6 + 6 = 12

12 = 12

The left and right sides of the equation are equal.

c. Verify that the ordered pair named in part (a) is a solution to y = \(\frac{5}{6}\) x – 2.

Answer:

3 = \(\frac{5}{6}\) (6) – 2

3 = 5 – 2

3 = 3

The left and right sides of the equation are equal.

d. Could the point (4, 4) be a solution to the system of linear equations? That is, would (4, 4) make both equations true? Why or why not?

Answer:

No. The graphs of the equations represent all of the possible solutions to the given equations. The point (4, 4) is a solution to the equation 2y + x = 12 because it is on the graph of that equation. However, the point (4, 4) is not on the graph of the equation y = \(\frac{5}{6}\) x – 2. Therefore, (4, 4) cannot be a solution to the system of equations.

Exercise 2.

Sketch the graphs of the linear system on a coordinate plane:

x + y = – 2

y = 4x + 3

Answer:

For the equation x + y = – 2:

0 + y = – 2

y = – 2

The y – intercept point is (0, – 2).

x + 0 = – 2

x = – 2

The x – intercept point is ( – 2, 0).

For the equation y = 4x + 3:

The slope is \(\frac{4}{1}\), and the y – intercept point is (0, 3).

a. Name the ordered pair where the graphs of the two linear equations intersect.

Answer:

( – 1, – 1)

b. Verify that the ordered pair named in part (a) is a solution to x + y = – 2.

Answer:

– 1 + ( – 1) = – 2

– 2 = – 2

The left and right sides of the equation are equal.

c. Verify that the ordered pair named in part (a) is a solution to y = 4x + 3.

Answer:

– 1 = 4( – 1) + 3

– 1 = – 4 + 3

– 1 = – 1

The left and right sides of the equation are equal.

d. Could the point ( – 4, 2) be a solution to the system of linear equations? That is, would ( – 4, 2) make both equations true? Why or why not?

Answer:

No. The graphs of the equations represent all of the possible solutions to the given equations. The point ( – 4, 2) is a solution to the equation x + y = – 2 because it is on the graph of that equation. However, the point ( – 4, 2) is not on the graph of the equation y = 4x + 3. Therefore, ( – 4, 2) cannot be a solution to the system of equations.

Exercise 3.

Sketch the graphs of the linear system on a coordinate plane:

3x + y = – 3

– 2x + y = 2

Answer:

For the equation 3x + y = – 3:

3(0) + y = – 3

y = – 3

The y – intercept point is (0, – 3).

3x + 0 = – 3

3x = – 3

x = – 1

The x – intercept point is ( – 1, 0).

For the equation – 2x + y = 2:

– 2(0) + y = 2

y = 2

The y – intercept point is (0, 2).

– 2x + 0 = 2

– 2x = 2

x = – 1

The x – intercept point is ( – 1, 0).

a. Name the ordered pair where the graphs of the two linear equations intersect.

Answer:

( – 1, 0)

b. Verify that the ordered pair named in part (a) is a solution to 3x + y = – 3.

Answer:

3( – 1) + 0 = – 3

– 3 = – 3

The left and right sides of the equation are equal.

c. Verify that the ordered pair named in part (a) is a solution to – 2x + y = 2.

Answer:

– 2( – 1) + 0 = 2

2 = 2

The left and right sides of the equation are equal.

d. Could the point (1, 4) be a solution to the system of linear equations? That is, would (1, 4) make both equations true? Why or why not?

Answer:

No. The graphs of the equations represent all of the possible solutions to the given equations. The point (1, 4) is a solution to the equation – 2x + y = 2 because it is on the graph of that equation. However, the point (1, 4) is not on the graph of the equation 3x + y = – 3. Therefore, (1, 4) cannot be a solution to the system of equations.

Exercise 4.

Sketch the graphs of the linear system on a coordinate plane:

2x – 3y = 18

2x + y = 2

Answer:

For the equation 2x – 3y = 18:

2(0) – 3y = 18

– 3y = 18

y = – 6

The y – intercept point is (0, – 6).

2x – 3(0) = 18

2x = 18

x = 9

The x – intercept point is (9, 0).

For the equation 2x + y = 2:

2(0) + y = 2

y = 2

The y – intercept point is (0, 2).

2x + 0 = 2

2x = 2

x = 1

The x – intercept point is (1, 0).

a. Name the ordered pair where the graphs of the two linear equations intersect.

Answer:

(3, – 4)

b. Verify that the ordered pair named in part (a) is a solution to 2x – 3y = 18.

Answer:

2(3) – 3( – 4) = 18

6 + 12 = 18

18 = 18

The left and right sides of the equation are equal.

c. Verify that the ordered pair named in part (a) is a solution to 2x + y = 2.

Answer:

2(3) + ( – 4) = 2

6 – 4 = 2

2 = 2

The left and right sides of the equation are equal.

d. Could the point (3, – 1) be a solution to the system of linear equations? That is, would (3, – 1) make both equations true? Why or why not?

Answer:

No. The graphs of the equations represent all of the possible solutions to the given equations. The point (3, – 1) is not on the graph of either line; therefore, it is not a solution to the system of linear equations.

Exercise 5.

Sketch the graphs of the linear system on a coordinate plane:

y – x = 3

y = – 4x – 2

Answer:

For the equation y – x = 3:

y – 0 = 3

y = 3

The y – intercept point is (0, 3).

0 – x = 3

– x = 3

x = – 3

The x – intercept point is ( – 3, 0).

For the equation y = – 4x – 2:

The slope is – \(\frac{4}{1}\), and the y – intercept point is (0, – 2).

a. Name the ordered pair where the graphs of the two linear equations intersect.

Answer:

( – 1, 2)

b. Verify that the ordered pair named in part (a) is a solution to y – x = 3.

Answer:

2 – ( – 1) = 3

3 = 3

The left and right sides of the equation are equal.

c. Verify that the ordered pair named in part (a) is a solution to y = – 4x – 2.

Answer:

2 = – 4( – 1) – 2

2 = 4 – 2

2 = 2

The left and right sides of the equation are equal.

d. Could the point ( – 2, 6) be a solution to the system of linear equations? That is, would ( – 2, 6) make both equations true? Why or why not?

Answer:

No. The graphs of the equations represent all of the possible solutions to the given equations. The point ( – 2, 6) is a solution to the equation y = – 4x – 2 because it is on the graph of that equation. However, the point ( – 2, 6) is not on the graph of the equation y – x = 3. Therefore, ( – 2, 6) cannot be a solution to the system of equations.

Exercise 6.

Write two different systems of equations with (1, – 2) as the solution.

Answer:

Answers will vary. Two sample solutions are provided:

### Eureka Math Grade 8 Module 4 Lesson 25 Problem Set Answer Key

Question 1.

Sketch the graphs of the linear system on a coordinate plane:

y = \(\frac{1}{3}\) x + 1

y = – 3x + 11

Answer:

For the equation y = \(\frac{1}{3}\) x + 1:

The slope is \(\frac{1}{3}\), and the y – intercept point is (0, 1).

For the equation y = – 3x + 11:

The slope is – \(\frac{3}{1}\), and the y – intercept point is (0, 11).

a. Name the ordered pair where the graphs of the two linear equations intersect.

Answer:

(3, 2)

b. Verify that the ordered pair named in part (a) is a solution to y = \(\frac{3}{1}\) x + 1.

Answer:

2 = \(\frac{3}{1}\) (3) + 1

2 = 1 + 1

2 = 2

The left and right sides of the equation are equal.

c. Verify that the ordered pair named in part (a) is a solution to y = – 3x + 11.

Answer:

2 = – 3(3) + 11

2 = – 9 + 11

2 = 2

The left and right sides of the equation are equal.

Question 2.

Sketch the graphs of the linear system on a coordinate plane:

y = \(\frac{1}{2}\) x + 4

x + 4y = 4

Answer:

For the equation y = \(\frac{1}{2}\) x + 4:

The slope is \(\frac{1}{2}\), and the y – intercept point is(0, 4).

For the equation x + 4y = 4:

0 + 4y = 4

4y = 4

y = 1

The y – intercept point is (0, 1).

x + 4(0) = 4

x = 4

The x – intercept point is (4, 0).

a. Name the ordered pair where the graphs of the two linear equations intersect.

Answer:

( – 4, 2)

b. Verify that the ordered pair named in part (a) is a solution to y = \(\frac{1}{2}\) x + 4.

Answer:

2 = \(\frac{1}{2}\) ( – 4) + 4

2 = – 2 + 4

2 = 2

The left and right sides of the equation are equal.

c. Verify that the ordered pair named in part (a) is a solution to x + 4y = 4.

Answer:

– 4 + 4(2) = 4

– 4 + 8 = 4

4 = 4

The left and right sides of the equation are equal.

Question 3.

Sketch the graphs of the linear system on a coordinate plane:

y = 2

x + 2y = 10

Answer:

For the equation x + 2y = 10:

0 + 2y = 10

2y = 10

y = 5

The y – intercept point is (0, 5).

x + 2(0) = 10

x = 10

The x – intercept point is (10, 0).

a. Name the ordered pair where the graphs of the two linear equations intersect.

Answer:

(6, 2)

b. Verify that the ordered pair named in part (a) is a solution to y = 2.

Answer:

2 = 2

The left and right sides of the equation are equal.

c. Verify that the ordered pair named in part (a) is a solution to x + 2y = 10.

Answer:

6 + 2(2) = 10

6 + 4 = 10

10 = 10

The left and right sides of the equation are equal.

Question 4.

Sketch the graphs of the linear system on a coordinate plane:

– 2x + 3y = 18

2x + 3y = 6

Answer:

For the equation – 2x + 3y = 18:

– 2(0) + 3y = 18

3y = 18

y = 6

The y – intercept point is (0, 6).

– 2x + 3(0) = 18

– 2x = 18

x = – 9

The x – intercept point is ( – 9, 0).

For the equation 2x + 3y = 6:

2(0) + 3y = 6

3y = 6

y = 2

The y – intercept point is (0, 2).

2x + 3(0) = 6

2x = 6

x = 3

The x – intercept point is (3, 0).

a. Name the ordered pair where the graphs of the two linear equations intersect.

Answer:

( – 3, 4)

b. Verify that the ordered pair named in part (a) is a solution to – 2x + 3y = 18.

Answer:

– 2( – 3) + 3(4) = 18

6 + 12 = 18

18 = 18

The left and right sides of the equation are equal.

c. Verify that the ordered pair named in part (a) is a solution to 2x + 3y = 6.

Answer:

2( – 3) + 3(4) = 6

– 6 + 12 = 6

6 = 6

The left and right sides of the equation are equal.

Question 5.

Sketch the graphs of the linear system on a coordinate plane:

x + 2y = 2

y = \(\frac{2}{3}\) x – 6

Answer:

For the equation x + 2y = 2:

0 + 2y = 2

2y = 2

y = 1

The y – intercept point is (0, 1).

x + 2(0) = 2

x = 2

The x – intercept point is (2, 0).

For the equation y = \(\frac{2}{3}\) x – 6:

The slope is \(\frac{2}{3}\), and the y – intercept point is (0, – 6).

a. Name the ordered pair where the graphs of the two linear equations intersect.

Answer:

(6, – 2)

b. Verify that the ordered pair named in part (a) is a solution to x + 2y = 2.

Answer:

6 + 2( – 2) = 2

6 – 4 = 2

2 = 2

The left and right sides of the equation are equal.

c. Verify that the ordered pair named in part (a) is a solution to y = \(\frac{2}{3}\) x – 6.

Answer:

– 2 = \(\frac{2}{3}\) (6) – 6

– 2 = 4 – 6

– 2 = – 2

The left and right sides of the equation are equal.

Question 6.

Without sketching the graph, name the ordered pair where the graphs of the two linear equations intersect.

x = 2

y = – 3

Answer:

(2, – 3)

### Eureka Math Grade 8 Module 4 Lesson 25 Exit Ticket Answer Key

Question 1.

Sketch the graphs of the linear system on a coordinate plane:

2x – y = – 1

y = 5x – 5

Answer:

a. Name the ordered pair where the graphs of the two linear equations intersect.

Answer:

(2, 5)

b. Verify that the ordered pair named in part (a) is a solution to 2x – y = – 1.

Answer:

2(2) – 5 = – 1

4 – 5 = – 1

– 1 = – 1

The left and right sides of the equation are equal.

c. Verify that the ordered pair named in part (a) is a solution to y = 5x – 5.

Answer:

5 = 5(2) – 5

5 = 10 – 5

5 = 5

The left and right sides of the equation are equal.