## Engage NY Eureka Math 8th Grade Module 4 Lesson 18 Answer Key

### Eureka Math Grade 8 Module 4 Lesson 19 Opening Exercise Answer Key

Examine each of the graphs and their equations. Identify the coordinates of the point where the line intersects the y-axis. Describe the relationship between the point and the equation y=mx+b.

a. y=\(\frac{1}{2}\) x+3

b. y=-3x+7

c. y=-\(\frac{2}{3}\) x-2

d. y=5x-4

Answer:

A point is noted in each graph above where the line intersects the y-axis:

y=\(\frac{1}{2}\) x+3, (0,3)

y=-3x+7, (0,7)

y=-\(\frac{2}{3}\) x-2, (0,-2)

y=5x-4, (0,-4)

In each equation, the number b was the y-coordinate of the point where the line intersected the y-axis.

### Eureka Math Grade 8 Module 4 Lesson 19 Example Answer Key

Graph an equation in the form of y=mx+b.

Example 1.

Graph the equation y=\(\frac{2}{3}\) x+1. Name the slope and y-intercept point.

Answer:

The slope is m=\(\frac{2}{3}\), and the y-intercept point is (0,1).

â†’ To graph the equation, we must begin with the known point. In this case, that is the y-intercept point. We cannot begin with the slope because the slope describes the rate of change between two points. That means we need a point to begin with. On a graph, we plot the point (0,1).

â†’ Next, we use the slope to find the second point. We know that m=\(\frac{|Q R|}{|P Q|}\) =\(\frac{2}{3}\). The slope tells us exactly how many units to go to the right of P to find point Q and then how many vertical units we need to go from Q to find point R. How many units will we go to the right in order to find point Q? How do you know?

â†’ We need to go 3 units to the right of point P to find Q. We go 3 units because |PQ|=3.

â†’ How many vertical units from point Q must we go to find point R? How do you know?

â†’ We need to go 2 units from point Q to find R. We go 2 units because |QR|=2.

â†’ Will we go up from point Q or down from point Q to find R? How do you know?

â†’ We need to go up because the slope is positive. That means that the line will be left-to-right inclining.

â†’ Since we know that the line joining two distinct points of the form y=mx+b has slope m, and we specifically constructed points P and R with the slope in mind, we can join the points with a line.

Example 2.

Graph the equation y=-\(\frac{3}{4}\) x-2. Name the slope and y-intercept point.

Answer:

The slope is m=-\(\frac{3}{4}\), and the y-intercept point is (0,-2).

â†’ How do we begin?

â†’ We must begin by putting a known point on the graph, (0,-2).

â†’ We know that m=\(\frac{|Q R|}{|P Q|}\) =-\(\frac{3}{4}\). How many units will we go to the right in order to find point Q? How do you know?

â†’ We need to go 4 units to the right of point P to find Q. We go 4 units because |PQ|=4.

â†’ How many units from point Q must we go to find point R? How do you know?

â†’ We need to go 3 units from point Q to find R. We go 3 units because |QR|=3.

â†’ Will we go up from point Q or down from point Q to find R? How do you know?

â†’ We need to go down from point Q to point R because the slope is negative. That means that the line will be left-to-right declining.

â†’ Now we draw the line through the points P and R.

Example 3

Graph the equation y=4x-7. Name the slope and y-intercept point.

Answer:

The slope is m=4, and the y-intercept point is (0,-7).

â†’ Graph the equation y=4x-7. Name the slope and y-intercept point.

â†’ The slope is m=4, and the y-intercept point is (0,-7).

â†’ How do we begin?

â†’ We must begin by putting a known point on the graph, (0,-7).

â†’ Notice this time that the slope is the integer 4. In the last two examples, our slopes have been in the form of a fraction so that we can use the information in the numerator and denominator to determine the lengths of |PQ| and |QR|. Since m=\(\frac{|Q R|}{|P Q|}\) =4, what fraction can we use to represent slope to help us graph?

â†’ The number 4 is equivalent to the fraction \(\frac{4}{1}\).

â†’ Using m=\(\frac{|Q R|}{|P Q|}\) =\(\frac{4}{1}\), how many units will we go to the right in order to find point Q? How do you know?

â†’ We need to go 1 unit to the right of point P to find Q. We go 1 unit because |PQ|=1.

â†’ How many vertical units from point Q must we go to find point R? How do you know?

â†’ We need to go 4 units from point Q to find R. We go 4 units because |QR|=4.

â†’ Will we go up from point Q or down from point Q to find R? How do you know?

â†’ We need to go up from point Q to point R because the slope is positive. That means that the line will be left-to-right inclining.

â†’ Now we join the points P and R to make the line.

### Eureka Math Grade 8 Module 4 Lesson 19 Exercise Answer Key

Exercises 1â€“4 (5 minutes)

Students complete Exercises 1â€“4 individually or in pairs.

Exercise 1.

Graph the equation y=\(\frac{5}{2}\) x -4.

a. Name the slope and the y-intercept point.

Answer:

The slope is m=\(\frac{5}{2}\), and the y-intercept point is (0,-4).

b. Graph the known point, and then use the slope to find a second point before drawing the line.

Answer:

Exercise 2.

Graph the equation y=-3x+6.

a. Name the slope and the y-intercept point.

Answer:

The slope is m=-3, and the y-intercept point is (0,6).

b. Graph the known point, and then use the slope to find a second point before drawing the line.

Answer:

Question 3.

The equation y=1x+0 can be simplified to y=x. Graph the equation y=x.

a. Name the slope and the y-intercept point.

Answer:

The slope is m=1, and the y-intercept point is (0,0).

b. Graph the known point, and then use the slope to find a second point before drawing the line.

Answer:

Question 4.

Graph the point (0,2).

Answer:

a. Find another point on the graph using the slope, m=\(\frac{2}{7}\).

b. Connect the points to make the line.

c. Draw a different line that goes through the point (0,2) with slope m=\(\frac{2}{7}\). What do you notice?

Answer:

Only one line can be drawn through the given point with the given slope.

Exercises 5â€“6

Students complete Exercises 5â€“6 individually or in pairs.

Exercise 5.

A bank put $10 into a savings account when you opened the account. Eight weeks later, you have a total of $24. Assume you saved the same amount every week.

a. If y is the total amount of money in the savings account and x represents the number of weeks, write an equation in the form y=mx+b that describes the situation.

Answer:

24=m(8)+10

14=8m

\(\frac{14}{8}\)=m

\(\frac{7}{4}\)=m

y=\(\frac{7}{4}\) x+10

b. Identify the slope and the y-intercept point. What do these numbers represent?

Answer:

The slope is \(\frac{7}{4}\), and the y-intercept point is (0,10). The y-intercept point represents the amount of money the bank gave me, in the amount of $10. The slope represents the amount of money I save each week, \(\frac{7}{4}\)=$1.75.

c. Graph the equation on a coordinate plane.

Answer:

d. Could any other line represent this situation? For example, could a line through point (0,10) with slope \(\frac{7}{5}\) represent the amount of money you save each week? Explain.

Answer:

No, a line through point (0,10) with slope \(\frac{7}{5}\) cannot represent this situation. That line would show that at the end of the 8 weeks I would have $21.20, but I was told that I would have $24 by the end of the 8 weeks.

Exercise 6.

A group of friends are on a road trip. After 120 miles, they stop to eat lunch. They continue their trip and drive at a constant rate of 50 miles per hour.

a. Let y represent the total distance traveled, and let x represent the number of hours driven after lunch. Write an equation to represent the total number of miles driven that day.

Answer:

y=50x+120

b. Identify the slope and the y-intercept point. What do these numbers represent?

Answer:

The slope is 50 and represents the rate of driving. The y-intercept point is 120 and represents the number of miles they had already driven before driving at the given constant rate.

c. Graph the equation on a coordinate plane.

Answer:

d. Could any other line represent this situation? For example, could a line through point (0,120) with slope 75 represent the total distance the friends drive? Explain.

Answer:

No, a line through point (0,120) with a slope of 75 could not represent this situation. That line would show that after an hour, the friends traveled a total distance of 195 miles. According to the information given, the friends would only have traveled 170 miles after one hour.

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

Mrs. Hodson said that the graphs of the equations below are incorrect. Find the studentâ€™s errors, and correctly graph the equations.

Question 1.

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

Error:

Answer:

The student should have gone up 1 unit when finding |QR| since the slope is positive.

Correct graph of the equation:

Answer:

Question 2.

Student graph of the equation y=-\(\frac{3}{5}\) x-1:

Error:

Answer:

The student did not find the y-intercept point correctly. It should be the point (0,-1).

Correct graph of the equation:

Answer:

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

Students practice graphing equations using y-intercept point and slope. Students need graph paper to complete the Problem Set. Optional Problem 11 has students show that there is only one line passing through a point with a given negative slope.

Graph each equation on a separate pair of x- and y-axes.

Question 1.

Graph the equation y=\(\frac{4}{5}\) x-5.

a. Name the slope and the y-intercept point.

Answer:

The slope is m=\(\frac{4}{5}\), and the y-intercept point is (0,-5).

Answer:

Question 2.

Graph the equation y=x+3.

a. Name the slope and the y-intercept point.

Answer:

The slope is m=1, and the y-intercept point is (0,3).

Answer:

Question 3.

Graph the equation y=-\(\frac{4}{3}\) x+4.

a. Name the slope and the y-intercept point.

Answer:

The slope is m=-\(\frac{4}{3}\), and the y-intercept point is (0,4).

Answer:

Question 4.

Graph the equation y=\(\frac{5}{2}\) x.

a. Name the slope and the y-intercept point.

Answer:

The slope is m=\(\frac{5}{2}\), and the y-intercept point is (0,0).

Answer:

Question 5.

Graph the equation y=2x-6.

a. Name the slope and the y-intercept point.

Answer:

The slope is m=2, and the y-intercept point is (0,-6).

Answer:

Question 6.

Graph the equation y=-5x+9.

a. Name the slope and the y-intercept point.

Answer:

The slope is m=-5, and the y-intercept point is (0,9).

Answer:

Question 7.

Graph the equation y=\(\frac{1}{3}\) x+1.

a. Name the slope and the y-intercept point.

Answer:

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

Answer:

Question 8.

Graph the equation 5x+4y=8. (Hint: Transform the equation so that it is of the form y=mx+b.)

a. Name the slope and the y-intercept point.

Answer:

5x+4y=8

5x-5x+4y=8-5x

4y=8-5x

\(\frac{4}{4}\) y=\(\frac{8}{4}\)–\(\frac{5}{4}\) x

y=2-\(\frac{5}{4}\) x

y=-\(\frac{5}{4}\) x+2

The slope is m=-\(\frac{5}{4}\), and the y-intercept point is (0,2).

Answer:

Question 9.

Graph the equation -2x+5y=30.

a. Name the slope and the y-intercept point.

Answer:

-2x+5y=30

2x+2x+5y=30+2x

5y=30+2x

\(\frac{5}{5}\) y=\(\frac{30}{5}\)+\(\frac{2}{5}\) x

y=6+\(\frac{2}{5}\) x

y=\(\frac{2}{5}\) x+6

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

Answer:

Question 10.

Let l and l’ be two lines with the same slope m passing through the same point P. Show that there is only one line with a slope m, where m<0, passing through the given point P. Draw a diagram if needed.

Answer:

First, assume that there are two different lines l and l’ with the same negative slope passing through P. From point P, I mark a point Q one unit to the right. Then, I draw a line parallel to the y-axis through point Q. The intersection of this line and line l and l’ are noted with points R and R’, respectively. By definition of slope, the lengths |QR| and |QR’ | represent the slopes of lines l and l’, respectively. We are given that the lines have the same slope, which means that lengths |QR| and |QR’| are equal. Since that is true, then points R and R’coincide and so do lines l and l’. Therefore, our assumption that they are different lines is false; l and l’ must be the same line. Therefore, there is only one line with slope m passing through the given point P.