## Engage NY Eureka Math 8th Grade Module 4 Lesson 3 Answer Key

### Eureka Math Grade 8 Module 4 Lesson 3 Exercise Answer Key

Exercise 1.

Is the equation a true statement when x=-3? In other words, is -3 a solution to the equation 6x+5=5x+8+2x? Explain.

Answer:

If we replace x with the number -3, then the left side of the equation is

6∙(-3)+5=-18+5

=-13″,”

and the right side of the equation is

5∙(-3)+8+2∙(-3)=-15+8-6

=-7-6

=-13.

Since -13=-13, then x=-3 is a solution to the equation 6x+5=5x+8+2x.

Note: Some students may have transformed the equation.

Exercise 2.

Does x=12 satisfy the equation 16-\(\frac{1}{2}\) x=\(\frac{3}{4}\) x+1? Explain.

Answer:

If we replace x with the number 12, then the left side of the equation is

16-\(\frac{1}{2}\) x=16-\(\frac{1}{2}\)∙(12)

=16-6

=10,

and the right side of the equation is

\(\frac{3}{4}\) x+1=\(\frac{3}{4}\)∙(12)+1

=9+1

=10.

Since 10=10, then x=12 is a solution to the equation 16-\(\frac{1}{2}\) x=\(\frac{3}{4}\) x+1.

Exercise 3.

Chad solved the equation 24x+4+2x=3(10x-1) and is claiming that x=2 makes the equation true. Is Chad correct? Explain.

Answer:

If we replace x with the number 2, then the left side of the equation is

24x+4+2x=24∙2+4+2∙2

=48+4+4

=56,

and the right side of the equation is

3(10x-1)=3(10∙2-1)

=3(20-1)

=3(19)

=57.

Since 56≠57, then x=2 is not a solution to the equation 24x+4+2x=3(10x-1), and Chad is not correct.

Exercise 4.

Lisa solved the equation x+6=8+7x and claimed that the solution is x=-\(\frac{1}{3}\). Is she correct? Explain.

Answer:

If we replace x with the number –\(\frac{1}{3}\), then the left side of the equation is

x+6=-\(\frac{1}{3}\)+6

=5 \(\frac{2}{3}\),

and the right side of the equation is

8+7x=8+7∙(-\(\frac{1}{3}\))

=8-\(\frac{7}{3}\)

=\(\frac{24}{3}\)–\(\frac{7}{3}\)

=\(\frac{17}{3}\).

Since 5 \(\frac{2}{3}\)=\(\frac{17}{3}\), then x=-\(\frac{1}{3}\) is a solution to the equation x+6=8+7x, and Lisa is correct.

Exercise 5.

Angel transformed the following equation from 6x+4-x=2(x+1) to 10=2(x+1). He then stated that the solution to the equation is x=4. Is he correct? Explain.

Answer:

No, Angel is not correct. He did not transform the equation correctly. The expression on the left side of the equation 6x+4-x=2(x+1) would transform to

6x+4-x=6x-x+4

=(6-1)x+4

=5x+4.

If we replace x with the number 4, then the left side of the equation is

5x+4=5∙4+4

=20+4

=24,

and the right side of the equation is

2(x+1)=2(4+1)

=2(5)

=10.

Since 24≠10, then x=4 is not a solution to the equation 6x+4-x=2(x+1), and Angel is not correct.

Exercise 6.

Claire was able to verify that x=3 was a solution to her teacher’s linear equation, but the equation got erased from the board. What might the equation have been? Identify as many equations as you can with a solution of x=3.

Answer;

Answers will vary. Ask students to share their equations and justifications as to how they knew x=3 would make a true number sentence.

Exercise 7.

Does an equation always have a solution? Could you come up with an equation that does not have a solution?

Answer:

Answers will vary. Expect students to write equations that are false. Ask students to share their equations and justifications as to how they knew the equation they wrote did not have a solution. The concept of “no solution” is introduced in Lesson 6 and solidified in Lesson 7.

### Eureka Math Grade 8 Module 4 Lesson 3 Problem Set Answer Key

Students practice determining whether or not a given number is a solution to the linear equation.

Question 1.

Given that 2x+7=27 and 3x+1=28, does 2x+7=3x+1? Explain.

Answer:

No, because a linear equation is a statement about equality. We are given that 2x+7=27, but 3x+1=28. Since each linear expression is equal to a different number, 2x+7≠3x+1.

Question 2.

Is -5 a solution to the equation 6x+5=5x+8+2x? Explain.

Answer:

If we replace x with the number -5, then the left side of the equation is

6∙(-5)+5=-30+5

=-25,

and the right side of the equation is

5∙(-5)+8+2∙(-5)=-25+8-10

=-17-10

=-27.

Since -25≠-27, then -5 is not a solution of the equation 6x+5=5x+8+2x.

Note: Some students may have transformed the equation.

Question 3.

Does x=1.6 satisfy the equation 6-4x=-\(\frac{x}{4}\)? Explain.

Answer:

If we replace x with the number 1.6, then the left side of the equation is

6-4∙1.6=6-6.4

=-0.4,

and the right side of the equation is

–\(\frac{-1.6}{4}\)=-0.4.

Since -0.4=-0.4, then x=1.6 is a solution of the equation 6-4x=-\(\frac{x}{4}\).

Question 4.

Use the linear equation 3(x+1)=3x+3 to answer parts (a)–(d).

a. Does x=5 satisfy the equation above? Explain.

Answer:

If we replace x with the number 5, then the left side of the equation is

3(5+1)=3(6)

=18,

and the right side of the equation is

3x+3=3∙5+3

=15+3

=18.

Since 18=18, then x=5 is a solution of the equation 3(x+1)=3x+3.

b. Is x=-8 a solution of the equation above? Explain.

Answer:

If we replace x with the number -8, then the left side of the equation is

3(-8+1)=3(-7)

=-21,

and the right side of the equation is

3x+3=3∙(-8)+3

=-24+3

=-21.

Since -21=-21, then x=-8 is a solution of the equation 3(x+1)=3x+3.

c. Is x=\(\frac{1}{2}\) a solution of the equation above? Explain.

Answer:

If we replace x with the number \(\frac{1}{2}\), then the left side of the equation is

3(\(\frac{1}{2}\)+1)=3(\(\frac{1}{2}\)+\(\frac{2}{2}\))

=3(\(\frac{3}{2}\))

=\(\frac{9}{2}\),

and the right side of the equation is

3x+3=3∙(\(\frac{1}{2}\))+3

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

=\(\frac{3}{2}\)+\(\frac{6}{2}\)

=\(\frac{9}{2}\).

Since \(\frac{9}{2}\)=\(\frac{9}{2}\), then x=\(\frac{1}{2}\) is a solution of the equation 3(x+1)=3x+3.

d. What interesting fact about the equation 3(x+1)=3x+3 is illuminated by the answers to parts (a), (b), and (c)? Why do you think this is true?

Answer:

Note to teacher: Ideally, students will notice that the equation 3(x+1)=3x+3 is an identity under the distributive law. The purpose of this problem is to prepare students for the idea that linear equations can have more than one solution, which is a topic of Lesson 7.

### Eureka Math Grade 8 Module 4 Lesson 3 Exit Ticket Answer Key

Question 1.

Is 8 a solution to \(\frac{1}{2}\) x+9=13? Explain.

Answer:

If we replace x with the number 8, then the left side is \(\frac{1}{2}\) (8)+9=4+9=13, and the right side is 13. Since 13=13, then x=8 is a solution.

Question 2.

Write three different equations that have x=5 as a solution.

Answer:

Answers will vary. Accept equations where x=5 makes a true number sentence.

Question 3.

Is -3 a solution to the equation 3x-5=4+2x? Explain.

Answer:

If we replace x with the number -3, then the left side is 3(-3)-5=-9-5=-14. The right side is 4+2(-3)=4-6=-2. Since -14≠-2, then -3 is not a solution of the equation.