## Engage NY Eureka Math 8th Grade Module 4 Lesson 13 Answer Key

### Eureka Math Grade 8 Module 4 Lesson 13 Exercise Answer Key

Exercise 1.

Find at least ten solutions to the linear equation 3x+y=-8, and plot the points on a coordinate plane.

Answer:

What shape is the graph of the linear equation taking?

Answer:

The graph appears to be the shape of a line.

Exercise 2.

Find at least ten solutions to the linear equation x-5y=11, and plot the points on a coordinate plane.

Answer:

What shape is the graph of the linear equation taking?

Answer:

The graph appears to be the shape of a line.

Exercise 3.

Compare the solutions you found in Exercise 1 with a partner. Add the partner’s solutions to your graph. Is the prediction you made about the shape of the graph still true? Explain.

Answer:

Yes. With the additional points, the graph still appears to be the shape of a line.

Exercise 4.

Compare the solutions you found in Exercise 2 with a partner. Add the partner’s solutions to your graph. Is the prediction you made about the shape of the graph still true? Explain.

Answer:

Yes. With the additional points, the graph still appears to be the shape of a line.

Exercise 5.

Joey predicts that the graph of -x+2y=3 will look like the graph shown below. Do you agree? Explain why or why not.

Answer:

No, I do not agree with Joey. The graph that Joey drew contains the point (0,0). If (0,0) is on the graph of the linear equation, then it will be a solution to the equation; however, it is not. Therefore, the point cannot be on the graph of the equation, which means Joey’s prediction is incorrect.

Exercise 6.

We have looked at some equations that appear to be lines. Can you write an equation that has solutions that do not form a line? Try to come up with one, and prove your assertion on the coordinate plane.

Answer:

Answers will vary. Any nonlinear equation that students write will graph as something other than a line. For example, the graph of y=x^{2} or the graph of y=x^{3} will not be a line.

### Eureka Math Grade 8 Module 4 Lesson 13 Problem Set Answer Key

In Problem 1, students graph linear equations by plotting points that represent solutions. For that reason, they need graph paper. Students informally explain why the graph of a linear equation is not curved by showing that a point on the curve is not a solution to the linear equation.

Question 1.

Find at least ten solutions to the linear equation \(\frac{1}{2}\) x+y=5, and plot the points on a coordinate plane.

Answer:

What shape is the graph of the linear equation taking?

Answer:

The graph appears to be the shape of a line.

Question 2.

Can the following points be on the graph of the equation x-y=0? Explain.

Answer:

The graph shown contains the point (0,-2). If (0,-2) is on the graph of the linear equation, then it will be a solution to the equation. It is not; therefore, the point cannot be on the graph of the equation, which means the graph shown cannot be the graph of the equation x-y=0.

Question 3.

Can the following points be on the graph of the equation x+2y=2? Explain.

Answer:

The graph shown contains the point (-4,1). If (-4,1) is on the graph of the linear equation, then it will be a solution to the equation. It is not; therefore, the point cannot be on the graph of the equation, which means the graph shown cannot be the graph of the equation x+2y=2.

Question 4.

Can the following points be on the graph of the equation x-y=7? Explain.

Answer:

Yes, because each point on the graph represents a solution to the linear equation x-y=7.

Question 5.

Can the following points be on the graph of the equation x+y=2? Explain.

Answer:

Yes, because each point on the graph represents a solution to the linear equation x+y=2.

Question 6.

Can the following points be on the graph of the equation 2x-y=9? Explain.

Answer:

Yes, because each point on the graph represents a solution to the linear equation 2x-y=9.

Question 7.

Can the following points be on the graph of the equation x-y=1? Explain.

Answer:

The graph shown contains the point (-2,-1). If (-2,-1) is on the graph of the linear equation, then it will be a solution to the equation. It is not; therefore, the point cannot be on the graph of the equation, which means the graph shown cannot be the graph of the equation x-y=1.

### Eureka Math Grade 8 Module 4 Lesson 13 Exit Ticket Answer Key

Question 1.

Ethan found solutions to the linear equation 3x-y=8 and graphed them. What shape is the graph of the linear equation taking?

Answer:

It appears to take the shape of a line.

Question 2.

Could the following points be on the graph of -x+2y=5?

Answer:

Students may have chosen any point to make the claim that this is not the graph of the equation -x+2y=5.

Although the graph appears to be a line, the graph contains the point (0,3). The point (0,3) is not a solution to the linear equation; therefore, this is not the graph of -x+2y=5.

Note to teacher: Accept any point as not being a solution to the linear equation.