Eureka Math

Engage NY Eureka Math 8th Grade Module 4 Lesson 16 Answer Key

Eureka Math Grade 8 Module 4 Lesson 16 Example Answer Key

Example 1.
Using what you learned in the last lesson, determine the slope of the line with the following graph.

Answer:
The slope of the line is 3.

Example 2.
Using what you learned in the last lesson, determine the slope of the line with the following graph.

Answer:
The slope of this line is 2.

Example 3.
What is different about this line compared to the last two examples?

Answer:
This time, if we choose two points on the line that have a horizontal distance at 1, we cannot precisely determine the slope of the line because the vertical change is not an integer. It is some fractional amount.

→ Make a conjecture about how you could find the slope of this line.

Have students write their conjectures and share their ideas about how to find the slope of the line in this example; then, continue with the Discussion that follows.

Eureka Math Grade 8 Module 4 Lesson 16 Exercise Answer Key

Exercise
Let’s investigate concretely to see if the claim that we can find slope between any two points is true.

a. Select any two points on the line to label as P and R.
Answer:
Sample points are selected on the graph.

b. Identify the coordinates of points P and R.
Answer:
Sample points are labeled on the graph.

c. Find the slope of the line using as many different points as you can. Identify your points, and show your work below.
Answer:
Points selected by students will vary, but the slope should always equal 2. Students could choose to use points (0,5), (-1,3), (-2,1), (-3,-1), (-4,-3), and (-5,-5).

Eureka Math Grade 8 Module 4 Lesson 16 Problem Set Answer Key

Students practice finding slope between any two points on a line. Students also see that m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\) yields the same result as m=\(\frac{r_{2}-p_{2}}{r_{1}-p_{1}}\).

Question 1.
Calculate the slope of the line using two different pairs of points.

m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{30-(-10)}{-10-30}\)
=\(\frac{40}{-40}\)
=\(\frac{1}{1}\)
=-1
m=\(\frac{q_{2}-r_{2}}{q_{1}-r_{1}}\)
=\(\frac{10-(-10)}{10-30}\)
=\(\frac{20}{-20}\)
=-1

Question 2.
Calculate the slope of the line using two different pairs of points.

Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{3-2}{-2-2}\)
=\(\frac{1}{-4}\)
=-\(\frac{1}{4}\)

m=\(\frac{q_{2}-r_{2}}{q_{1}-r_{1}}\)
=\(\frac{1-2}{6-2}\)
=\(\frac{-1}{4}\)
=-\(\frac{1}{4}\)

Question 3.
Calculate the slope of the line using two different pairs of points.

Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{1-5}{5-6}\)
=\(\frac{-4}{-1}\)
=\(\frac{4}{1}\)
=4

m=\(\frac{q_{2}-r_{2}}{q_{1}-r_{1}}\)
=\(\frac{-3-5}{4-6}\)
=\(\frac{-8}{-2}\)
=4

Question 4.
Calculate the slope of the line using two different pairs of points.

Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{5-1}{3-5}\)
=\(\frac{4}{-2}\)
=\(-\frac{2}{1}\)
=-2

m=\(\frac{q_{2}-r_{2}}{q_{1}-r_{1}}\)
=\(\frac{3-1}{4-5}\)
=\(\frac{2}{-1}\)
=-2

Question 5.
Calculate the slope of the line using two different pairs of points.

Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{2-1}{1-6}\)
=\(\frac{1}{-5}\)
=-\(\frac{1}{5}\)

m=\(\frac{q_{2}-r_{2}}{q_{1}-r_{1}}\)
=\(\frac{0-1}{11-6}\)
=\(\frac{-1}{5}\)
=-\(\frac{1}{5}\)

Question 6.
Calculate the slope of the line using two different pairs of points.

a. Select any two points on the line to compute the slope.
Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{-5-(-2)}{-7-(-5)}\)
=\(\frac{-3}{-2}\)
=\(\frac{3}{2}\)

b. Select two different points on the line to calculate the slope.
Answer:
Let the two new points be (-3,1) and (-1,4).
m=\(\frac{q_{2}-s_{2}}{q_{1}-s_{1}}\)
=\(\frac{1-4}{-3-(-1)}\)
=\(\frac{-3}{-2}\)
=\(\frac{3}{2}\)

c. What do you notice about your answers in parts (a) and (b)? Explain.
Answer:
The slopes are equal in parts (a) and (b). This is true because of what we know about similar triangles.
The slope triangle that is drawn between the two points selected in part (a) is similar to the slope triangle that is drawn between the two points in part (b) by the AA criterion. Then, because the corresponding sides of similar triangles are equal in ratio, the slopes are equal.

Question 7.
Calculate the slope of the line in the graph below.

Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{-4-2}{-7-2}\)
=\(\frac{-6}{-9}\)
=\(\frac{2}{3}\)

Question 8.
Your teacher tells you that a line goes through the points (-6, \(\frac{1}{2}\)) and (-4,3).
a. Calculate the slope of this line.
Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{\frac{1}{2}-3}{-6-(-4)}\)
=\(\frac{-\frac{5}{2}}{-2}\)
= \(\frac{\frac{5}{2}}{2}\)
=\(\frac{5}{2}\)÷2
=\(\frac{5}{2}\)×\(\frac{1}{2}\)
= \(\frac{5}{4}\)

b. Do you think the slope will be the same if the order of the points is reversed? Verify by calculating the slope, and explain your result.
The slope should be the same because we are joining the same two points.
Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
= \(\frac{3-\frac{1}{2}}{-4-(-6)}\)
= \(\frac{\frac{5}{2}}{2}\)
= \(\frac{5}{4}\)
Since the slope of a line can be computed using any two points on the same line, it makes sense that it does not matter which point we name as P and which point we name as R.

Question 9.
Use the graph to complete parts (a)–(c).

a. Select any two points on the line to calculate the slope.
Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{1-(-3)}{-3-0}\)
=\(\frac{4}{-3}\)
=-\(\frac{4}{3}\)

b. Compute the slope again, this time reversing the order of the coordinates.
Answer:
m=\(\frac{\left(r_{2}-p_{2}\right)}{\left(r_{1}-p_{1}\right)}\)
=\(\frac{-3-1}{0-(-3)}\)
=\(\frac{-4}{3}\)
=-\(\frac{4}{3}\)

c. What do you notice about the slopes you computed in parts (a) and (b)?
Answer:
The slopes are equal.

d. Why do you think m=\(\frac{\left(p_{2}-r_{2}\right)}{\left(p_{1}-r_{1}\right)}\)=\(\frac{\left(r_{2}-p_{2}\right)}{\left(r_{1}-p_{1}\right)}\).?
Answer:
If I multiply the first fraction by \(\frac{-1}{-1}\), then I get the second fraction:
\(\frac{-1}{-1}\)×\(\left(\frac{\left(p_{2}-r_{2}\right)}{\left(p_{1}-r_{1}\right)}\right)\)=\(\frac{\left(r_{2}-p_{2}\right)}{\left(r_{1}-p_{1}\right)}\).
I can do the same thing to the second fraction to obtain the first:
\(\frac{-1}{-1}\)×\(\left(\frac{\left(r_{2}-p_{2}\right)}{\left(r_{1}-p_{1}\right)}\right)\)=\(\frac{\left(p_{2}-r_{2}\right)}{\left(p_{1}-r_{1}\right)}\)
Also, since I know that I can find the slope between any two points, it should not matter which point I pick first.

Question 10.
Each of the lines in the lesson was non-vertical. Consider the slope of a vertical line, x=2. Select two points on the line to calculate slope. Based on your answer, why do you think the topic of slope focuses only on non-vertical lines?

Answer:
Students can use any points on the line x=2 to determine that the slope is undefined. The computation of slope using the formula leads to a fraction with zero as its denominator. Since the slope of a vertical line is undefined, there is no need to focus on them.

Challenge:

Question 11.
A certain line has a slope of \(\frac{1}{2}\). Name two points that may be on the line.
Answer:
Answers will vary. Accept any answers that have a difference in y-values equal to 1 and a difference of x-values equal to 2. Points (6,4) and (4,3) may be on the line, for example.

Eureka Math Grade 8 Module 4 Lesson 16 Exit Ticket Answer Key

Find the rate of change of the line by completing parts (a) and (b).

a. Select any two points on the line to label as P and R. Name their coordinates.
Answer:
Answers will vary. Other points on the graph may have been chosen.
P(-1,0) and R(5,3)

b. Compute the rate of change of the line.
Answer:
m = \(\frac{\left(p_{2}-r_{2}\right)}{\left(p_{1}-r_{1}\right)}\)
m=\(\frac{0-3}{-1-5}\)
=\(\frac{-3}{-6}\)
=\(\frac{1}{2}\)

Engage NY Eureka Math 8th Grade Module 4 Lesson 14 Answer Key

Eureka Math Grade 8 Module 4 Lesson 14 Exercise Answer Key

Exercise 1.
Find at least four solutions to graph the linear equation 1x+2y=5.
Answer:

Exercise 2.
Find at least four solutions to graph the linear equation 1x+0y=5.
Answer:

Exercise 3.
What was different about the equations in Exercises 1 and 2? What effect did this change have on the graph?
Answer:
In the first equation, the coefficient of y was 2. In the second equation, the coefficient of y was 0. The graph changed from being slanted to a vertical line.

Exercises 4–6
Students complete Exercises 4–6 independently. Students need graph paper to complete the exercises.

Exercise 4.
Graph the linear equation x=-2.
Answer:

Exercise 5.
Graph the linear equation x=3.
Answer:

Exercise 6.
What will the graph of x=0 look like?
Answer:
The graph of x=0 will look like a vertical line that goes through the point (0,0). It will be the same as the y-axis.

Exercises 7–9
Students complete Exercises 7–9 independently or in pairs in preparation for the discussion that follows. Students need graph paper to complete the exercises.

Exercise 7.
Find at least four solutions to graph the linear equation 2x+1y=2.
Answer:

Exercise 8.
Find at least four solutions to graph the linear equation 0x+1y=2.
Answer:

Exercise 9.
What was different about the equations in Exercises 7 and 8? What effect did this change have on the graph?
Answer:
In the first equation, the coefficient of x was 2. In the second equation, the coefficient of x was 0. The graph changed from being a slanted line to a horizontal line.

Exercises 10–12
Students complete Exercises 10–12 independently. Students need graph paper to complete the exercises.

Exercise 10.
Graph the linear equation y=-2.
Answer:

Exercise 11.
Graph the linear equation y=3.
Answer:

Exercise 12.
What will the graph of y=0 look like?
Answer;
The graph of y=0 will look like a horizontal line that goes through the point (0,0). It will be the same as the x-axis.

Eureka Math Grade 8 Module 4 Lesson 14 Exit Ticket Answer Key

Question 1.
Graph the linear equation ax+by=c, where a=0, b=1, and c=1.5.

Answer:

Question 2.
Graph the linear equation ax+by=c, where a=1, b=0, and c=-\(\frac{5}{2}\).

Answer:

Question 3.
What linear equation represents the graph of the line that coincides with the x-axis?
Answer:
y=0

Question 4.
What linear equation represents the graph of the line that coincides with the y-axis?
Answer:
x=0

Eureka Math Grade 8 Module 4 Lesson 14 Problem Set Answer Key

Students need graph paper to complete the Problem Set.

Question 1.
Graph the two-variable linear equation ax+by=c, where a=0, b=1, and c=-4.
Answer:

Question 2.
Graph the two-variable linear equation ax+by=c, where a=1, b=0, and c=9.
Answer:

Question 3.
Graph the linear equation y=7.
Answer:

Question 4.
Graph the linear equation x=1.
Answer:

Question 5.
Explain why the graph of a linear equation in the form of y=c is the horizontal line, parallel to the x-axis passing through the point (0,c).
Answer:
The graph of y=c passes through the point (0,c), which means the graph of y=c cannot be parallel to the y-axis because the graph intersects it. For that reason, the graph of y=c must be the horizontal line parallel to the
x-axis.

Question 6.
Explain why there is only one line with the equation y=c that passes through the point (0,c).
Answer:
There can only be one line parallel to another that goes through a given point. Since the graph of y=c is parallel to the x-axis and it goes through the point (0,c), then it must be the only line that does. Therefore, there is only one line that is the graph of the equation y=c that passes through (0,c).

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=x² or the graph of y=x³ 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.

Engage NY Eureka Math 8th Grade Module 4 Lesson 12 Answer Key

Eureka Math Grade 8 Module 4 Lesson 12 Opening Exercise Answer Key

Emily tells you that she scored 32 points in a basketball game. Write down all the possible ways she could have scored 32 with only two- and three-point baskets. Use the table below to organize your work.

Answer:

Let x be the number of two-pointers and y be the number of three-pointers that Emily scored. Write an equation to represent the situation.
Answer:
2x+3y=32

Eureka Math Grade 8 Module 4 Lesson 12 Exercise Answer Key

Exploratory Challenge/Exercises

Question 1.
Find five solutions for the linear equation x+y=3, and plot the solutions as points on a coordinate plane.

Answer:

Question 2.
Find five solutions for the linear equation 2x-y=10, and plot the solutions as points on a coordinate plane.

Answer:

Question 3.
Find five solutions for the linear equation x+5y=21, and plot the solutions as points on a coordinate plane.

Answer:

Question 4.
Consider the linear equation \(\frac{2}{5}\) x+y=11.
a. Will you choose to fix values for x or y? Explain.
Answer:
If I fix values for x, it will make the computations easier. Solving for y can be done in one step.

b. Are there specific numbers that would make your computational work easier? Explain.
Answer:
Values for x that are multiples of 5 will make the computations easier. When I multiply \(\frac{2}{5}\) by a multiple of 5, I will get an integer.

c. Find five solutions to the linear equation \(\frac{2}{5}\) x+y=11, and plot the solutions as points on a coordinate plane.
x Linear Equation:

Answer:

Question 5.
At the store, you see that you can buy a bag of candy for $2 and a drink for $1. Assume you have a total of $35 to spend. You are feeling generous and want to buy some snacks for you and your friends.
a. Write an equation in standard form to represent the number of bags of candy, x, and the number of drinks, y, that you can buy with $35.
Answer:
2x+y=35

b. Find five solutions to the linear equation from part (a), and plot the solutions as points on a coordinate plane.

Answer:

Eureka Math Grade 8 Module 4 Lesson 12 Exit Ticket Answer Key

Question 1.
Is the point (1,3) a solution to the linear equation 5x-9y=32? Explain.
Answer:
No, (1,3) is not a solution to 5x-9y=32 because 5(1)-9(3)=5-27=-22, and -22≠32.

Question 2.
Find three solutions for the linear equation 4x-3y=1, and plot the solutions as points on a coordinate plane.

Answer:

Eureka Math Grade 8 Module 4 Lesson 12 Problem Set Answer Key

Students practice finding and graphing solutions for linear equations that are in standard form.

Question 1.
Consider the linear equation x-\(\frac{3}{2}\) y=-2.
a. Will you choose to fix values for x or y? Explain.
Answer:
If I fix values for y, it will make the computations easier. Solving for x can be done in one step.

b. Are there specific numbers that would make your computational work easier? Explain.
Answer:
Values for y that are multiples of 2 will make the computations easier. When I multiply \(\frac{3}{2}\) by a multiple of 2, I will get a whole number.

c. Find five solutions to the linear equation x-\(\frac{3}{2}\) y=-2, and plot the solutions as points on a coordinate plane.

Answer:

Question 2.
Find five solutions for the linear equation \(\frac{1}{3}\) x+y=12, and plot the solutions as points on a coordinate plane.
Answer:

Question 3.
Find five solutions for the linear equation -x+\(\frac{3}{4}\) y=-6, and plot the solutions as points on a coordinate plane.
Answer:

Question 4.
Find five solutions for the linear equation 2x+y=5, and plot the solutions as points on a coordinate plane.
Answer:

Question 5.
Find five solutions for the linear equation 3x-5y=15, and plot the solutions as points on a coordinate plane.
Answer:

Engage NY Eureka Math 8th Grade Module 4 Lesson 10 Answer Key

Eureka Math Grade 8 Module 4 Lesson 10 Example Answer Key

Example 1.
Consider the word problem below. We can do several things to answer this problem, but let’s begin to organize our work using a table for time and distance:

Example 1.
Paul walks 2 miles in 25 minutes. How many miles can Paul walk in 137.5 minutes?

Answer:

As students answer the questions below, fill in the table.
→ How many miles would Paul be able to walk in 50 minutes? Explain.
→ Paul could walk 4 miles in 50 minutes because 50 minutes is twice the time we were given, so we can calculate twice the distance, which is 4.
→ How many miles would Paul be able to walk in 75 minutes? Explain.
→ Paul could walk 6 miles in 75 minutes because 75 minutes is three times the number of minutes we were given, so we can calculate three times the distance, which is 6.
→ How many miles would Paul be able to walk in 100 minutes?
→ He could walk 8 miles.
→ How many miles would he walk in 125 minutes?
→ He could walk 10 miles.
→ How could we determine the number of miles Paul could walk in 137.5 minutes?

Provide students time to think about the answer to this question. They may likely say that they can write a proportion to figure it out. Allow them to share and demonstrate their solutions. Then, proceed with the discussion below, if necessary.
→ Since the relationship between the distance Paul walks and the time it takes him to walk that distance is proportional, we let y represent the distance Paul walks in 137.5 minutes and write the following:
Answer:
\(\frac{25}{2}\)=\(\frac{13.75}{y}\)
25y=137.5(2)
25y=275
y=11
Therefore, Paul can walk 11 miles in 137.5 minutes.

→ How many miles, y, can Paul walk in x minutes?
Provide students time to think about the answer to this question. Allow them to share their ideas, and then proceed with the discussion below, if necessary.
→ We know for a fact that Paul can walk 2 miles in 25 minutes, so we can write the ratio \(\frac{25}{2}\) as we did with the proportion. We can write another ratio for the number of miles, y, Paul walks in x minutes. It is \(\frac{x}{y}\). For the same reason we could write the proportion before, we can write one now with these two ratios:
\(\frac{25}{2}\)=\(\frac{x}{y}\).
Does this remind you of something we have done recently? Explain.
→ This is a linear equation in disguise. All we need to do is multiply each numerator by the other fraction’s denominator, and then we will have a linear equation.
25y=2x
→ Recall our original question: How many miles, y, can Paul walk in x minutes? We need to solve this equation for y.
25y=2x
y=\(\frac{2}{25}\) x
y=0.08x
Paul can walk 0.08x miles in x minutes. This equation will allow us to answer all kinds of questions about Paul with respect to any given number of minutes or miles.

→ Let’s go back to the table and look for y=0.08x or its equivalent y=\(\frac{2}{25}\) x. What do you notice?

→ The fraction \(\frac{2}{25}\) came from the first row in the table. It is the distance traveled divided by the time it took to travel that distance. It is also in between each row of the table. For example, the difference between 4 miles and 2 miles is 2, and the difference between the associated times 50 and 25 is 25. The pattern repeats throughout the table.

Show on the table the +2 between each distance interval and the +25 between each time interval. Remind students that they have done work like this before, specifically finding a unit rate for a proportional relationship. Make clear that the unit rate found in the table was exactly the same as the unit rate found using the proportion and that the unit rate is the rate at which Paul walks.
→ Let’s look at another problem where only a table is provided.

We want to know how many miles, y, can be traveled in any number of hours x. Using our previous work, what should we do?
→ We can write and solve a proportion that contains both x and y or use the table to help us determine the unit rate.
→ How many miles, y, can be traveled in any number of hours x?
→ Student work:
\(\frac{123}{3}\)=\(\frac{y}{x}\)
123x=3y
\(\frac{123}{3}\) x=y
41x=y
→ What does the equation y=41x mean?
→ It means that the distance traveled, y, is equal to the rate of 41 multiplied by the number of hours x traveled at that rate.

Example 2.
The point of this problem is to make clear to students that constant rate must be assumed in order to write linear equations in two variables and to use those equations to answer questions about distance, time, and rate.
→ Consider the following word problem: Alexxa walked from Grand Central Station (GCS) on 42^nd Street to Penn Station on 7^th Avenue. The total distance traveled was 1.1 miles. It took Alexxa 25 minutes to make the walk. How many miles did she walk in the first 10 minutes?
→ Give students a minute to think and/or work on the problem. Expect them to write a proportion and solve the problem. The next part of the discussion gets them to think about what is meant by “constant” speed or, rather, lack of it.
→ She walked 0.44 miles. (Assuming students used a proportion to solve.)
→ Are you sure about your answer? How often do you walk at a constant speed? Notice the problem did not even mention that she was walking at the same rate throughout the entire 1.1 miles. What if you have more information about her walk: Alexxa walked from GCS along 42^nd Street to an ATM 0.3 miles away in 8 minutes. It took her 2 minutes to get some money out of the machine. Do you think your answer is still correct?
→ Probably not since we now know that she had to stop at the ATM.
→ Let’s continue with Alexxa’s walk: She reached the 7^th Avenue junction 13 minutes after she left GCS, a distance of 0.6 miles. There, she met her friend Karen with whom she talked for 2 minutes. After leaving her friend, she finally got to Penn Station 25 minutes after her walk began.
→ Is this a more realistic situation than believing that she walked the exact same speed throughout the entire trip? What other events typically occur during walks in the city?
→ Stoplights at crosswalks, traffic, maybe a trip/fall, or running an errand
→ This is precisely the reason we need to take a critical look at what we call proportional relationships and constant speed, in general.
→ The following table shows an accurate picture of Alexxa’s walk:

With this information, we can answer the question. Alexxa walked exactly 0.3 miles in 10 minutes.
→ Now that we have an idea of what could go wrong when we assume a person walks at a constant rate or that a proportion can give us the correct answer all of the time, let’s define what is called average speed.
→ Suppose a person walks a distance of d (miles) in a given time interval t (minutes). Then, the average speed in the given time interval is \(\frac{d}{t}\) in miles per minute.
→ With this definition, we can calculate Alexxa’s average speed: The distance that Alexxa traveled divided by the time interval she walked is \(\frac{1.1}{25}\) miles per minute.
→ If we assume that someone can actually walk at the same average speed over any time interval, then we say that the person is walking at a constant speed.
Suppose the average speed of a person is the same constant C for any given time interval. Then, we say that the person is walking at a constant speed C.
→ If the original problem included information specifying constant speed, then we could write the following:
Alexxa’s average speed for 25 minutes is \(\frac{1.1}{25}\).
Let y represent the distance Alexxa walked in 10 minutes. Then, her average speed for 10 minutes is \(\frac{y}{10}\).
Since Alexxa is walking at a constant speed of C miles per minute, then we know that
\(\frac{1.1}{25}\)=C, and \(\frac{y}{10}\)=C.
Since both fractions are equal to C, then we can write
\(\frac{1.1}{25}\) = \(\frac{y}{10}\)
With the assumption of constant speed, we now have a proportional relationship, which would make the answer you came up with in the beginning correct.
We can go one step further and write a statement in general. If Alexxa walks y miles in x minutes, then
\(\)=C, and \(\)=\(\).
To find how many miles y Alexxa walks in x miles, we solve the equation for y:
\(\frac{1.1}{25}\)=\(\frac{y}{x}\)
25y=1.1x
\(\frac{25}{2}\)5 y=\(\frac{1.1}{25}\) x
y=\(\frac{1.1}{25}\) x,
where the last equation is an example of a linear equation in two variables x and y. With this general equation, we can find the distance y Alexxa walks in any given time x. Since we have more information about Alexxa’s walk, where and when she stopped, we know that the equation cannot accurately predict the distance she walks after a certain number of minutes. To do so requires us to assume that she walks at a constant rate. This is an assumption we generally take for granted when solving problems about rate.

Eureka Math Grade 8 Module 4 Lesson 10 Exercise Answer Key

Question 1.
Wesley walks at a constant speed from his house to school 1.5 miles away. It took him 25 minutes to get to school.
a. What fraction represents his constant speed, C?
Answer:
\(\frac{1.5}{25}\)=C

b. You want to know how many miles he has walked after 15 minutes. Let y represent the distance he traveled after 15 minutes of walking at the given constant speed. Write a fraction that represents the constant speed, C, in terms of y.
Answer:
\(\frac{y}{15}\)=C

c. Write the fractions from parts (a) and (b) as a proportion, and solve to find how many miles Wesley walked after 15 minutes.
Answer:
\(\frac{1.5}{25}\)=\(\frac{y}{15}\)
25y=22.5
\(\frac{25}{25}\) y=\(\frac{22.5}{25}\)
y=0.9
Wesley walks 0.9 miles in 15 minutes.

d. Let y be the distance in miles that Wesley traveled after x minutes. Write a linear equation in two variables that represents how many miles Wesley walked after x minutes.
Answer:
\(\frac{1.5}{25}\)=\(\frac{y}{x}\)
25y=1.5x
\(\frac{25}{25}\)y=\(\frac{1.5}{25}\) x
y=\(\frac{1.5}{25}\) x

Question 2.
Stefanie drove at a constant speed from her apartment to her friend’s house 20 miles away. It took her 45 minutes to reach her destination.
a. What fraction represents her constant speed, C?
Answer:
\(\frac{20}{45}\)=C

b. What fraction represents constant speed, C, if it takes her x number of minutes to get halfway to her friend’s house?
Answer:
\(\frac{10}{x}\)=C

c. Write and solve a proportion using the fractions from parts (a) and (b) to determine how many minutes it takes her to get to the halfway point.
Answer:
\(\frac{20}{45}\)=\(\frac{10}{x}\)
20x=450
\(\frac{20}{20}\) x=\(\frac{450}{20}\)
x=22.5
Stefanie gets halfway to her friend’s house, 10 miles away, after 22.5 minutes.

d. Write a two-variable equation to represent how many miles Stefanie can drive over any time interval.
Answer:
Let y represent the distance traveled over any time interval x. Then,
\(\frac{20}{45}\)=\(\frac{y}{x}\)
20x=45y
\(\frac{20}{45}\) x=\(\frac{45}{45}\) y
\(\frac{4}{9}\) x=y.

Exercise 3.
The equation that represents how many miles, y, Dave travels after x hours is y=50x+15. Use the equation to complete the table below.

Answer:

Eureka Math Grade 8 Module 4 Lesson 10 Exit Ticket Answer Key

Alex skateboards at a constant speed from his house to school 3.8 miles away. It takes him 18 minutes.
a. What fraction represents his constant speed, C?
Answer:
\(\frac{3.8}{18}\)=C

b. After school, Alex skateboards at the same constant speed to his friend’s house. It takes him 10 minutes. Write the fraction that represents constant speed, C, if he travels a distance of y.
Answer:
\(\frac{y}{10}\)=C

c. Write the fractions from parts (a) and (b) as a proportion, and solve to find out how many miles Alex’s friend’s house is from school. Round your answer to the tenths place.
\(\frac{3.8}{18}\)=\(\frac{y}{10}\)
3.8(10)=18y
38=18y
\(\frac{38}{18}\)=y
2.1≈y
Alex’s friend lives about 2.1 miles from school.

Eureka Math Grade 8 Module 4 Lesson 10 Problem Set Answer Key

Students practice writing and solving proportions to solve constant speed problems. Students write two variable equations to represent situations, generally.

Question 1.
Eman walks from the store to her friend’s house, 2 miles away. It takes her 35 minutes.
a. What fraction represents her constant speed, C?
Answer:
\(\frac{2}{35}\)=C

b. Write the fraction that represents her constant speed, C, if she walks y miles in 10 minutes.
Answer:
\(\frac{y}{10}\)=C

c. Write and solve a proportion using the fractions from parts (a) and (b) to determine how many miles she walks after 10 minutes. Round your answer to the hundredths place.
Answer:
\(\frac{2}{35}\)=\(\frac{y}{10}\)
35y=20
\(\frac{35}{35}\) y=\(\frac{20}{35}\)
y=0.57142…
Eman walks about 0.57 miles after 10 minutes.

d. Write a two-variable equation to represent how many miles Eman can walk over any time interval.
Answer:
Let y represent the distance Eman walks in x minutes.
\(\frac{2}{35}\)=\(\frac{y}{x}\)
35y=2x
\(\frac{35}{35}\) y=\(\frac{2}{35}\) x
y=\(\frac{2}{35}\) x

Question 2.
Erika drives from school to soccer practice 1.3 miles away. It takes her 7 minutes.
a. What fraction represents her constant speed, C?
Answer:
\(\frac{1.3}{7}\)=C

b. What fraction represents her constant speed, C, if it takes her x minutes to drive exactly 1 mile?
Answer:
\(\frac{1}{x}\)=C

c. Write and solve a proportion using the fractions from parts (a) and (b) to determine how much time it takes her to drive exactly 1 mile. Round your answer to the tenths place.
Answer:
\(\frac{1.3}{7}\)=\(\frac{1}{x}\)
1.3x=7
\(\frac{1.3}{1.3}\) x=7/(1.3)
x=5.38461…
It takes Erika about 5.4 minutes to drive exactly 1 mile.

d. Write a two-variable equation to represent how many miles Erika can drive over any time interval.
Answer:
Let y be the number of miles Erika travels in x minutes.
\(\frac{1.3}{7}\)=\(\frac{y}{x}\)
7y=1.3x
\(\frac{7}{7}\) y=\(\frac{1.3}{7}\) x
y=\(\frac{1.3}{7}\) x

Question 3.
Darla drives at a constant speed of 45 miles per hour.
a. If she drives for y miles and it takes her x hours, write the two-variable equation to represent the number of miles Darla can drive in x hours.
Answer;
\(\frac{y}{x}\)=45
y=45x

b. Darla plans to drive to the market 14 miles from her house, then to the post office 3 miles from the market, and then return home, which is 15 miles from the post office. Assuming she drives at a constant speed the entire time, how long will it take her to run her errands and get back home? Round your answer to the hundredths place.
Answer:
Altogether, Darla plans to drive 32 miles because 14+3+15=32.
32=45x
\(\frac{32}{45}\)=\(\frac{45}{45}\) x
0.71111…=x
It will take Darla about 0.71 hours to run her errands and get back home.

Question 4.
Aaron walks from his sister’s house to his cousin’s house, a distance of 4 miles, in 80 minutes. How far does he walk in 30 minutes?
Answer:
I cannot say for sure how far Aaron walks in 30 minutes because I do not know if he is walking at a constant speed. Maybe he stopped at his friend’s house for 20 minutes.

Question 5.
Carlos walks 4 miles every night for exercise. It takes him exactly 63 minutes to finish his walk.
a. Assuming he walks at a constant rate, write an equation that represents how many miles, y, Carlos can walk in x minutes.
Answer:
Since \(\frac{4}{63}\)=C and \(\frac{y}{x}\)=C, then
\(\frac{4}{63}\)=\(\frac{y}{x}\)
63y=4x
\(\frac{63}{63}\)y=\(\frac{4}{63}\) x
y=\(\frac{4}{63}\) x”.”

b. Use your equation from part (a) to complete the table below. Use a calculator, and round all values to the hundredths place.

Answer:

Engage NY Eureka Math 8th Grade Module 4 Lesson 11 Answer Key

Eureka Math Grade 8 Module 4 Lesson 11 Example Answer Key

Example 1.
Pauline mows a lawn at a constant rate. Suppose she mows a 35-square-foot lawn in 2.5 minutes. What area, in square feet, can she mow in 10 minutes? t minutes?

→ What is Pauline’s average rate in 2.5 minutes?
→ Pauline’s average rate in 2.5 minutes is \(\frac{35}{2.5}\) square feet per minute.
→ What is Pauline’s average rate in 10 minutes?
→ Let A represent the square feet of the area mowed in 10 minutes. Pauline’s average rate in 10 minutes is \(\frac{A}{10}\) square feet per mintute.
→ Let C be Pauline’s constant rate in square feet per minute; then, \(\frac{35}{2.5}\)=C, and \(\frac{A}{10}\)=C. Therefore,
Answer;
\(\frac{35}{2.5}\)=\(\frac{A}{10}\)
350=2.5A
\(\frac{350}{2.5}\)=\(\frac{2.5}{2.5}\) A
140=A
Pauline mows 140 square feet of lawn in 10 minutes.

→ If we let y represent the number of square feet Pauline can mow in t minutes, then Pauline’s average rate in t minutes is \(\frac{y}{t}\) square feet per minute.
→ Write the two-variable equation that represents the area of lawn, y, Pauline can mow in t minutes.
Answer:
\(\frac{35}{2.5}\)=\(\frac{y}{t}\)
2.5y=35t
\(\frac{2.5}{2.5}\) y=\(\frac{35}{2.5}\) t
y=\(\frac{35}{2.5}\) t

→ What is the meaning of \(\frac{35}{2.5}\) in the equation y=\(\frac{35}{2.5}\) t?
→ The number \(\frac{35}{2.5}\) represents the constant rate at which Pauline can mow a lawn.
→ We can organize the information in a table.

Answer:

→ On a coordinate plane, we will let the x-axis represent time t, in minutes, and the y-axis represent the area of mowed lawn in square feet. Then we have the following graph.

Answer:

→ Because Pauline mows at a constant rate, we would expect the square feet of mowed lawn to continue to rise as the time, in minutes, increases.

Example 2.
Water flows at a constant rate out of a faucet. Suppose the volume of water that comes out in three minutes is 10.5 gallons. How many gallons of water come out of the faucet in t minutes?
→ Write the linear equation that represents the volume of water, V, that comes out in t minutes.
Answer:
Let C represent the constant rate of water flow.
\(\frac{10.5}{3}\)=C, and \(\frac{V}{t}\)=C; then, \(\frac{10.5}{3}\)=\(\frac{V}{t}\).
\(\frac{10.5}{3}\)=\(\frac{V}{t}\)
3V=10.5t
\(\frac{3}{3}\) V=\(\frac{10.5}{3}\) t
V=\(\frac{10.5}{3}\) t

→ What is the meaning of the number \(\frac{10.5}{3}\) in the equation V=\(\frac{10.5}{3}\) t?
→ The number \(\frac{10.5}{3}\) represents the constant rate at which water flows from a faucet.
→ Using the linear equation V=\(\frac{10.5}{3}\) t, complete the table.

Answer:

→ On a coordinate plane, we will let the x-axis represent time t in minutes and the y-axis represent the volume of water. Graph the data from the table.

Answer:

→ Using the graph, about how many gallons of water do you think would flow after 1 \(\frac{1}{2}\) minutes? Explain.
→ After 1 \(\frac{1}{2}\) minutes, between 3 \(\frac{1}{2}\) and 7 gallons of water will flow. Since the water is flowing at a constant rate, we can expect the volume of water to rise between 1 and 2 minutes. The number of gallons that flow after 1 \(\frac{1}{2}\) minutes then would have to be between the number of gallons that flow out after 1 minute and 2 minutes.
→ Using the graph, about how long would it take for 15 gallons of water to flow out of the faucet? Explain.
→ It would take between 4 and 5 minutes for 15 gallons of water to flow out of the faucet. It takes 4 minutes for 14 gallons to flow; therefore, it must take more than 4 minutes for 15 gallons to come out. It must take less than 5 minutes because 3 \(\frac{1}{2}\) gallons flow out every minute.
→ Graphing proportional relationships like these last two constant rate problems provides us more information than simply solving an equation and calculating one value. The graph provides information that is not so obvious in an equation.

Eureka Math Grade 8 Module 4 Lesson 11 Exercise Answer Key

Exercise 1.
Juan types at a constant rate. He can type a full page of text in 3 \(\frac{1}{2}\) minutes. We want to know how many pages, p, Juan can type after t minutes.
a. Write the linear equation in two variables that represents the number of pages Juan types in any given time interval.
Let C represent the constant rate that Juan types in pages per minute. Then,
Answer:
\(\frac{1}{3.5}\)=C, and \(\frac{p}{t}\)=C; therefore, \(\frac{1}{3.5}\)=\(\frac{p}{t}\).
\(\frac{1}{3.5}\)=\(\frac{p}{t}\)
3.5p=t
\(\frac{3.5}{3.5}\) p=\(\frac{1}{3.5}\) t
p=\(\frac{1}{3.5}\) t

b. Complete the table below. Use a calculator, and round your answers to the tenths place.

Answer:

c. Graph the data on a coordinate plane.

Answer:

d. About how long would it take Juan to type a 5-page paper? Explain.
Answer;
It would take him between 15 and 20 minutes. After 15 minutes, he will have typed 4.3 pages. In 20 minutes, he can type 5.7 pages. Since 5 pages is between 4.3 and 5.7, then it will take him between 15 and 20 minutes.

Question 2.
Emily paints at a constant rate. She can paint 32 square feet in 5 minutes. What area, A, in square feet, can she paint in t minutes?
a. Write the linear equation in two variables that represents the number of square feet Emily can paint in any given time interval.
Answer:
Let C be the constant rate that Emily paints in square feet per minute. Then,
\(\frac{32}{5}\)=C, and \(\frac{A}{t}\)=C; therefore, \(\frac{32}{5}\)=\(\frac{A}{t}\).
\(\frac{32}{5}\)=\(\frac{A}{t}\)
5A=32t
\(\frac{5}{5}\) A=\(\frac{32}{5}\) t
A=\(\frac{32}{5}\) t

b. Complete the table below. Use a calculator, and round answers to the tenths place.

Answer:

c. Graph the data on a coordinate plane.

Answer:

d. About how many square feet can Emily paint in 2 \(\frac{1}{2}\) minutes? Explain.
Answer:
Emily can paint between 12.8 and 19.2 square feet in 2 \(\frac{1}{2}\) minutes. After 2 minutes, she paints 12.8 square feet, and after 3 minutes, she will have painted 19.2 square feet.

Question 3.
Joseph walks at a constant speed. He walked to a store that is one-half mile away in 6 minutes. How many miles, m, can he walk in t minutes?
a. Write the linear equation in two variables that represents the number of miles Joseph can walk in any given time interval, t.
Answer:
Let C be the constant rate that Joseph walks in miles per minute. Then,
\(\frac{0.5}{6}\)=C, and \(\frac{m}{t}\)=C; therefore, \(\frac{0.5}{6}\)=\(\frac{m}{t}\).
\(\frac{0.5}{6}\)=\(\frac{m}{t}\)
6m=0.5t
\(\frac{6}{6}\) m=\(\frac{0.5}{6}\) t
m=\(\frac{0.5}{6}\) t

b. Complete the table below. Use a calculator, and round answers to the tenths place.

Answer:

c. Graph the data on a coordinate plane.

Answer:

d. Joseph’s friend lives 4 miles away from him. About how long would it take Joseph to walk to his friend’s house? Explain.
Answer:
It will take Joseph a little less than an hour to walk to his friend’s house. Since it takes 30 minutes for him to walk 2.5 miles and 60 minutes to walk 5 miles, and 4 is closer to 5 than 2.5, it will take Joseph less than an hour to walk the 4 miles.

Eureka Math Grade 8 Module 4 Lesson 11 Exit Ticket Answer Key

Vicky reads at a constant rate. She can read 5 pages in 9 minutes. We want to know how many pages, p, Vicky can read after t minutes.

a. Write a linear equation in two variables that represents the number of pages Vicky reads in any given time interval.
Answer:
Let C represent the constant rate that Vicky reads in pages per minute. Then,
\(\frac{5}{9}\)=C, and \(\frac{p}{t}\)=C; therefore, \(\frac{5}{9}\)=\(\frac{p}{t}\).
\(\frac{5}{9}\)=\(\frac{p}{t}\)
9p=5t
\(\frac{9}{9}\) p=\(\frac{5}{9}\) t
p=\(\frac{5}{9}\) t

b. Complete the table below. Use a calculator, and round answers to the tenths place.

Answer:

c. About how long would it take Vicky to read 25 pages? Explain.
Answer:
It would take her a little over 40 minutes. After 40 minutes, she can read about 22.2 pages, and after 1 hour, she can read about 33.3 pages. Since 25 pages is between 22.2 and 33.3, it will take her between 40 and 60 minutes to read 25 pages.

Eureka Math Grade 8 Module 4 Lesson 11 Problem Set Answer Key

Students practice writing two-variable equations that represent a constant rate.

Question 1.
A train travels at a constant rate of 45 miles per hour.
a. What is the distance, d, in miles, that the train travels in t hours?
Answer:
Let C be the constant rate the train travels. Then, \(\frac{45}{1}\)=C, and \(\frac{d}{t}\)=C; therefore, \(\frac{45}{1}\)=\(\frac{d}{t}\).
\(\frac{45}{1}\)=\(\frac{d}{t}\)
d=45t

b. How many miles will it travel in 2.5 hours?
Answer:
d=45(2.5)
=112.5
The train will travel 112.5 miles in 2.5 hours.

Question 2.
Water is leaking from a faucet at a constant rate of \(\frac{1}{3}\) gallons per minute.
a. What is the amount of water, w, in gallons per minute, that is leaked from the faucet after t minutes?
Answer:
Let C be the constant rate the water leaks from the faucet in gallons per minute. Then,
\(\frac{\frac{1}{3}}{1}\)=C, and \(\frac{w}{t}\)=C; therefore, \(\frac{\frac{1}{3}}{1}\) = \(\frac{w}{t}\).
\(\frac{\frac{1}{3}}{1}\)=\(\frac{w}{t}\)
w=\(\frac{1}{3}\) t

b. How much water is leaked after an hour?
Answer:
w=\(\frac{1}{3}\) t
=\(\frac{1}{3}\) (60)
=20
The faucet will leak 20 gallons in one hour.

Question 3.
A car can be assembled on an assembly line in 6 hours. Assume that the cars are assembled at a constant rate.
a. How many cars, y, can be assembled in t hours?
Answer:
Let C be the constant rate the cars are assembled in cars per hour. Then,
\(\frac{1}{6}\)=C, and \(\frac{y}{t}\)=C; therefore, \(\frac{1}{6}\)=\(\frac{y}{t}\).
\(\frac{1}{6}\)=\(\frac{y}{t}\)
6y=t
\(\frac{6}{6}\) y=\(\frac{1}{6}\)t
y=\(\frac{1}{6}\) t

b. How many cars can be assembled in a week?
Answer:
A week is 24×7=168 hours. So, y=\(\frac{1}{6}\) (168)=28. Twenty-eight cars can be assembled in a week.

Question 4.
A copy machine makes copies at a constant rate. The machine can make 80 copies in 2 \(\frac{1}{2}\) minutes.
a. Write an equation to represent the number of copies, n, that can be made over any time interval in minutes, t.
Answer:
Let C be the constant rate that copies can be made in copies per minute. Then,
\(\frac{80}{2 \frac{1}{2}}\))=C, and \(\frac{n}{t}\)=C; therefore, \(\frac{80}{2 \frac{1}{2}}\)= \(\frac{n}{t}\).
\(\frac{80}{2 \frac{1}{2}}\)=\(\frac{n}{t}\)
2 \(\frac{1}{2}\) n=80t
\(\frac{5}{2}\) n=80t
\(\frac{2}{5}\)∙\(\frac{5}{2}\) n=\(\frac{2}{5}\)∙80t
n=32t

b. Complete the table below.

Answer:

c. Graph the data on a coordinate plane.

Answer:

d. The copy machine runs for 20 seconds and then jams. About how many copies were made before the jam occurred? Explain.
Answer:
Since 20 seconds is approximately 0.3 of a minute, then the number of copies made will be between 8 and 16 because 0.3 is between 0.25 and 0.5.

Question 5.
Connor runs at a constant rate. It takes him 34 minutes to run 4 miles.
a. Write the linear equation in two variables that represents the number of miles Connor can run in any given time interval in minutes, t.
Answer:
Let C be the constant rate that Connor runs in miles per minute, and let m represent the number of miles he ran in t minutes. Then,
\(\frac{4}{34}\)=C, and \(\frac{m}{t}\)=C; therefore, \(\frac{4}{34}\)=\(\frac{m}{t}\).
\(\frac{4}{34}\)=\(\frac{m}{t}\)
34m=4t
\(\frac{34}{34}\) m=\(\frac{4}{34}\) t
m=\(\frac{4}{34}\) t
m=\(\frac{2}{17}\) t

b. Complete the table below. Use a calculator, and round answers to the tenths place.

Answer:

c. Graph the data on a coordinate plane.

Answer:

d. Connor ran for 40 minutes before tripping and spraining his ankle. About how many miles did he run before he had to stop? Explain.
Answer:
Since Connor ran for 40 minutes, he ran more than 3.5 miles but less than 5.3 miles. Since 40 is between 30 and 45, then we can use those reference points to make an estimate of how many miles he ran in 40 minutes, probably about 5 miles.

Engage NY Eureka Math 8th Grade Module 4 Lesson 9 Answer Key

Eureka Math Grade 8 Module 4 Lesson 9 Exercise Answer Key

Exercise 1.
Write the equation for the 15^th step.
Answer:
S₁₅-7+7∙5¹⁵=5S₁₅

Exercise 2.
How many people would see the photo after 15 steps? Use a calculator if needed.
Answer:
S₁₅-7+7∙5¹⁵=5S₁₅
S₁₅-5S₁₅-7++7∙5¹⁵=5S₁₅-5S₁₅
S₁₅ (1-5)-7+7∙5¹⁵=0
S₁₅ (1-5)-7+7+7∙5¹⁵-7∙5¹⁵=7-7∙5¹⁵
S₁₅ (1-5)=7(1-5¹⁵ )
S₁₅=\(\frac{7\left(1-5^{15}\right)}{(1-5)}\)
S₁₅=53 405 761 717

Exercises 3–11 as an Alternative to Discussion
Students should be able to complete the following problems independently as they are an application of skills learned to this point, namely, transcription and solving linear equations in one variable. Have students work on the problems one at a time and share their work with the whole class, or assign the entire set and allow students to work at their own pace. Provide correct solutions at the end of the lesson.

Exercise 3.
Marvin paid an entrance fee of $5 plus an additional $1.25 per game at a local arcade. Altogether, he spent $26.25. Write and solve an equation to determine how many games Marvin played.
Let x represent the number of games he played.
5+1.25x=26.25
1.25x=21.25
x=\(\frac{21.25}{1.25}\)
x=17
Marvin played 17 games.

Exercise 4.
The sum of four consecutive integers is -26. What are the integers?
Answer:
Let x be the first integer.
x+(x+1)+(x+2)+(x+3)=-26
4x+6=-26
4x=-32
x=-8
The integers are -8, -7, -6, and -5.

Exercise 5.
A book has x pages. How many pages are in the book if Maria read 45 pages of a book on Monday, \(\frac{1}{2}\) the book on Tuesday, and the remaining 72 pages on Wednesday?
Answer:
Let x be the number of pages in the book.
x=45+\(\frac{1}{2}\) x+72
x=117+\(\frac{1}{2}\) x
\(\frac{1}{2}\) x=117
x=234
The book has 234 pages.

Exercise 6.
A number increased by 5 and divided by 2 is equal to 75. What is the number?
Answer:
Let x be the number.
\(\frac{x+5}{2}\)=75
x+5=150
x=145
The number is 145.

Exercise 7.
The sum of thirteen and twice a number is seven less than six times a number. What is the number?
Answer:
Let x be the number.
13+2x=6x-7
20+2x=6x
20=4x
5=x
The number is 5.

Exercise 8.
The width of a rectangle is 7 less than twice the length. If the perimeter of the rectangle is 43.6 inches, what is the area of the rectangle?
Answer:
Let x represent the length of the rectangle.
2x+2(2x-7)=43.6
2x+4x-14=43.6
6x-14=43.6
6x=57.6
x=\(\frac{57.6}{6}\)
x=9.6
The length of the rectangle is 9.6 inches, and the width is 12.2 inches, so the area is 117.12 in².

Exercise 9.
Two hundred and fifty tickets for the school dance were sold. On Monday, 35 tickets were sold. An equal number of tickets were sold each day for the next five days. How many tickets were sold on one of those days?
Answer:
Let x be the number of tickets sold on one of those days.
250=35+5x
215=5x
43=x
43 tickets were sold on each of the five days.

Exercise 10.
Shonna skateboarded for some number of minutes on Monday. On Tuesday, she skateboarded for twice as many minutes as she did on Monday, and on Wednesday, she skateboarded for half the sum of minutes from Monday and Tuesday. Altogether, she skateboarded for a total of three hours. How many minutes did she skateboard each day?
Answer:
Let x be the number of minutes she skateboarded on Monday.
x+2x+\(\frac{2 x+x}{2}\)=180
\(\frac{2 x}{2}\)+\(\frac{4 x}{2}\)+\(\frac{2 x+x}{2}\)=180
\(\frac{9 x}{2}\)=180
9x=360
x=40
Shonna skateboarded 40 minutes on Monday, 80 minutes on Tuesday, and 60 minutes on Wednesday.

Exercise 11.
In the diagram below, △ABC ~△A^’ B^’ C^’. Determine the length of \(\overline{A C}\) and \(\overline{B C}\).

Answer:
\(\frac{18}{10.3-x}\)=\(\frac{38}{5 x-0.3}\)
18(5x-0.3)=38(10.3-x)
90x-5.4=391.4-38x
128x-5.4=391.4
128x=396.8
x=\(\frac{396.8}{128}\)
x=3.1
The length of \(\overline{A C}\) is 7.2 mm, and the length of \(\overline{B C}\) is 15.2 mm.

Eureka Math Grade 8 Module 4 Lesson 9 Exit Ticket Answer Key

Question 1.
Rewrite the equation that would represent the sum in the fifth step of the Facebook problem:
S₅=7+7∙5+7∙5²+7∙5³+7∙5⁴
S₅=7+7∙5+7∙5²+7∙5³+7∙5⁴
S₅-7=7∙5+7∙5²+7∙5³+7∙5⁴
S₅-7+7∙5⁵=7∙5+7∙5²+7∙5³+7∙5⁴+7∙5⁵
S₅-7+7∙5⁵=5(7+7∙5+7∙5²+7∙5³+7∙5⁴ )
S₅-7+7∙5⁵=5(S₅ )
S₅-5S₅-7+7∙5⁵=0
S₅-5S₅=7-(7∙5⁵ )
(1-5) S₅=7-(7∙5⁵ )
(1-5) S₅=7(1-5⁵ )
S₅=\(\frac{7\left(1-5^{5}\right)}{1-5}\)

Question 2.
The sum of four consecutive integers is 74. Write an equation, and solve to find the numbers.
Answer:
Let x be the first number.
x+(x+1)+(x+2)+(x+3)=74
4x+6=74
4x=68
x=17
The numbers are 17, 18, 19, and 20.

Eureka Math Grade 8 Module 4 Lesson 9 Problem Set Answer Key

Assign the problems that relate to the elements of the lesson that were used with students.

Question 1.
You forward an e-card that you found online to three of your friends. They liked it so much that they forwarded it on to four of their friends, who then forwarded it on to four of their friends, and so on. The number of people who saw the e-card is shown below. Let S₅ represent the number of people who saw the e-card after one step, let S₂ represent the number of people who saw the e-card after two steps, and so on.
S₅=3
S₂=3+3∙4
S₃=3+3∙4+3∙4²
S₄=3+3∙4+3∙4²+3∙4³

a. Find the pattern in the equations.
Answer:
S₂=3+3∙4
S₂-3=3∙4
S₂-3+3∙4²=3∙4+3∙4²
S₂-3+3∙4²=4(3+3∙4)
S₂-3+3∙4²=4S₂

S₃=3+3∙4+3∙4²
S₃-3=3∙4+3∙4²
S₃-3+3∙4³=3∙4+3∙4²+3∙4³
S₃-3+3∙4³=4(3+3∙4+3∙4² )
S₃-3+3∙4³=4S₃

S₄=3+3∙4+3∙4²+3∙4³
S₄-3=3∙4+3∙4²+3∙4³
S₄-3+3∙4⁴=3∙4+3∙4²+3∙4³+3∙4⁴
S₄-3+3∙4⁴=4(3+3∙4+3∙4²+3∙4³ )
S₄-3+3∙4⁴=4S₄

b. Assuming the trend continues, how many people will have seen the e-card after 10 steps?
Answer:
S₅0-3+3∙4¹⁰=4S₁₀
S₁₀-4S₁₀-3+3∙4¹⁰=0
S₁₀(1-4)=3-3∙4¹⁰
S₁₀(1-4)=3(1-4¹⁰)
S₁₀=\(\frac{3\left(1-4^{10}\right)}{(1-4)}\)
S₁₀=1 048 575
After 10 steps, 1,048,575 people will have seen the e-card.

c. How many people will have seen the e-card after n steps?
Answer:
S_n=\(\frac{3\left(1-4^{n}\right)}{(1-4)}\)

For each of the following questions, write an equation, and solve to find each answer.

Question 2.
Lisa has a certain amount of money. She spent $39 and has \(\frac{3}{4}\) of the original amount left. How much money did she have originally?
Answer:
Let x be the amount of money Lisa had originally.
x-39=\(\frac{3}{4}\) x
-39=-\(\frac{1}{4}\) x
156=x
Lisa had $156 originally.

Question 3.
The length of a rectangle is 4 more than 3 times the width. If the perimeter of the rectangle is 18.4 cm, what is the area of the rectangle?
Answer:
Let x represent the width of the rectangle.
2(4+3x)+2x=18.4
8+6x+2x=18.4
8+8x=18.4
8x=10.4
x=\(\frac{10.4}{8}\)
x=1.3
The width of the rectangle is 1.3 cm, and the length is 7.9 cm, so the area is 10.27 cm².

Question 4.
Eight times the result of subtracting 3 from a number is equal to the number increased by 25. What is the number?
Answer:
Let x be the number.
8(x-3)=x+25
8x-24=x+25
7x-24=25
7x=49
x=7
The number is 7.

Question 5.
Three consecutive odd integers have a sum of 3. What are the numbers?
Answer:
Let x be the first odd number.
x+(x+2)+(x+4)=3
3x+6=3
3x=-3
x=-1
The three consecutive odd integers are -1, 1, and 3.

Question 6.
Each month, Liz pays $35 to her phone company just to use the phone. Each text she sends costs her an additional $0.05. In March, her phone bill was $72.60. In April, her phone bill was $65.85. How many texts did she send each month?
Answer:
Let x be the number of texts she sent in March.
35+0.05x=72.60
0.05x=37.6
x=\(\frac{37.6}{0.05}\)
x=752
She sent 752 texts in March.
Let y be the number of texts she sent in April.
35+0.05y=65.85
0.05y=30.85
y=\(\frac{30.85}{0.05}\)
y=617
She sent 617 texts in April.

Question 7.
Claudia is reading a book that has 360 pages. She read some of the book last week. She plans to read 46 pages today. When she does, she will be \(\frac{4}{5}\) of the way through the book. How many pages did she read last week?
Answer:
Let x be the number of pages she read last week.
x+46=\(\frac{4}{5}\) (360)
x+46=288
x=242
Claudia read 242 pages last week.

Question 8.
In the diagram below, △ABC ~△A^’ B^’ C^’. Determine the measure of ∠A.

Answer:
7x-4=x+32
6x-4=32
6x=36
x=6
The measure of ∠A is 38°.

Question 9.
In the diagram below, △ABC ~△A^’ B^’ C^’. Determine the measure of ∠A.

Answer:
10x-9=4x+57
6x-9=57
6x=66
x=11
The measure of ∠A is 101°.

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

Eureka Math Grade 8 Module 4 Lesson 8 Example Answer Key

Example 1.
→ Given a linear equation in disguise, we will try to solve it. To do so, we must first assume that the following equation is true for some number x.
\(\frac{x-1}{2}\) = \(\frac{x+\frac{1}{3}}{4}\)
We want to make this equation look like the linear equations we are used to. For that reason, we will multiply both sides of the equation by 2 and 4, as we normally do with proportions:
2(x+\(\frac{1}{3}\))=4(x-1).
→ Is this a linear equation? How do you know?
→ Yes, this is a linear equation because the expressions on the left and right of the equal sign are linear expressions.
→ Notice that the expressions that contained more than one term were put in parentheses. We do that so we do not make a mistake and forget to use the distributive property.
→ Now that we have a linear equation, we will use the distributive property and solve as usual.
2(x+\(\frac{1}{3}\))=4(x-1)
2x+\(\frac{2}{3}\)=4x-4
2x-2x+\(\frac{2}{3}\)=4x-2x-4
\(\frac{2}{3}\)=2x-4
\(\frac{2}{3}\)+4=2x-4+4
\(\frac{14}{3}\)=2x
\(\frac{1}{2}\)∙\(\frac{14}{3}\)=\(\frac{1}{2}\)∙2x
\(\frac{7}{3}\)=x
→ How can we verify that \(\frac{7}{3}\) is the solution to the equation?
→ We can replace x with \(\frac{7}{3}\) in the original equation.

Since \(\frac{7}{3}\) made the equation true, we know it is a solution to the equation.

Example 2.
→ Can we solve the following equation? Explain.
\(\frac{\frac{1}{5}-x}{7}\) = \(\frac{2 x+9}{3}\)
→ We need to multiply each numerator with the other fraction’s denominator.
→ So,
\(\frac{\frac{1}{5}-x}{7}\) = \(\frac{2 x+9}{3}\)
7(2x+9)=3(\(\frac{1}{5}\)-x).
→ What would be the next step?
→ Use the distributive property.
→ Now we have
7(2x+9)=3(\(\frac{1}{5}\)-x)
14x+63=\(\frac{3}{5}\)-3x
14x+3x+63=\(\frac{3}{5}\)-3x+3x
17x+63=\(\frac{3}{5}\)
17x+63-63=\(\frac{3}{5}\)-63
17x=\(\frac{3}{5}\)–\(\frac{315}{5}\)
17x=-\(\frac{312}{5}\)
\(\frac{1}{17}\) (17x)=(-\(\frac{312}{5}\))\(\frac{1}{17}\)
x=-\(\frac{312}{85}\).
→ Is this a linear equation? How do you know?
→ Yes, this is a linear equation because the left and right side are linear expressions.

Example 3.
Can this equation be solved?
\(\frac{6+x}{7 x+\frac{2}{3}}\)=\(\frac{3}{8}\)
Give students a few minutes to work. Provide support to students as needed.
→ Yes, we can solve the equation because we can multiply each numerator with the other fraction’s denominator and then use the distributive property to begin solving it.
Answer:
\(\frac{6+x}{7 x+\frac{2}{3}}\)=\(\frac{3}{8}\)
(6+x)8=(7x+\(\frac{2}{3}\))3
48+8x=21x+2
48+8x-8x=21x-8x+2
48=13x+2
48-2=13x+2-2
46=13x
\(\frac{46}{13}\)=x

Example 4.
Can this equation be solved?
\(\frac{7}{3 x+9}\)=\(\frac{1}{8}\)
Give students a few minutes to work. Provide support to students as needed.
→ Yes, we can solve the equation because we can multiply each numerator with the other fraction’s denominator and then use the distributive property to begin solving it.
Answer:
\(\frac{7}{3 x+9}\)=\(\frac{1}{8}\)
7(8)=(3x+9)1
56=3x+9
56-9=3x+9-9
47=3x
\(\frac{47}{3}\)=x

Example 5.
In the diagram below, △ABC~ △A’ B’ C’. Using what we know about similar triangles, we can determine the value of x.

→ Begin by writing the ratios that represent the corresponding sides.
Answer:
\(\frac{x-2}{9.5}\) = \(\frac{x+2}{12}\)

It is possible to write several different proportions in this case. If time, discuss this fact with students.
→ Now that we have the ratios, solve for x and find the lengths of \(\overline{A B}\) and \(\overline{A C}\).
Answer:
\(\frac{x-2}{9.5}\) = \(\frac{x+2}{12}\)
(x-2)12=9.5(x+2)
12x-24=9.5x+19
12x-24+24=9.5x+19+24
12x=9.5x+43
12x-9.5x=9.5x-9.5x+43
2.5x=43
x=\(\frac{43}{2.5}\)
x=17.2
|AB|=15.2 cm, and |AC|=19.2 cm.

Eureka Math Grade 8 Module 4 Lesson 8 Exercise Answer Key

Solve the following equations of rational expressions, if possible.

Question 1.
\(\frac{2 x+1}{9}\)=\(\frac{1-x}{6}\)
Answer:
\(\frac{2 x+1}{9}\)=\(\frac{1-x}{6}\)
9(1-x)=(2x+1)6
9-9x=12x+6
9-9x+9x=12x+9x+6
9=21x+6
9-6=21x+6-6
3=21x
\(\frac{3}{21}\)=\(\frac{21}{2}\)1 x
\(\frac{1}{7}\)=x

Question 2.
\(\frac{5+2 x}{3 x-1}\)=\(\frac{6}{7}\)
Answer:
\(\frac{5+2 x}{3 x-1}\)=\(\frac{6}{7}\)
(5+2x)7=(3x-1)6
35+14x=18x-6
35-35+14x=18x-6-35
14x=18x-41
14x-18x=18x-18x-41
-4x=-41
\(\frac{-4}{-4}\) x=\(\frac{-41}{-4}\)
x=\(\frac{41}{4}\)

Question 3.
\(\frac{x+9}{12}\)=\(\frac{-2 x-\frac{1}{2}}{3}\)
Answer:
\(\frac{x+9}{12}\)=\(\frac{-2 x-\frac{1}{2}}{3}\)
12(-2x-\(\frac{1}{2}\))=(x+9)3
-24x-6=3x+27
-24x+24x-6=3x+24x+27
-6=27x+27
-6-27=27x+27-27
-33=27x
\(\frac{-33}{27}\)=\(\frac{27}{27}\) x
–\(\frac{11}{9}\)=x

Question 8.
\(\frac{8}{3-4 x}\) = \(\frac{5}{2 x+\frac{1}{4}}\)
Answer:
\(\frac{8}{3-4 x}\) = \(\frac{5}{2 x+\frac{1}{4}}\)
8(2x+\(\frac{1}{4}\))=(3-4x)5
16x+2=15-20x
16x+2-2=15-2-20x
16x=13-20x
16x+20x=13-20x+20x
36x=13
\(\frac{36}{36}\) x=\(\frac{13}{36}\)
x=\(\frac{13}{36}\)

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

Students practice solving equations with rational expressions, if a solution is possible.

Solve the following equations of rational expressions, if possible. If an equation cannot be solved, explain why.

Question 1.
\(\frac{5}{6 x-2}\) = \(\frac{-1}{x+1}\)
Answer:
\(\frac{5}{6 x-2}\) = \(\frac{-1}{x+1}\)
5(x+1)=-1(6x-2)
5x+5=-6x+2
5x+5-5=-6x+2-5
5x=-6x-3
5x+6x=-6x+6x-3
11x=-3
x=-\(\frac{3}{11}\)

Question 2.
\(\frac{4-x}{8}\) = \(\frac{7 x-1}{3}\)
Answer:
\(\frac{4-x}{8}\) = \(\frac{7 x-1}{3}\)
8(7x-1)=(4-x)3
56x-8=12-3x
56x-8+8=12+8-3x
56x=20-3x
56x+3x=20-3x+3x
59x=20
\(\frac{59}{59}\) x=\(\frac{20}{59}\)
x=\(\frac{20}{59}\)

Question 3.
\(\frac{3 x}{x+2}\) = \(\frac{5}{9}\)
Answer:
\(\frac{3 x}{x+2}\) = \(\frac{5}{9}\)
9(3x)=(x+2)5
27x=5x+10
27x-5x=5x-5x+10
22x=10
\(\frac{22}{22}\) x=\(\frac{10}{22}\)
x=\(\frac{5}{11}\)

Question 4.
\(\frac{\frac{1}{2} x+6}{3}\) = \(\frac{x-3}{2}\)
Answer:
\(\frac{\frac{1}{2} x+6}{3}\) = \(\frac{x-3}{2}\)
3(x-3)=2(\(\frac{1}{2}\) x+6)
3x-9=x+12
3x-9+9=x+12+9
3x=x+21
3x-x=x-x+21
2x=21
x=\(\frac{21}{2}\)

Question 5.
\(\frac{7-2 x}{6}\) = \(\frac{x-5}{1}\)
Answer:
\(\frac{7-2 x}{6}\) = \(\frac{x-5}{1}\)
6(x-5)=(7-2x)1
6x-30=7-2x
6x-30+30=7+30-2x
6x=37-2x
6x+2x=37-2x+2x
8x=37
\(\frac{8}{8}\) x=\(\frac{37}{8}\)
x=\(\frac{37}{8}\)

Question 6.
\(\frac{2 x+5}{2}\) = \(\frac{3 x-2}{6}\)
Answer:
\(\frac{2 x+5}{2}\) = \(\frac{3 x-2}{6}\)
2(3x-2)=6(2x+5)
6x-4=12x+30
6x-4+4=12x+30+4
6x=12x+34
6x-12x=12x-12x+34
-6x=34
x=-\(\frac{34}{6}\)
x=-\(\frac{17}{3}\)

Question 7.
\(\frac{6 x+1}{3}\) = \(\frac{9-x}{7}\)
Answer:
\(\frac{6 x+1}{3}\) = \(\frac{9-x}{7}\)
(6x+1)7=3(9-x)
42x+7=27-3x
42x+7-7=27-7-3x
42x=20-3x
42x+3x=20-3x+3x
45x=20
\(\frac{45}{45}\)x=\(\frac{20}{45}\)
x=\(\frac{4}{9}\)

Question 8.
\(\frac{\frac{1}{3} x-8}{12}\) = \(\frac{-2-x}{15}\)
Answer:
\(\frac{\frac{1}{3} x-8}{12}\) = \(\frac{-2-x}{15}\)
12(-2-x)=(\(\frac{1}{3}\) x-8)15
-24-12x=5x-120
-24-12x+12x=5x+12x-120
-24=17x-120
-24+120=17x-120+120
96=17x
\(\frac{96}{17}\)=\(\frac{17}{17}\) x
\(\frac{96}{17}\)=x

Question 9.
\(\frac{3-x}{1-x}\)=\(\frac{3}{2}\)
Answer:
\(\frac{3-x}{1-x}\)=\(\frac{3}{2}\)
(3-x)2=(1-x)3
6-2x=3-3x
6-2x+2x=3-3x+2x
6=3-x
6-3=3-3-x
3=-x
-3=x

Question 10.
In the diagram below, △ABC~ △A’ B’ C’. Determine the lengths of \(\overline{A C}\) and \(\overline{B C}\).

Answer:
\(\frac{x+4}{4.5}\) = \(\frac{3 x-2}{9}\)
9(x+4)=4.5(3x-2)
9x+36=13.5x-9
9x+36+9=13.5x-9+9
9x+45=13.5x
9x-9x+45=13.5x-9x
45=4.5x
10=x
|AC|=14 cm, and |BC|=28 cm.

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

Solve the following equations for x.

Question 1.
\(\frac{5 x-8}{3}\) = \(\frac{11 x-9}{5}\)
Answer:
\(\frac{5 x-8}{3}\) = \(\frac{11 x-9}{5}\)
5(5x-8)=3(11x-9)
25x-40=33x-27
25x-25x-40=33x-25x-27
-40=8x-27
-40+27=8x-27+27
-13=8x
–\(\frac{13}{8}\)=x

Question 2.
\(\frac{x+11}{7}\)=\(\frac{2 x+1}{-8}\)
Answer:
\(\frac{x+11}{7}\)=\(\frac{2 x+1}{-8}\)
7(2x+1)=-8(x+11)
14x+7=-8x-88
14x+7-7=-8x-88-7
14x=-8x-95
14x+8x=-8x+8x-95
22x=-95
x=-\(\frac{95}{22}\)

Question 3.
\(\frac{-x-2}{-4}\) = \(\frac{3 x+6}{2}\)
Answer:
\(\frac{-x-2}{-4}\) = \(\frac{3 x+6}{2}\)
-4(3x+6)=2(-x-2)
-12x-24=-2x-4
-12x-24+24=-2x-4+24
-12x=-2x+20
-12x+2x=-2x+2x+20
-10x=20
x=-2

Engage NY Eureka Math 8th Grade Module 4 Mid Module Assessment Answer Key

Eureka Math Grade 8 Module 4 Mid Module Assessment Task Answer Key

Question 1.
Write and solve each of the following linear equations.
a. Ofelia has a certain amount of money. If she spends $12, then she has \(\frac{1}{5}\) of the original amount left. How much money did Ofelia have originally?
Answer:
Let x be the amount of money ofelia had
x – 12 = \(\frac{1}{5}\)
x – \(\frac{1}{5}\)x – 12 + 12 = \(\frac{1}{5}\) x – \(\frac{1}{5}\) x +12
\(\frac{4}{5}\) x = 12
x= 12 ∙ \(\frac{5}{4}\) = \(\frac{60}{4}\)
Ofelia had $15.00 originally.

b. Three consecutive integers have a sum of 234. What are the three integers?
Answer:
Let x be the first integer
x + x + 1 + x + 2 = 234
3x + 3 = 234
3x = 234 – 3
3x = 231
x = 77
The integers are 77, 78, and 79

c. Gil is reading a book that has 276 pages. He already read some of it last week. He plans to read 20 pages tomorrow. By then, he will be \(\frac{2}{3}\)of the way through the book. How many pages did Gil read last week?
Answer:
Let x be the number of pages gil read last week.
x + 20 = \(\frac{2}{3}\)(276)
x + 20 = 184
x + 20 – 20 = 184 – 20
x = 164
Gil read 164 pages last week

Question 2.
a. Without solving, identify whether each of the following equations has a unique solution, no solution,
or infinitely many solutions.
i. 3x + 5 = – 2
Answer:
Unique

ii. 6(x – 11) = 15 – 4x
Answer:
Unique

iii. 12x + 9 = 8x + 1 + 4x
Answer:
No solution

iv. 2(x – 3) = 10x – 6 – 8x
Answer:
Infinitely many solutions

v. 5x + 6 = 5x – 4
Answer:
No solution

b. Solve the following equation for a number x. Verify that your solution is correct.
– 15 = 8x + 1
Answer:

-2 = x

-15 = 8(-2) + 1
-15 = -16 + 1
-15 = -15

c. Solve the following equation for a number x. Verify that your solution is correct.
7(2x + 5) = 4x – 9 – x
Answer:
7(2x + 5) = 4x – 9 – x
14x + 35 = 4x – 9 – x
14x + 35 = 3x – 9
14x – 3x + 35 = 3x – 3x – 9
11x + 35 = -9
11x + 35 – 35 = -9 – 35
11x = -44
x = -4

7(2(-4) + 5) = 4(-4) – 9 – (-4)
7(-8 + 5) = -16 – 9 + 4
7(-3) = -25 + 4
-21 = -21

Question 3.
a. Parker paid $4.50 for three pounds of gummy candy. Assuming each pound of gummy candy costs the same amount, complete the table of values representing the cost of gummy candy in pounds.

Answer:

b. Graph the data on the coordinate plane.

Answer:

c. On the same day, Parker’s friend, Peggy, was charged $5 for 1 \(\frac{1}{2}\) lb. of gummy candy. Explain in terms of the graph why this must be a mistake.
Answer:
Even though 1\(\frac{1}{2}\) pounds of candy isn’t a point on the graph, it is reasonable to believe it will fall in line with the other points. The cost of 1\(\frac{1}{2}\) pounds of candy does not fit the pattern.

Engage NY Eureka Math 8th Grade Module 4 Lesson 19 Answer Key

Eureka Math Grade 8 Module 4 Lesson 19 Exercise Answer Key

Exercises 1 – 11
THEOREM: The graph of a linear equation y = mx + b is a non – vertical line with slope m and passing through (0, b), where b is a constant.

Exercise 1.
Prove the theorem by completing parts (a)–(c). Given two distinct points, P and Q, on the graph of y = mx + b, and let l be the line passing through P and Q. You must show the following:
(1) Any point on the graph of y = mx + b is on line l, and
(2) Any point on the line l is on the graph of y = mx + b.
a. Proof of (1): Let R be any point on the graph of y = mx + b. Show that R is on l. Begin by assuming it is not. Assume the graph looks like the diagram below where R is on l’.

What is the slope of line l?
Answer:
Since the points P and Q are on the graph of y = mx + b, then we know that the slope of the line passing through those points must have slope m. Therefore, line l has slope m.

What is the slope of line l’?
Answer:
We know that point R is on the graph of y = mx + b. Then the coordinates of point R are (r₁, mr₁ + b) because R is a solution to y = mx + b, and r₂ = mr₁ + b. Similarly, the coordinates of P are (p₁, mp₁ + b).
Answer:

What can you conclude about lines l and l’? Explain.
Answer:
Lines l and l’are the same line. Both lines go through point P and have slope m. There can be only one line with a given slope going through a given point; therefore, line l is the same as l’.

b. Proof of (2): Let S be any point on line l, as shown.

Show that S is a solution to y = mx + b. Hint: Use the point (0, b).
Answer:
Point S is on line l. Let S = (s₁, s₂).
slope of l = \(\frac{s_{2} – b}{s_{1} – 0}\)
We know the slope of l is m.
m = \(\frac{s_{2} – b}{s_{1} – 0}\)
m(s₁ – 0) = s₂ – b
ms₁ = s₂ – b
ms₁ + b = s₂ – b + b
ms₁ + b = s₂
s₂ = ms₁ + b,
which shows S is a solution to y = mx + b.

c. Now that you have shown that any point on the graph of y = mx + b is on line l in part (a), and any point on line l is on the graph of y = mx + b in part (b), what can you conclude about the graphs of linear equations?
Answer:
The graph of a linear equation is a line.

Exercise 2.
Use x = 4 and x = – 4 to find two solutions to the equation x + 2y = 6. Plot the solutions as points on the coordinate plane, and connect the points to make a line.
Answer:
The solutions are (4, 1) and ( – 4, 5).

a. Identify two other points on the line with integer coordinates. Verify that they are solutions to the equation x + 2y = 6.
Answer:
The choice of points and verifications will vary. Several possibilities are noted in the graph above.

b. When x = 1, what is the value of y? Does this solution appear to be a point on the line?
Answer:
1 + 2y = 6
2y = 5
y = \(\frac{5}{2}\)
Yes, (1, \(\frac{5}{2}\)) does appear to be a point on the line.

c. When x = – 3, what is the value of y? Does this solution appear to be a point on the line?
Answer:
– 3 + 2y = 6
2y = 9
y = \(\frac{9}{2}\)
Yes, ( – 3, \(\frac{9}{2}\)) does appear to be a point on the line.

d. Is the point (3, 2) on the line?
Answer:
No, (3, 2) is not a point on the line.

e. Is the point (3, 2) a solution to the linear equation x + 2y = 6?
Answer:
No, (3, 2) is not a solution to x + 2y = 6.
3 + 2(2) = 6
3 + 4 = 6
7≠6

Exercise 3.
Use x = 4 and x = 1 to find two solutions to the equation 3x – y = 9. Plot the solutions as points on the coordinate plane, and connect the points to make a line.
Answer:
The solutions are (4, 3) and (1, – 6).

a. Identify two other points on the line with integer coordinates. Verify that they are solutions to the equation 3x – y = 9.
Answer:
The choice of points and verifications will vary. Several possibilities are noted in the graph above.

b. When x = 4.5, what is the value of y? Does this solution appear to be a point on the line?
Answer:
3(4.5) – y = 9
13.5 – y = 9
– y = – 4.5
y = 4.5
Yes, (4.5, 4.5) does appear to be a point on the line.

c. When x = \(\frac{1}{2}\), what is the value of y? Does this solution appear to be a point on the line?
Answer:
3(\(\frac{1}{2}\)) – y = 9
\(\frac{3}{2}\) – y = 9
– y = \(\frac{15}{2}\)
y = – \(\frac{15}{2}\)
Yes, (\(\frac{1}{2}\), – \(\frac{15}{2}\)) does appear to be a point on the line.

d. Is the point (2, 4) on the line?
Answer:
No, (2, 4) is not a point on the line.

e. Is the point (2, 4) a solution to the linear equation 3x – y = 9?
Answer:
No, (2, 4) is not a solution to 3x – y = 9.
3(2) – 4 = 9
6 – 4 = 9
2≠9

Exercise 4.
Use x = 3 and x = – 3 to find two solutions to the equation 2x + 3y = 12. Plot the solutions as points on the coordinate plane, and connect the points to make a line.
Answer:
The solutions are ( – 3, 6) and (3, 2).

a. Identify two other points on the line with integer coordinates. Verify that they are solutions to the equation 2x + 3y = 12.
Answer:
The choice of points and verifications will vary. Several possibilities are noted in the graph above.

b. When x = 2, what is the value of y? Does this solution appear to be a point on the line?
Answer:
2(2) + 3y = 12
4 + 3y = 12
3y = 8
y = \(\frac{8}{3}\)
Yes, (2, \(\frac{8}{3}\)) does appear to be a point on the line.

c. When x = – 2, what is the value of y? Does this solution appear to be a point on the line?
Answer:
2( – 2) + 3y = 12
– 4 + 3y = 12
3y = 16
y = \(\frac{16}{3}\)
Yes, ( – 2, \(\frac{16}{3}\)) does appear to be a point on the line.

d. Is the point (8, – 3) on the line?
Answer:
No, (8, – 3) is not a point on the line.

e. Is the point (8, – 3) a solution to the linear equation 2x + 3y = 12?
Answer:
No, (8, – 3) is not a solution to 2x + 3y = 12.
2(8) + 3( – 3) = 12
16 – 9 = 12
7≠12

Exercise 5.
Use x = 4 and x = – 4 to find two solutions to the equation x – 2y = 8. Plot the solutions as points on the coordinate plane, and connect the points to make a line.
Answer:
The solutions are (4, – 2) and ( – 4, – 6).

a. Identify two other points on the line with integer coordinates. Verify that they are solutions to the equation x – 2y = 8.
Answer:
The choice of points and verifications will vary. Several possibilities are noted in the graph above.

b. When x = 7, what is the value of y? Does this solution appear to be a point on the line?
Answer:
7 – 2y = 8
– 2y = 1
y = – \(\frac{1}{2}\)
Yes, (7, – \(\frac{1}{2}\)) does appear to be a point on the line.

c. When x = – 3, what is the value of y? Does this solution appear to be a point on the line?
Answer:
– 3 – 2y = 8
– 2y = 11
y = – \(\frac{11}{2}\)
Yes, ( – 3, – \(\frac{11}{2}\)) does appear to be a point on the line.

d. Is the point ( – 2, – 3) on the line?
Answer:
No, ( – 2, – 3) is not a point on the line.

e. Is the point ( – 2, – 3) a solution to the linear equation x – 2y = 8?
Answer:
No, ( – 2, – 3) is not a solution to x – 2y = 8.
– 2 – 2( – 3) = 8
– 2 + 6 = 8
4≠8

Exercise 6.
Based on your work in Exercises 2–5, what conclusions can you draw about the points on a line and solutions to a linear equation?
Answer:
It appears that all points on the line represent a solution to the equation. In other words, any point identified on the line is a solution to the linear equation.

Exercise 7.
Based on your work in Exercises 2–5, will a point that is not a solution to a linear equation be a point on the graph of a linear equation? Explain.
Answer:
No. Each time we were given a point off the line in part (d), we verified that it was not a solution to the equation in part (e). For that reason, I would expect that all points not on the line would not be a solution to the equation.

Exercise 8.
Based on your work in Exercises 2–5, what conclusions can you draw about the graph of a linear equation?
Answer:
The graph of a linear equation is a line.

Exercise 9.
Graph the equation – 3x + 8y = 24 using intercepts.
Answer:
– 3x + 8y = 24
– 3(0) + 8y = 24
8y = 24
y = 3
The y – intercept point is (0, 3).
– 3x + 8y = 24
– 3x + 8(0) = 24
– 3x = 24
x = – 8
The x – intercept point is ( – 8, 0).

Exercise 10.
Graph the equation x – 6y = 15 using intercepts.
Answer:
x – 6y = 15
0 – 6y = 15
– 6y = 15
y = – \(\frac{15}{6}\)
y = – \(\frac{5}{2}\)
The y – intercept point is (0, – \(\frac{5}{2}\)).
x – 6y = 15
x – 6(0) = 15
x = 15
The x – intercept point is (15, 0).

Exercise 11.
Graph the equation 4x + 3y = 21 using intercepts.
Answer:
4x + 3y = 21
4(0) + 3y = 21
3y = 21
y = 7
The y – intercept point is (0, 7).
4x + 3y = 21
4x + 3(0) = 21
4x = 21
x = 21/4
The x – intercept point is (21/4, 0).

Eureka Math Grade 8 Module 4 Lesson 19 Problem Set Answer Key

Question 1.
Graph the equation: y = – 6x + 12.
Answer:

Question 2.
Graph the equation: 9x + 3y = 18.
Answer:
9(0) + 3y = 18
3y = 18
y = 6
The y – intercept point is (0, 6).
9x + 3(0) = 18
9x = 18
x = 2
The x – intercept point is (2, 0).

Question 3.
Graph the equation: y = 4x + 2.
Answer:

Question 4.
Graph the equation: y = – \(\frac{5}{7}\) x + 4.
Answer:

Question 5.
Graph the equation: \(\frac{3}{4}\) x + y = 8.
Answer:
\(\frac{3}{4}\) (0) + y = 8
y = 8
The y – intercept point is (0, 8).
\(\frac{3}{4}\) x + 0 = 8
\(\frac{3}{4}\) x = 8
x = \(\frac{32}{3}\)
The x – intercept point is (\(\frac{32}{3}\), 0).

Question 6.
Graph the equation: 2x – 4y = 12.
Answer:
2(0) – 4y = 12
– 4y = 12
y = – 3
The y – intercept point is (0, – 3).
2x – 4(0) = 12
2x = 12
x = 6
The x – intercept point is (6, 0).

Question 7.
Graph the equation: y = 3. What is the slope of the graph of this line?
Answer:
The slope of this line is zero.

Question 8.
Graph the equation: x = – 4. What is the slope of the graph of this line?
Answer:
The slope of this line is undefined.

Question 9.
Is the graph of 4x + 5y = \(\frac{3}{7}\) a line? Explain.
Answer:
Yes, the graph of 4x + 5y = \(\frac{3}{7}\) is a line because it is a linear equation comprising linear expressions on both sides of the equal sign.

Question 10.
Is the graph of 6x² – 2y = 7 a line? Explain.
Answer:
Maybe. The equation 6x² – 2y = 7 is not a linear equation because the expression on the left side of the equal sign is not a linear expression. If this were a linear equation, then I would be sure that it graphs as a line, but because it is not, I am not sure what the graph of this equation would look like.

Eureka Math Grade 8 Module 4 Lesson 19 Exit Ticket Answer Key

Question 1.
Graph the equation y = \(\frac{5}{4}\) x – 10 using the y – intercept point and slope.

Answer:

Question 2.
Graph the equation 5x – 4y = 40 using intercepts.

Answer:

Question 3.
What can you conclude about the equations y = \(\frac{5}{4}\) x – 10 and 5x – 4y = 40?
Answer:
Since the points (0, – 10), (4, – 5), and (8, 0) are common to both graphs, then the lines must be the same. There is only one line that can pass through two points. If you transform the equation y = \(\frac{5}{4}\) x – 10 so that it is in standard form, it is the equation 5x – 4y = 40.

Engage NY Eureka Math 8th Grade Module 4 End of Module Assessment Answer Key

Eureka Math Grade 8 Module 4 End of Module Assessment Task Answer Key

Question 1.
Use the graph below to answer parts (a)–(c).

a. Use any pair of points to calculate the slope of the line.
Answer:
m = \(\frac{6-3}{0-2}=\frac{3}{-2}\) = –\(\frac{3}{2}\)

b. Use a different pair of points to calculate the slope of the line.
Answer:
m = \(\frac{6-0}{0-4}=\frac{6}{-4}\) = –\(\frac{3}{2}\)

c. Explain why the slopes you calculated in parts (a) and (b) are equal.
Answer:
The slopes are equal because the slope triangle are similar, ∆ABC ~ ∆AB’C’. Each triangle has 90° angle at ∠ABC & ∠AB’C’, respectively. They are 90° because they are at intersection of the grid lines. Both triangles share ∠BAC. By the AA criterion ∆ABC ~ ∆AB’C’, which means their corresponding sides are equal in ratio.
\(\frac{\left|B^{\prime} C^{\prime}\right|}{|B C|}=\frac{\left|A B^{\prime}\right|}{|A B|}\) which is the same as \(\frac{\left|B^{\prime} C^{\prime}\right|}{\left|A B^{\prime}\right|}=\frac{|B C|}{|A B|}\) where –\(\frac{\left|B^{\prime} C^{\prime}\right|}{\left|A B^{\prime}\right|}\) is the slope in (b) and –\(\frac{|B C|}{|A B|}\) is the slope in (a).

Question 2.
Jeremy rides his bike at a rate of 12 miles per hour. Below is a table that represents the number of hours and miles Kevin rides. Assume both bikers ride at a constant rate.

a. Which biker rides at a greater speed? Explain your reasoning.
Answer:
Let Y be the distance triangled and X be the number of hours,
Then for jeremy, \(\frac{y}{x}\) = \(\frac{12}{1}\) ⇒ 12x
For kevin, \(\frac{46-23}{4-2}\) = \(\frac{23}{2}\) = 11.5, then y = 11.5x
When you compare their rates, 12 > 11.5, therefore jeremy rides at a greater speed.
Graphically:

b. Write an equation for a third biker, Lauren, who rides twice as fast as Kevin. Use y to represent the number of miles Lauren travels in x hours. Explain your reasoning.
Answer:
“Twice as Fast” means lauren goes twice the distance in the same time. Then in 2 hours she rides 46 miles and in 4 hours, 92 miles. If y is the total distance in x hours, y = \(\frac{46}{2}\) x
y = 23x

c. Create a graph of the equation in part (b).

Answer:

d. Calculate the slope of the line in part (c), and interpret its meaning in this situation.
Answer:
m = \(\frac{46-23}{2-1}\) = \(\frac{23}{1}\)
The slope is the rate that lauren rides, 23 miles per hour.

Question 3.
The cost of five protractors is $14.95 at Store A. The graph below compares the cost of protractors at Store A with the cost at Store B.

Estimate the cost of one protractor at Store B. Use evidence from the graph to justify your answer.
Answer:
The cost of protractors at store B is probably about $2.99 per protractor and it looks like the slope for store B is about half of the slope for store A.

Question 4.
Given the equation 3x+9y=-8, write a second linear equation to create a system that:
a. Has exactly one solution. Explain your reasoning.
Answer:
4x + 9y = -10
This equation has a slope different from 3x + 9y = -8. So the graphs of the equations will intersect.

b. Has no solution. Explain your reasoning.
Answer:
x + 3y = 10
This equation has the same slopes as 3x + 9y = -8, And no common points (solutions) Therefore the graphs of the equation are parallel lines.

c. Has infinitely many solutions. Explain your reasoning.
Answer:
6x + 18y = -16
This equation defines the same line as 3x + 9y = -8 and the graphs of the equations will coincide.

d. Interpret the meaning of the solution, if it exists, in the context of the graph of the following system of equations.
-5x+2y=10
10x-4y=-20
Answer:
-5x+2y=10 m = \(\frac{5}{2}\) (0, 5)
10x-4y=-20 m = \(\frac{5}{2}\) (0, 5)
This system will have infinitely many solutions because the graphs of these linear equations are the same line. Each equation has a slope of m = \(\frac{5}{2}\) and a y-intercept at (0, 5). There exists only one line through a point and a given slope. Therefore this system graphs as the same line and has infinitely many solutions.

Question 5.
Students sold 275 tickets for a fundraiser at school. Some tickets are for children and cost $3, while the rest are adult tickets that cost $5. If the total value of all tickets sold was $1,025, how many of each type of ticket was sold?
Answer:
Let X be the # of kids tickets
Let Y be the # of adults tickets
x + y = 275
3x + 5y = 1025

y = 100
x + 100 = 275
x = 175
(175, 100)
175 children’s tickets and 100 adults tickets were sold.

Question 6.
a. Determine the equation of the line connecting the points (0,-1) and (2,3).
Answer:
m = \(\frac{3-(-1)}{2 – 0}\) = \(\frac{4}{2}\) = 2
y = 2x – 1

b. Will the line described by the equation in part (a) intersect the line passing through the points (-2,4) and (-3,3)? Explain why or why not.
Answer:
m = \(\frac{4-3}{-2-(3)}\) = \(\frac{1}{1}\)
Yes, The lines will intersect because they have different slopes they will eventually intersect.

Question 7.
Line l₁ and line l₂ are shown on the graph below. Use the graph to answer parts (a)–(f).

a. What is the y-intercept of l₁?
Answer:
(0, 4)

b. What is the y-intercept of l₂?
Answer:
(0, 2)

c. Write a system of linear equations representing lines l₁ and l₂.
Answer:
l₁ : y = \(\frac{1}{2}\)x + 4
l₂ : y = x + 2

d. Use the graph to estimate the solution to the system.
Answer:
(1.2, 3.3)

e. Solve the system of linear equations algebraically.
Answer:
y = \(\frac{1}{2}\)x + 4
y = x + 2
–\(\frac{1}{2}\) x + 4 = x + 2
4 = \(\frac{3}{2}\) x + 2
2 = \(\frac{3}{2}\) x
\(\frac{4}{3}\) = x

y = \(\frac{4}{3}\) + 2
= \(\frac{10}{3}\)
(\(\frac{4}{3}\), \(\frac{10}{3}\))

f. Show that your solution from part (e) satisfies both equations.
Answer:
\(\frac{10}{3}\) = –\(\frac{1}{2}\)(\(\frac{4}{3}\)) + 4
\(\frac{10}{3}\) = –\(\frac{2}{3}\) + 4
\(\frac{10}{3}\) = \(\frac{10}{3}\)

\(\frac{10}{3}\) = \(\frac{4}{3}\) + 2
\(\frac{10}{3}\) = \(\frac{10}{3}\)

Engage NY Eureka Math 8th Grade Module 4 Lesson 21 Answer Key

Eureka Math Grade 8 Module 4 Lesson 21 Example Answer Key

Example 1.
Let a line l be given in the coordinate plane. What linear equation is the graph of line l?

Answer:
We can pick two points to determine the slope, but the precise location of the y – intercept point cannot be determined from the graph.
Calculate the slope of the line.
Using points ( – 2, 2) and (5, 4), the slope of the line is
m = \(\frac{2 – 4}{ – 2 – 5}\)
= \(\frac{ – 2}{ – 7}\)
= \(\frac{2}{7}\)
→ Now we need to determine the y – intercept point of the line. We know that it is a point with coordinates (0, b), and we know that the line goes through points ( – 2, 2) and (5, 4) and has slope m = \(\frac{2}{7}\). Using this information, we can determine the coordinates of the y – intercept point and the value of b that we need in order to write the equation of the line.

→ Recall what it means for a point to be on a line; the point is a solution to the equation. In the equation y = mx + b, (x, y) is a solution, and m is the slope. Can we find the value of b? Explain.
Yes. We can substitute one of the points and the slope into the equation and solve for b.

→ Do you think it matters which point we choose to substitute into the equation? That is, will we get a different equation if we use the point ( – 2, 2) compared to (5, 4)?
No, because there can be only one line with a given slope that goes through a point.

→ Verify this claim by using m = \(\frac{2}{7}\) and ( – 2, 2) to find the equation of the line and then by using m = \(\frac{2}{7}\) and (5, 4) to see if the result is the same equation.
Sample student work:
2 = \(\frac{2}{7}\) ( – 2) + b
2 = – \(\frac{4}{7}\) + b
2 + \(\frac{4}{7}\) = – \(\frac{4}{7}\) + \(\frac{4}{7}\) + b
\(\frac{18}{7}\) = b

4 = \(\frac{2}{7}\) (5) + b
4 = \(\frac{10}{7}\) + b
4 – \(\frac{10}{7}\) = \(\frac{10}{7}\) – \(\frac{10}{7}\) + b
\(\frac{18}{7}\) = b
The y – intercept point is at (0, \(\frac{18}{7}\)), and the equation of the line is y = \(\frac{2}{7}\) x + \(\frac{18}{7}\).

→ The equation of the line is
y = \(\frac{2}{7}\) x + \(\frac{18}{7}\).
→ Write it in standard form.
Sample student work:
(y = \(\frac{2}{7}\) x + \(\frac{18}{7}\))7
7y = 2x + 18
– 2x + 7y = 2x – 2x + 18
– 2x + 7y = 18
– 1( – 2x + 7y = 18)
2x – 7y = – 18

Example 2.
Let a line l be given in the coordinate plane. What linear equation is the graph of line l?

Answer:
Determine the slope of the line.
Using points ( – 1, 4) and (4, 1), the slope of the line is
m = \(\frac{4 – 1}{ – 1 – 4}\)
= \(\frac{3}{ – 5}\)
= – \(\frac{3}{5}\).

Determine the y – intercept point of the line.
Sample student work:
4 = ( – \(\frac{3}{5}\))( – 1) + b
4 = \(\frac{3}{5}\) + b
4 – \(\frac{3}{5}\) = \(\frac{3}{5}\) – \(\frac{3}{5}\) + b
\(\frac{17}{5}\) = b
The y – intercept point is at (0, \(\frac{17}{5}\)).

→ Now that we know the slope, m = – \(\frac{3}{5}\), and the y – intercept point, (0, \(\frac{17}{5}\)), write the equation of the line l in slope – intercept form.
y = – \(\frac{3}{5}\) x + \(\frac{17}{5}\)
→ Transform the equation so that it is written in standard form.
Sample student work:
y = – \(\frac{3}{5}\) x + \(\frac{17}{5}\)
(y = – \(\frac{3}{5}\) x + \(\frac{17}{5}\))5
5y = – 3x + 17
3x + 5y = – 3x + 3x + 17
3x + 5y = 17

Example 3.
Let a line l be given in the coordinate plane. What linear equation is the graph of line l?

Answer:
→ Using points (12, 2) and (13, 7), the slope of the line is
m = \(\frac{2 – 7}{12 – 13}\)
= \(\frac{ – 5}{ – 1}\)
= 5.

→ Now, determine the y – intercept point of the line, and write the equation of the line in slope – intercept form.
Sample student work:
2 = 5(12) + b
2 = 60 + b
b = – 58
The y – intercept point is at (0, – 58), and the equation of the line is y = 5x – 58.

Now that we know the slope, m = 5, and the y – intercept point, (0, – 58), write the equation of the line l in standard form.
Sample student work:
y = 5x – 58
– 5x + y = 5x – 5x – 58
– 5x + y = – 58
– 1( – 5x + y = – 58)
5x – y = 58

Example 4.
Let a line l be given in the coordinate plane. What linear equation is the graph of line l?

Answer:
Using points (3, 1) and ( – 3, – 1), the slope of the line is
m = \(\frac{ – 1 – 1}{ – 3 – 3}\)
= \(\frac{ – 2}{ – 6}\)
= \(\frac{1}{3}\)
The y – intercept point is at (0, 0), and the equation of the line is y = \(\frac{1}{3}\) x.

Eureka Math Grade 8 Module 4 Lesson 21 Exercise Answer Key

Exercises

Exercise 1.
Write the equation for the line l shown in the figure.

Answer:
Using the points ( – 1, – 3) and (2, – 2), the slope of the line is
m = \(\frac{ – 3 – ( – 2)}{ – 1 – 2}\)
= \(\frac{ – 1}{ – 3}\)
= \(\frac{1}{3}\) .
– 2 = \(\frac{1}{3}\) (2) + b
– 2 = \(\frac{2}{3}\) + b
– 2 – \(\frac{2}{3}\) = \(\frac{2}{3}\) – \(\frac{2}{3}\) + b
– \(\frac{8}{3}\) = b
The equation of the line is y = \(\frac{1}{3}\) x – \(\frac{8}{3}\).

Exercise 2.
Write the equation for the line l shown in the figure.

Answer:
Using the points ( – 3, 7) and (2, 8), the slope of the line is
m = \(\frac{7 – 8}{ – 3 – 2}\)
= \(\frac{ – 1}{ – 5}\)
= \(\frac{1}{5}\).
8 = \(\frac{1}{5}\) (2) + b
8 = \(\frac{2}{5}\) + b
8 – \(\frac{2}{5}\) = \(\frac{2}{5}\) – \(\frac{2}{5}\) + b
\(\frac{38}{5}\) = b
The equation of the line is y = \(\frac{1}{5}\) x + \(\frac{38}{5}\).

Exercise 3.
Determine the equation of the line that goes through points ( – 4, 5) and (2, 3).
Answer:
The slope of the line is
m = \(\frac{5 – 3}{ – 4 – 2}\)
= \(\frac{2}{ – 6}\)
= – \(\frac{1}{3}\)
The y – intercept point of the line is
3 = – \(\frac{1}{3}\) (2) + b
3 = – \(\frac{2}{3}\) + b
\(\frac{11}{3}\) = b.
The equation of the line is y = – \(\frac{1}{3}\) x + \(\frac{11}{3}\).

Exercise 4.
Write the equation for the line l shown in the figure.

Answer:
Using the points ( – 7, 2) and ( – 6, – 2), the slope of the line is
m = \(\frac{2 – ( – 2)}{ – 7 – ( – 6)}\)
= \(\frac{4}{ – 1}\)
= – 4.
– 2 = – 4( – 6) + b
– 2 = 24 + b
– 26 = b
The equation of the line is y = – 4x – 26.

Exercise 5.
A line goes through the point (8, 3) and has slope m = 4. Write the equation that represents the line.
Answer:
3 = 4(8) + b
3 = 32 + b
– 29 = b
The equation of the line is y = 4x – 29.

Eureka Math Grade 8 Module 4 Lesson 21 Problem Set Answer Key

Question 1.
Write the equation for the line l shown in the figure.

Answer:
Using the points ( – 3, 2) and (2, – 2), the slope of the line is
m = \(\frac{2 – ( – 2)}{ – 3 – 2}\)
= \(\frac{4}{ – 5}\)
= – \(\frac{4}{5}\)
2 = ( – \(\frac{4}{5}\))( – 3) + b
2 = \(\frac{12}{5}\) + b
2 – \(\frac{12}{5}\) = \(\frac{12}{5}\) – \(\frac{12}{5}\) + b
– \(\frac{2}{5}\) = b
The equation of the line is y = – \(\frac{4}{5}\) x – \(\frac{2}{5}\).

Question 2.
Write the equation for the line l shown in the figure.

Answer:
Using the points ( – 6, 2) and ( – 5, 5), the slope of the line is
m = \(\frac{2 – 5}{ – 6 – ( – 5)}\)
= \(\frac{ – 3}{ – 1}\)
= 3.
5 = 3( – 5) + b
5 = – 15 + b
20 = b
The equation of the line is y = 3x + 20.

Question 3.
Write the equation for the line l shown in the figure.

Answer:
Using the points ( – 3, 1) and (2, 2), the slope of the line is
m = \(\frac{1 – 2}{ – 3 – 2}\)
= \(\frac{ – 1}{ – 5}\)
= \(\frac{1}{5}\)
2 = \(\frac{1}{5}\) (2) + b
2 = \(\frac{2}{5}\) + b
2 – \(\frac{2}{5}\) = \(\frac{2}{5}\) – \(\frac{2}{5}\) + b
\(\frac{8}{5}\) = b
The equation of the line is y = \(\frac{1}{5}\) x + \(\frac{8}{5}\).

Question 4.
Triangle ABC is made up of line segments formed from the intersection of lines L_AB, L_BC, and L_AC. Write the equations that represent the lines that make up the triangle.

Answer:
A( – 3, – 3), B(3, 2), C(5, – 2)
The slope of L_AB:
m = \(\frac{ – 3 – 2}{ – 3 – 3}\)
= \(\frac{ – 5}{ – 6}\)
= \(\frac{5}{6}\)
2 = \(\frac{5}{6}\) (3) + b
2 = \(\frac{5}{2}\) + b
2 – \(\frac{5}{2}\) = \(\frac{5}{2}\) – \(\frac{5}{2}\) + b
– \(\frac{1}{2}\) = b
The equation of L_AB is y = \(\frac{5}{6}\) x – \(\frac{1}{2}\).

The slope of L_BC:
m = \( = \frac{2 – ( – 2)}{3 – 5}\)
= \(\frac{4}{ – 2}\)
= – 2
2 = – 2(3) + b
2 = – 6 + b
8 = b
The equation of L_BC is y = – 2x + 8.

The slope of L_AC:
m = \(\frac{ – 3 – ( – 2)}{ – 3 – 5}\)
= \(\frac{ – 1}{ – 8}\)
= \(\frac{1}{8}\)
– 2 = \(\frac{1}{8}\) (5) + b
– 2 = \(\frac{5}{8}\) + b
– 2 – \(\frac{5}{8}\) = \(\frac{5}{8}\) – \(\frac{5}{8}\) + b
– 2\(\frac{1}{8}\) = b
The equation of L_AC is y = \(\frac{1}{8}\) x – 2\(\frac{1}{8}\).

Question 5.
Write the equation for the line that goes through point ( – 10, 8) with slope m = 6.
Answer:
8 = 6( – 10) + b
8 = – 60 + b
68 = b
The equation of the line is y = 6x + 68.

Question 6.
Write the equation for the line that goes through point (12, 15) with slope m = – 2.
Answer:
15 = – 2(12) + b
15 = – 24 + b
39 = b
The equation of the line is y = – 2x + 39.

Question 7.
Write the equation for the line that goes through point (1, 1) with slope m = – 9.
Answer:
1 = – 9(1) + b
1 = – 9 + b
10 = b
The equation of the line is y = – 9x + 10.

Question 8.
Determine the equation of the line that goes through points (1, 1) and (3, 7).
Answer:
The slope of the line is
m = \(\frac{1 – 7}{1 – 3}\)
= \(\frac{ – 6}{ – 2}\)
= 3.
The y – intercept point of the line is
7 = 3(3) + b
7 = 9 + b
– 2 = b.
The equation of the line is y = 3x – 2.

Eureka Math Grade 8 Module 4 Lesson 21 Exit Ticket Answer Key

Question 1.
Write the equation for the line l shown in the figure below.

Answer:
Using the points ( – 3, 1) and (6, 5), the slope of the line is
m = \(\frac{5 – 1}{6 – ( – 3)}\)
m = \(\frac{4}{9}\)
5 = \(\frac{4}{9}\) (6) + b
5 = \(\frac{8}{3}\) + b
5 – \(\frac{8}{3}\) = \(\frac{8}{3}\) – \(\frac{8}{3}\) + b
\(\frac{7}{3}\) = b
The equation of the line is y = \(\frac{4}{9}\) x + \(\frac{7}{3}\).

Question 2.
A line goes through the point (5, – 7) and has slope m = – 3. Write the equation that represents the line.
Answer:
– 7 = – 3(5) + b
– 7 = – 15 + b
8 = b
The equation of the line is y = – 3x + 8.

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 ???? 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

Engage NY Eureka Math 8th Grade Module 4 Lesson 22 Answer Key

Eureka Math Grade 8 Module 4 Lesson 22 Exercise Answer Key

Exercises

Exercise 1.
Peter paints a wall at a constant rate of 2 square feet per minute. Assume he paints an area y, in square feet, after x minutes.
a. Express this situation as a linear equation in two variables.
Answer:
\(\frac{y}{x}\) = \(\frac{2}{1}\)
y = 2x

b. Sketch the graph of the linear equation.

Answer:

c. Using the graph or the equation, determine the total area he paints after 8 minutes, 1 \(\frac{1}{2}\) hours, and 2 hours. Note that the units are in minutes and hours.
Answer:
In 8 minutes, he paints 16 square feet.
y = 2(90)
= 180

In 1 \(\frac{1}{2}\) hours, he paints 180 square feet.
y = 2(120)
= 240

In 2 hours, he paints 240 square feet.

Exercise 2.
The figure below represents Nathan’s constant rate of walking.

a. Nicole just finished a 5-mile walkathon. It took her 1.4 hours. Assume she walks at a constant rate. Let y represent the distance Nicole walks in x hours. Describe Nicole’s walking at a constant rate as a linear equation in two variables.
Answer:
\(\frac{y}{x}\) = \(\frac{5}{1.4}\)
y = \(\frac{25}{7}\) x

b. Who walks at a greater speed? Explain.
Answer:
Nathan walks at a greater speed. The slope of the graph for Nathan is 4, and the slope or rate for Nicole is \(\frac{25}{7}\). When you compare the slopes, you see that 4 > \(\frac{25}{7}\).

Exercise 3.
a. Susan can type 4 pages of text in 10 minutes. Assuming she types at a constant rate, write the linear equation that represents the situation.
Answer:
Let y represent the total number of pages Susan can type in x minutes. We can write \(\frac{y}{x}\) = \(\frac{4}{10}\) and y = \(\frac{2}{5}\) x.

b. The table of values below represents the number of pages that Anne can type, y, in a few selected x minutes. Assume she types at a constant rate.

Answer:
Anne types faster. Using the table, we can determine that the slope that represents Anne’s constant rate of typing is \(\frac{2}{3}\). The slope or rate for Nicole is \(\frac{2}{5}\). When you compare the slopes, you see that \(\frac{2}{3}\) > \(\frac{2}{5}\).

Exercise 4.
a. Phil can build 3 birdhouses in 5 days. Assuming he builds birdhouses at a constant rate, write the linear equation that represents the situation.
Answer:
Let y represent the total number of birdhouses Phil can build in x days. We can write \(\frac{y}{x}\) = \(\frac{3}{5}\) and y = \(\frac{3}{5}\) x.

b. The figure represents Karl’s constant rate of building the same kind of birdhouses.

Who builds birdhouses faster? Explain.
Answer:
Karl can build birdhouses faster. The slope of the graph for Karl is \(\frac{3}{4}\), and the slope or rate of change for Phil is \(\frac{3}{5}\). When you compare the slopes, \(\frac{3}{4}\) > \(\frac{3}{5}\).

Exercise 5.
Explain your general strategy for comparing proportional relationships.
Answer:
When comparing proportional relationships, we look specifically at the rate of change for each situation. The relationship with the greater rate of change will end up producing more, painting a greater area, or walking faster when compared to the same amount of time with the other proportional relationship.

Eureka Math Grade 8 Module 4 Lesson 22 Problem Set Answer Key

Question 1.
a. Train A can travel a distance of 500 miles in 8 hours. Assuming the train travels at a constant rate, write the linear equation that represents the situation.
Answer:
Let y represent the total number of miles Train A travels in x minutes. We can write \(\frac{y}{x}\) = \(\frac{500}{8}\) and y = \(\frac{125}{2}\) x.

b. The figure represents the constant rate of travel for Train B.

Which train is faster? Explain.
Answer:
Train B is faster than Train A. The slope or rate for Train A is \(\frac{125}{2}\), and the slope of the line for Train B is \(\frac{200}{3}\). When you compare the slopes, you see that \(\frac{200}{3}\) > \(\frac{125}{2}\).

Question 2.
a. Natalie can paint 40 square feet in 9 minutes. Assuming she paints at a constant rate, write the linear equation that represents the situation.
Answer:
Let y represent the total square feet Natalie can paint in x minutes. We can write \(\frac{y}{x}\) = \(\frac{40}{9}\), and y = \(\frac{40}{9}\) x.

b. The table of values below represents the area painted by Steven for a few selected time intervals. Assume Steven is painting at a constant rate.

Answer:
Who paints faster? Explain.
Natalie paints faster. Using the table of values, I can find the slope that represents Steven’s constant rate of painting: \(\frac{10}{3}\). The slope or rate for Natalie is \(\frac{40}{9}\). When you compare the slopes, you see that \(\frac{40}{9}\) > \(\frac{10}{3}\).

Question 3.
a. Bianca can run 5 miles in 41 minutes. Assuming she runs at a constant rate, write the linear equation that represents the situation.
Answer:
Let y represent the total number of miles Bianca can run in x minutes. We can write \(\frac{y}{x}\) = \(\frac{5}{41}\), and y = \(\frac{5}{41}\) x.

b. The figure below represents Cynthia’s constant rate of running.

Who runs faster? Explain.
Answer:
Cynthia runs faster. The slope of the graph for Cynthia is \(\frac{1}{7}\), and the slope or rate for Nicole is \(\frac{5}{41}\). When you compare the slopes, you see that \(\frac{1}{7}\) > \(\frac{5}{41}\).

Question 4.
a. Geoff can mow an entire lawn of 450 square feet in 30 minutes. Assuming he mows at a constant rate, write the linear equation that represents the situation.
Answer:
Let y represent the total number of square feet Geoff can mow in x minutes. We can write \(\frac{y}{x}\) = \(\frac{450}{30}\), and y = 15x.

b. The figure represents Mark’s constant rate of mowing a lawn.

Who mows faster? Explain.
Answer:
Geoff mows faster. The slope of the graph for Mark is \(\frac{14}{2}\) = 7, and the slope or rate for Geoff is \(\frac{450}{30}\) = 15. When you compare the slopes, you see that 15 > 7.

Question 5.
a. Juan can walk to school, a distance of 0.75 mile, in 8 minutes. Assuming he walks at a constant rate, write the linear equation that represents the situation.
Answer:
Let y represent the total distance in miles that Juan can walk in x minutes. We can write \(\frac{y}{x}\) = \(\frac{0.75}{8}\), and y = \(\frac{3}{32}\) x.

b. The figure below represents Lena’s constant rate of walking.

Who walks faster? Explain.
Answer:
Lena walks faster. The slope of the graph for Lena is \(\frac{1}{9}\), and the slope of the equation for Juan is \(\frac{0.75}{8}\), or \(\frac{3}{32}\). When you compare the slopes, you see that \(\frac{1}{9}\) > \(\frac{3}{32}\).

Eureka Math Grade 8 Module 4 Lesson 22 Exit Ticket Answer Key

Question 1.
Water flows out of Pipe A at a constant rate. Pipe A can fill 3 buckets of the same size in 14 minutes. Write a linear equation that represents the situation.
Answer:
Let y represent the total number of buckets that Pipe A can fill in x minutes. We can write \(\frac{y}{x}\) = \(\frac{3}{14}\) and y = \(\frac{3}{14}\) x.

Question 2.
The figure below represents the rate at which Pipe B can fill the same-sized buckets.

Which pipe fills buckets faster? Explain.
Answer:
Pipe A fills the same-sized buckets faster than Pipe B. The slope of the graph for Pipe B is \(\frac{1}{5}\), and the slope or rate for Pipe A is \(\frac{3}{14}\). When you compare the slopes, you see that \(\frac{3}{14}\) > \(\frac{1}{5}\).

Engage NY Eureka Math 8th Grade Module 4 Lesson 23 Answer Key

Eureka Math Grade 8 Module 4 Lesson 23 Exercise Answer Key

Exploratory Challenge/Exercises 1–3

Exercise 1.
Sketch the graph of the equation 9x + 3y = 18 using intercepts. Then, answer parts (a)–(f) that follow.
Answer:
9(0) + 3y = 18
3y = 18
y = 6
The y – intercept point is (0,6).
9x + 3(0) = 18
9x = 18
x = 2
The x – intercept point is (2,0).

a. Sketch the graph of the equation y = – 3x + 6 on the same coordinate plane.
Answer:

b. What do you notice about the graphs of 9x + 3y = 18 and y = – 3x + 6? Why do you think this is so?
Answer:
The graphs of the equations produce the same line. Both equations go through the same two points, so they are the same line.

c. Rewrite y = – 3x + 6 in standard form.
Answer:
y = – 3x + 6
3x + y = 6

d. Identify the constants a, b, and c of the equation in standard form from part (c).
Answer:
a = 3, b = 1, and c = 6

e. Identify the constants of the equation 9x + 3y = 18. Note them as a’, b’, and c’.
Answer:
a’ = 9, b’ = 3, and c’ = 18

f. What do you notice about \(\frac{a^{\prime}}{a}\), \(\frac{b^{\prime}}{b}\), and \(\frac{c^{\prime}}{c}\)?
Answer:
\(\frac{a^{\prime}}{a}\) = \(\frac{9}{3}\) = 3, \(\frac{b^{\prime}}{b}\) = \(\frac{3}{1}\) = 3, and \(\frac{c^{\prime}}{c}\) = \(\frac{18}{6}\) = 3
Each fraction is equal to the number 3.

Exercise 2.
Sketch the graph of the equation y = \(\frac{1}{2}\) x + 3 using the ????y – intercept point and the slope. Then, answer parts (a)–(f) that follow.
a. Sketch the graph of the equation 4x – 8y = – 24 using intercepts on the same coordinate plane.
Answer:
4(0) – 8y = – 24
– 8y = – 24
y = 3
The y – intercept point is (0,3).
4x – 8(0) = – 24
4x = – 24
x = – 6
The x – intercept point is ( – 6,0).

b. What do you notice about the graphs of y = \(\frac{1}{2}\) x + 3 and 4x – 8y = – 24? Why do you think this is so?
Answer:
The graphs of the equations produce the same line. Both equations go through the same two points, so they are the same line.

c. Rewrite y = \(\frac{1}{2}\) x + 3 in standard form.
Answer:
y = \(\frac{1}{2}\) x + 3
(y = \(\frac{1}{2}\) x + 3)2
2y = x + 6
– x + 2y = 6
– 1( – x + 2y = 6)
x – 2y = – 6

d. Identify the constants a, b, and c of the equation in standard form from part (c).
Answer:
a = 1, b = – 2, and c = – 6

e. Identify the constants of the equation 4x – 8y = – 24. Note them as a’, b’, and c’.
Answer:
a’ = 4, b’ = – 8, and c’ = – 24

f. What do you notice about \(\frac{a^{\prime}}{a}\), \(\frac{b^{\prime}}{b}\), and \(\frac{c^{\prime}}{c}\)?
Answer:
\(\frac{a^{\prime}}{a}\) = \(\frac{4}{1}\) = 4, \(\frac{b^{\prime}}{b}\) = \(\frac{ – 8}{ – 2}\) = 4, and \(\frac{c^{\prime}}{c}\) = \(\frac{ – 24}{ – 6}\) = 4
Each fraction is equal to the number 4.

Exercise 3.
The graphs of the equations y = \(\frac{2}{3}\) x – 4 and 6x – 9y = 36 are the same line.
a. Rewrite y = \(\frac{2}{3}\) x – 4 in standard form.
Answer:
y = \(\frac{2}{3}\) x – 4
(y = \(\frac{2}{3}\) x – 4)3
3y = 2x – 12
– 2x + 3y = – 12
– 1( – 2x + 3y = – 12)
2x – 3y = 12

b. Identify the constants a, b, and c of the equation in standard form from part (a).
Answer:
a = 2, b = – 3, and c = 12

c. Identify the constants of the equation 6x – 9y = 36. Note them as a’, b’, and c’.
Answer:
a’ = 6, b’ = – 9, and c’ = 36

d. What do you notice about \(\frac{a^{\prime}}{a}\), \(\frac{b^{\prime}}{b}\), and \(\frac{c^{\prime}}{c}\)?
Answer:
\(\frac{a^{\prime}}{a}\) = \(\frac{6}{2}\) = 3, \(\frac{b^{\prime}}{b}\) = \(\frac{ – 9}{ – 3}\) = 3, and \(\frac{c^{\prime}}{c}\) = \(\frac{36}{12}\) = 3
Each fraction is equal to the number 3.

e. You should have noticed that each fraction was equal to the same constant. Multiply that constant by the standard form of the equation from part (a). What do you notice?
Answer:
2x – 3y = 12
3(2x – 3y = 12)
6x – 9y = 36
After multiplying the equation from part (a) by 3, I noticed that it is the exact same equation that was given.

Exercises 4–8
Exercise 4.
Write three equations whose graphs are the same line as the equation 3x + 2y = 7.
Answer:
Answers will vary. Verify that students have multiplied a, b, and c by the same constant when they write the new equation.

Exercise 5.
Write three equations whose graphs are the same line as the equation x – 9y = \(\frac{3}{4}\).
Answer:
Answers will vary. Verify that students have multiplied a, b, and c by the same constant when they write the new equation.

Exercise 6.
Write three equations whose graphs are the same line as the equation – 9x + 5y = – 4.
Answer:
Answers will vary. Verify that students have multiplied a, b, and c by the same constant when they write the new equation.

Exercise 7.
Write at least two equations in the form ax + by = c whose graphs are the line shown below.

Answer:
Answers will vary. Verify that students have the equation – x + 4y = – 3 in some form.

Exercise 8.
Write at least two equations in the form ax + by = c whose graphs are the line shown below.

Answer:
Answers will vary. Verify that students have the equation 4x + 3y = 2 in some form.

Eureka Math Grade 8 Module 4 Lesson 23 Problem Set Answer Key

Question 1.
Do the equations x + y = – 2 and 3x + 3y = – 6 define the same line? Explain.
Yes, these equations define the same line. When you compare the constants from each equation, you get
Answer:
\(\frac{a^{\prime}}{a}\) = \(\frac{3}{1}\) = 3, \(\frac{b^{\prime}}{b}\) = \(\frac{3}{1}\) = 3, and \(\frac{c^{\prime}}{c}\) = \(\frac{ – 6}{ – 2}\) = 3
When I multiply the first equation by 3, I get the second equation.
(x + y = – 2)3
3x + 3y = – 6
Therefore, these equations define the same line.

Question 2.
Do the equations y = – \(\frac{5}{4}\) x + 2 and 10x + 8y = 16 define the same line? Explain.
Answer:
Yes, these equations define the same line. When you rewrite the first equation in standard form, you get
y = – \(\frac{5}{4}\) x + 2
(y = – \(\frac{5}{4}\) x + 2)4
4y = – 5x + 8
5x + 4y = 8″.”
When you compare the constants from each equation, you get
\(\frac{a^{\prime}}{a}\) = \(\frac{10}{5}\) = 2, \(\frac{b^{\prime}}{b}\) = \(\frac{8}{4}\) = 2, and \(\frac{c^{\prime}}{c}\) = \(\frac{16}{8}\) = 2.
When I multiply the first equation by 2, I get the second equation.
(5x + 4y = 8)2
10x + 8y = 16
Therefore, these equations define the same line.

Question 3.
Write an equation that would define the same line as 7x – 2y = 5.
Answer:
Answers will vary. Verify that students have written an equation that defines the same line by showing that the fractions \(\frac{a^{\prime}}{a}\) = \(\frac{b^{\prime}}{b}\) = \(\frac{c^{\prime}}{c}\) = s, where s is some constant.

Question 4.
Challenge: Show that if the two lines given by ax + by = c and a’x + b’y = c’ are the same when b = 0 (vertical lines), then there exists a nonzero number s so that a’ = sa, b’ = sb, and c’ = sc.
Answer:
When b = 0, then the equations are ax = c and a’x = c’. We can rewrite the equations as x = \(\frac{c}{a}\) and x = \(\frac{c^{\prime}}{a^{\prime}}\). Because the equations graph as the same line, then we know that
\(\frac{c}{a} = \frac{c^{\prime}}{a^{\prime}}\)
and we can rewrite those fractions as
\(\frac{a^{\prime}}{a} = \frac{c^{\prime}}{c}\)
These fractions are equal to the same number. Let that number be s. Then \(\frac{a^{\prime}}{a}\) = s and \(\frac{c^{\prime}}{c}\) = s. Therefore, a’ = sa and c’ = sc.

Question 5.
Challenge: Show that if the two lines given by ax + by = c and a’x + b’y = c’ are the same when a = 0 (horizontal lines), then there exists a nonzero number s so that a’ = sa, b’ = sb, and c’ = sc.
Answer:
When a = 0, then the equations are by = c and b’y = c’. We can rewrite the equations as y = \(\frac{c}{b}\) and y = \(\frac{c^{\prime}}{b^{\prime}}\). Because the equations graph as the same line, then we know that their slopes are the same.
\(\frac{c}{b} = \frac{c^{\prime}}{b^{\prime}}\)
We can rewrite the proportion.
\(\frac{b^{\prime}}{b} = \frac{c^{\prime}}{c}\)
These fractions are equal to the same number. Let that number be s. Then \(\frac{b^{\prime}}{b}\) = s and \(\frac{c^{\prime}}{c}\) = s. Therefore, b’ = sb and c’ = sc.

Eureka Math Grade 8 Module 4 Lesson 23 Exit Ticket Answer Key

Question 1.
Do the graphs of the equations – 16x + 12y = 33 and – 4x + 3y = 8 graph as the same line? Why or why not?
Answer:
No. In the first equation, a = – 16, b = 12, and c = 33, and in the second equation, a’ = – 4, b’ = 3, and c’ = 8. Then,
\(\frac{a^{\prime}}{a}\) = \(\frac{ – 4}{ – 16}\) = \(\frac{1}{4}\), \(\frac{b^{\prime}}{b}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\), and \(\frac{c^{\prime}}{c}\) = \(\frac{8}{33}\) = \(\frac{8}{33}\)
Since each fraction does not equal the same number, then they do not have the same graph.

Question 2.
Given the equation 3x – y = 11, write another equation that will have the same graph. Explain why.
Answer:
Answers will vary. Verify that students have written an equation that defines the same line by showing that the fractions \(\frac{a^{\prime}}{a}\) = \(\frac{b^{\prime}}{b}\) = \(\frac{c^{\prime}}{c}\) = s, where s is some constant.