## Engage NY Eureka Math 8th Grade Module 4 Lesson 27 Answer Key

### Eureka Math Grade 8 Module 4 Lesson 27 Exercise Answer Key

Exercises
Determine the nature of the solution to each system of linear equations.

Exercise 1.
3x + 4y = 5
y = – $$\frac{3}{4}$$ x + 1
The slopes of these two distinct equations are the same, which means the graphs of these two equations are parallel lines. Therefore, this system will have no solution.

Exercise 2.
7x + 2y = – 4
x – y = 5
The slopes of these two equations are different. That means the graphs of these two equations are distinct nonparallel lines and will intersect at one point. Therefore, this system has one solution.

Exercise 3.
9x + 6y = 3
3x + 2y = 1
The lines defined by the graph of this system of equations are the same line because they have the same slope and the same y – intercept point.

Determine the nature of the solution to each system of linear equations. If the system has a solution, find it algebraically, and then verify that your solution is correct by graphing.
Exercise 4.
3x + 3y = – 21
x + y = – 7
These equations define the same line. Therefore, this system will have infinitely many solutions.

Exercise 5.
y = $$\frac{3}{2}$$ x – 1
3y = x + 2
The slopes of these two equations are unique. That means they graph as distinct lines and will intersect at one point. Therefore, this system has one solution.
3(y = $$\frac{3}{2}$$ x – 1)
3y = $$\frac{9}{2}$$ x – 3
x + 2 = $$\frac{9}{2}$$ x – 3
2 = $$\frac{7}{2}$$ x – 3
5 = $$\frac{7}{2}$$ x
$$\frac{10}{7}$$ = x
y = $$\frac{3}{2}$$ ($$\frac{10}{7}$$) – 1
y = $$\frac{15}{7}$$ – 1
y = $$\frac{8}{7}$$
The solution is ($$\frac{10}{7}$$, $$\frac{8}{7}$$).

Exercise 6.
x = 12y – 4
x = 9y + 7
The slopes of these two equations are unique. That means they graph as distinct lines and will intersect at one point. Therefore, this system has one solution.
12y – 4 = 9y + 7
3y – 4 = 7
3y = 11
y = $$\frac{11}{3}$$
x = 9($$\frac{11}{3}$$) + 7
x = 33 + 7
x = 40
The solution is (40, $$\frac{11}{3}$$).

Exercise 7.
Write a system of equations with (4, – 5) as its solution.
Answers will vary. Verify that students have written a system of equations where (4, – 5) is a solution to each equation in the system.
Sample solution:

### Eureka Math Grade 8 Module 4 Lesson 27 Problem Set Answer Key

Determine the nature of the solution to each system of linear equations. If the system has a solution, find it algebraically, and then verify that your solution is correct by graphing.
Question 1.
y = $$\frac{3}{7}$$ x – 8
3x – 7y = 1
The slopes of these two equations are the same, and the y – intercept points are different, which means they graph as parallel lines. Therefore, this system will have no solution.

Question 2.
2x – 5 = y
– 3x – 1 = 2y
(2x – 5 = y)2
4x – 10 = 2y

4x – 10 = 2y
– 3x – 1 = 2y

4x – 10 = – 3x – 1
7x – 10 = – 1
7x = 9
x = $$\frac{9}{7}$$
y = 2($$\frac{9}{7}$$) – 5
y = $$\frac{18}{7}$$ – 5
y = – $$\frac{17}{7}$$
The solution is ($$\frac{9}{7}$$, – $$\frac{17}{7}$$).

Question 3.
x = 6y + 7
x = 10y + 2
6y + 7 = 10y + 2
7 = 4y + 2
5 = 4y
$$\frac{5}{4}$$ = y
x = 6($$\frac{5}{4}$$) + 7
x = $$\frac{15}{2}$$ + 7
x = $$\frac{29}{2}$$
The solution is ($$\frac{29}{2}$$, $$\frac{5}{4}$$).

Question 4.
5y = $$\frac{15}{4}$$ x + 25
y = $$\frac{3}{4}$$ x + 5
These equations define the same line. Therefore, this system will have infinitely many solutions.

Question 5.
x + 9 = y
x = 4y – 6
4y – 6 + 9 = y
4y + 3 = y
3 = – 3y
– 1 = y x + 9 = – 1
x = – 10
The solution is ( – 10, – 1).

Question 6.
3y = 5x – 15
3y = 13x – 2
5x – 15 = 13x – 2
– 15 = 8x – 2
– 13 = 8x
– $$\frac{13}{8}$$ = x
3y = 5( – $$\frac{13}{8}$$) – 15
3y = – $$\frac{65}{8}$$ – 15
3y = – $$\frac{185}{8}$$
y = – $$\frac{185}{24}$$
The solution is ( – $$\frac{13}{8}$$, – $$\frac{185}{24}$$).

Question 7.
6x – 7y = $$\frac{1}{2}$$
12x – 14y = 1
These equations define the same line. Therefore, this system will have infinitely many solutions.

Question 8.
5x – 2y = 6
– 10x + 4y = – 14
The slopes of these two equations are the same, and the y – intercept points are different, which means they graph as parallel lines. Therefore, this system will have no solution.

Question 9.
y = $$\frac{3}{2}$$ x – 6
2y = 7 – 4x
2(y = $$\frac{3}{2}$$ x – 6)
2y = 3x – 12

2y = 3x – 12
2y = 7 – 4x
3x – 12 = 7 – 4x
7x – 12 = 7
7x = 19
x = $$\frac{19}{7}$$
y = $$\frac{3}{2}$$ ($$\frac{19}{7}$$) – 6
y = $$\frac{57}{14}$$ – 6
y = – $$\frac{27}{14}$$
The solution is ($$\frac{19}{7}$$, – $$\frac{27}{14}$$).

Question 10.
7x – 10 = y
y = 5x + 12
7x – 10 = 5x + 12
2x – 10 = 12
2x = 22
x = 11
y = 5(11) + 12
y = 55 + 12
y = 67
The solution is (11, 67).

Question 11.
Write a system of linear equations with ( – 3, 9) as its solution.
Answers will vary. Verify that students have written a system of equations where ( – 3, 9) is a solution to each equation in the system. Sample solution:

### Eureka Math Grade 8 Module 4 Lesson 27 Exit Ticket Answer Key

Determine the nature of the solution to each system of linear equations. If the system has a solution, then find it without graphing.
Question 1.
y = $$\frac{1}{2}$$ x + $$\frac{5}{2}$$
x – 2y = 7
The slopes of these two equations are the same, and the y – intercept points are different, which means they graph as parallel lines. Therefore, this system will have no solution.

Question 2.
y = $$\frac{2}{3}$$ x + 4
2y + $$\frac{1}{2}$$ x = 2
The slopes of these two equations are unique. That means they graph as distinct lines and will intersect at one point. Therefore, this system has one solution.
2($$\frac{2}{3}$$ x + 4) + $$\frac{1}{2}$$ x = 2
$$\frac{4}{3}$$ x + 8 + $$\frac{1}{2}$$ x = 2
$$\frac{11}{6}$$ x + 8 = 2
$$\frac{11}{6}$$ x = – 6
x = – $$\frac{36}{11}$$
y = $$\frac{2}{3}$$ ( – $$\frac{36}{11}$$) + 4
y = – $$\frac{24}{11}$$ + 4
y = $$\frac{20}{11}$$
The solution is ( – $$\frac{36}{11}$$, $$\frac{20}{11}$$).

Question 3.
y = 3x – 2
– 3x + y = – 2