## 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:  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.
(6, 3)

b. Verify that the ordered pair named in part (a) is a solution to 2y + x = 12.
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.
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?
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 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.
( – 1, – 1)

b. Verify that the ordered pair named in part (a) is a solution to x + y = – 2.
– 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.
– 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?
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 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.
( – 1, 0)

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

b. Verify that the ordered pair named in part (a) is a solution to 2x – 3y = 18.
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.
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?
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 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.
( – 1, 2)

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

b. Verify that the ordered pair named in part (a) is a solution to y = $$\frac{3}{1}$$ x + 1.
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.
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
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.
( – 4, 2)

b. Verify that the ordered pair named in part (a) is a solution to y = $$\frac{1}{2}$$ x + 4.
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.
– 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
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.
(6, 2)

b. Verify that the ordered pair named in part (a) is a solution to y = 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 x + 2y = 10.
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
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.
( – 3, 4)

b. Verify that the ordered pair named in part (a) is a solution to – 2x + 3y = 18.
– 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.
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
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.
(6, – 2)

b. Verify that the ordered pair named in part (a) is a solution to x + 2y = 2.
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.
– 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
(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  a. Name the ordered pair where the graphs of the two linear equations intersect.
(2, 5)

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