Eureka Math Grade 8 Module 4 Lesson 6 Answer Key

Engage NY Eureka Math 8th Grade Module 4 Lesson 6 Answer Key

Eureka Math Grade 8 Module 4 Lesson 6 Exercise Answer Key

Exercises
Find the value of x that makes the equation true.

Exercise 1.
17-5(2x-9)=-(-6x+10)+4
Answer:
17-5(2x-9)=-(-6x+10)+4
17-10x+45=6x-10+4
62-10x=6x-6
62-10x+10x=6x+10x-6
62=16x-6
62+6=16x-6+6
68=16x
\(\frac{68}{16}\)=\(\frac{16}{16}\) x
\(\frac{68}{16}\) =x
\(\frac{17}{4}\) =x

Exercise 2.
-(x-7)+\(\frac{5}{3}\) =2(x+9)
Answer;
-(x-7)+\(\frac{5}{3}\) =2(x+9)
-x+7+\(\frac{5}{3}\) =2x+18
-x+\(\frac{26}{3}\) =2x+18
-x+x+\(\frac{26}{3}\) =2x+x+18
\(\frac{26}{3}\) =3x+18
\(\frac{26}{3}\) -18=3x+18-18
–\(\frac{28}{3}\) =3x
\(\frac{1}{3}\)∙\(\frac{-28}{3}\) =\(\frac{1}{3}\)∙3x
–\(\frac{28}{9}\) =x

Question 3.
\(\frac{4}{9}\) +4(x-1)=\(\frac{28}{9}\) -(x-7x)+1
Answer;
\(\frac{4}{9}\) +4(x-1)=\(\frac{28}{9}\) -(x-7x)+1
\(\frac{4}{9}\) –\(\frac{4}{9}\) +4(x-1)=\(\frac{28}{9}\) –\(\frac{4}{9}\) -(x-7x)+1
4x-4=2\(\frac{4}{9}\) -x+7x+1
4x-4=\(\frac{33}{9}\) +6x
4x-4+4=\(\frac{33}{9}\) +36/9+6x
4x=\(\frac{69}{9}\) +6x
4x-6x=\(\frac{69}{9}\) +6x-6x
-2x=\(\frac{23}{3}\)
\(\frac{1}{-2}\) ∙-2x=\(\frac{1}{-2}\)∙\(\frac{23}{3}\)
x=-\(\frac{23}{6}\)

Question 4.
5(3x+4)-2x=7x-3(-2x+11)
Answer:
5(3x+4)-2x=7x-3(-2x+11)
15x+20-2x=7x+6x-33
13x+20=13x-33
13x-13x+20=13x-13x-33
20≠-33
This equation has no solution.

Question 5.
7x-(3x+5)-8=\(\frac{1}{2}\) (8x+20)-7x+5
Answer:
7x-(3x+5)-8=\(\frac{1}{2}\) (8x+20)-7x+5
7x-3x-5-8=4x+10-7x+5
4x-13=-3x+15
4x-13+13=-3x+15+13
4x=-3x+28
4x+3x=-3x+3x+28
7x=28
x=4

Question 6.
Write at least three equations that have no solution.
Answer:
Answers will vary. Verify that the equations written have no solution.

Eureka Math Grade 8 Module 4 Lesson 6 Problem Set Answer Key

Students practice using the distributive property to transform equations and solve.

Transform the equation if necessary, and then solve it to find the value of x that makes the equation true.

Question 1.
x-(9x-10)+11=12x+3(-2x+\(\frac{1}{3}\))
x-(9x-10)+11=12x+3(-2x+\(\frac{1}{3}\))
x-9x+10+11=12x-6x+1
-8x+21=6x+1
-8x+8x+21=6x+8x+1
21=14x+1
21-1=14x+1-1
20=14x
\(\frac{20}{14}\)=\(\frac{14}{14}\)
\(\frac{10}{7}\)=x

Question 2.
7x+8(x+\(\frac{1}{4}\) )=3(6x-9)-8
Answer:
7x+8(x+\(\frac{1}{4}\) )=3(6x-9)-8
7x+8x+2=18x-27-8
15x+2=18x-35
15x-15x+2=18x-15x-35
2=3x-35
2+35=3x-35+35
37=3x
\(\frac{37}{3}\) =\(\frac{3}{3}\) x
\(\frac{37}{3}\) =x

Question 3.
-4x-2(8x+1)=-(-2x-10)
Answer:
-4x-2(8x+1)=-(-2x-10)
-4x-16x-2=2x+10
-20x-2=2x+10
-20x+20x-2=2x+20x+10
-2=22x+10
-2-10=22x+10-10
-12=22x
–\(\frac{12}{22}\) =\(\frac{22}{22}\) x
–\(\frac{6}{11}\) =x

Question 4.
11(x+10)=132
Answer:
11(x+10)=132
(\(\frac{1}{11}\) )11(x+10)=(\(\frac{1}{11}\) )132
x+10=12
x+10-10=12-10
x=2

Question 5.
37x+\(\frac{1}{2}\) -(x+\(\frac{1}{4}\) )=9(4x-7)+5
Answer:
37x+\(\frac{1}{2}\) -(x+\(\frac{1}{4}\) )=9(4x-7)+5
37x+\(\frac{1}{2}\) -x-\(\frac{1}{4}\) =36x-63+5
36x+\(\frac{1}{4}\) =36x-58
36x-36x+\(\frac{1}{4}\) =36x-36x-58
\(\frac{1}{4}\) ≠-58
This equation has no solution.

Question 6.
3(2x-14)+x=15-(-9x-5)
Answer:
3(2x-14)+x=15-(-9x-5)
6x-42+x=15+9x+5
7x-42=20+9x
7x-7x-42=20+9x-7x
-42=20+2x
-42-20=20-20+2x
-62=2x
-31=x

Question 7.
8(2x+9)=56
Answer:
8(2x+9)=56
(\(\frac{1}{8}\) )8(2x+9)=(\(\frac{1}{8}\) )56
2x+9=7
2x+9-9=7-9
2x=-2
(\(\frac{1}{2}\) )2x=(\(\frac{1}{2}\) )-2
x=-1

Eureka Math Grade 8 Module 4 Lesson 6 Exit Ticket Answer Key

Transform the equation if necessary, and then solve to find the value of x that makes the equation true.

Question 1.
5x-(x+3)=\(\frac{1}{3}\) (9x+18)-5
Answer:
5x-(x+3)=\(\frac{1}{3}\) (9x+18)-5
5x-x-3=3x+6-5
4x-3=3x+1
4x-3x-3=3x-3x+1
x-3=1
x-3+3=1+3
x=4

Question 2.
5(3x+9)-2x=15x-2(x-5)
Answer:
5(3x+9)-2x=15x-2(x-5)
15x+45-2x=15x-2x+10
13x+45=13x+10
13x-13x+45=13x-13x+10
45≠10
Since 45≠10, the equation has no solution.

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