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

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

For each problem, show your work, and check that your solution is correct.

Exercise 1.
Solve the linear equation x+x+2+x+4+x+6=-28. State the property that justifies your first step and why you chose it.
The left side of the equation can be transformed from x+x+2+x+4+x+6 to 4x+12 using the commutative and distributive properties. Using these properties decreases the number of terms of the equation. Now we have the equation:
4x+12=-28
4x+12-12=-28-12
4x=-40
$$\frac{1}{4}$$âˆ™4x=-40âˆ™$$\frac{1}{4}$$
x=-10.
The left side of the equation is equal to (-10)+(-10)+2+(-10)+4+(-10)+6, which is -28. Since the left side is equal to the right side, then x=-10 is the solution to the equation.
Note: Students could use the division property in the last step to get the answer.

Exercise 2.
Solve the linear equation 2(3x+2)=2x-1+x. State the property that justifies your first step and why you chose it.
Both sides of equation can be rewritten using the distributive property. I have to use it on the left side to expand the expression. I have to use it on the right side to collect like terms.
The left side is
2(3x+2)=6x+4.
The right side is
2x-1+x=2x+x-1
=3x-1.
The equation is
6x+4=3x-1
6x+4-4=3x-1-4
6x=3x-5
6x-3x=3x-3x-5
(6-3)x=(3-3)x-5
3x=-5
$$\frac{1}{3}$$âˆ™3x=$$\frac{1}{3}$$âˆ™(-5)
x=-$$\frac{5}{3}$$.
The left side of the equation is 2(3x+2). Replacing x with –$$\frac{5}{3}$$ gives 2(3(-$$\frac{5}{3}$$)+2)=2(-5+2)=2(-3)=-6. The right side of the equation is 2x-1+x. Replacing x with –$$\frac{5}{3}$$ gives 2(-$$\frac{5}{3}$$)-1+(-$$\frac{5}{3}$$)=-$$\frac{10}{3}$$-1-$$\frac{5}{3}$$=-6. Since both sides are equal to -6, then x=-$$\frac{5}{3}$$ is a solution to 2(3x+2)=2x-1+x.
Note: Students could use the division property in the last step to get the answer.

Exercise 3.
Solve the linear equation x-9=$$\frac{3}{5}$$ x. State the property that justifies your first step and why you chose it.
I chose to use the subtraction property of equality to get all terms with an x on one side of the equal sign.
x-9=$$\frac{3}{5}$$ x
x-x-9=$$\frac{3}{5}$$ x-x
(1-1)x-9=($$\frac{3}{5}$$-1)x
-9=-$$\frac{2}{5}$$ x
–$$\frac{5}{2}$$âˆ™(-9)=-$$\frac{5}{2}$$âˆ™-$$\frac{2}{5}$$ x
4$$\frac{5}{2}$$=x
The left side of the equation is 4$$\frac{5}{2}$$– $$\frac{18}{2}$$=$$\frac{27}{2}$$. The right side is $$\frac{3}{5}$$âˆ™4$$\frac{5}{2}$$=$$\frac{3}{1}$$âˆ™$$\frac{9}{2}$$=$$\frac{27}{2}$$. Since both sides are equal to the same number, then x=4$$\frac{5}{2}$$ is a solution to x-9=$$\frac{3}{5}$$ x.

Exercise 4.
Solve the linear equation 29-3x=5x+5. State the property that justifies your first step and why you chose it.
I chose to use the addition property of equality to get all terms with an x on one side of the equal sign.
29-3x=5x+5
29-3x+3x=5x+3x+5
29=8x+5
29-5=8x+5-5
24=8x
$$\frac{1}{8}$$âˆ™24=$$\frac{1}{8}$$âˆ™8x
3=x
The left side of the equal sign is 29-3(3)=29-9=20. The right side is equal to 5(3)+5=15+5=20. Since both sides are equal, x=3 is a solution to 29-3x=5x+5.
Note: Students could use the division property in the last step to get the answer.

Exercise 5.
Solve the linear equation $$\frac{1}{3}$$ x-5+171=x. State the property that justifies your first step and why you chose it.
I chose to combine the constants -5 and 171. Then, I used the subtraction property of equality to get all terms with an x on one side of the equal sign.
$$\frac{1}{3}$$ x-5+171=x
$$\frac{1}{3}$$ x+166=x
$$\frac{1}{3}$$ x-$$\frac{1}{3}$$ x+166=x-$$\frac{1}{3}$$ x
166=$$\frac{2}{3}$$ x
166âˆ™$$\frac{3}{2}$$=$$\frac{3}{2}$$âˆ™$$\frac{2}{3}$$ x
83âˆ™3=x
249=x
The left side of the equation is $$\frac{1}{3}$$âˆ™249-5+171=83-5+171=78+171=249, which is exactly equal to the right side. Therefore, x=249 is a solution to $$\frac{1}{3}$$ x-5+171=x.

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

Question 6.
Guess a number for x that would make the equation true. Check your solution.
5x-2=8
When x=2, the left side of the equation is 8, which is the same as the right side. Therefore, x=2 is the solution to the equation.

Question 7.
Use the properties of equality to solve the equation 7x-4+x=12. State which property justifies your first step and why you chose it. Check your solution.
I used the commutative and distributive properties on the left side of the equal sign to simplify the expression to fewer terms.
7x-4+x=12
7x+x-4=12
(7+1)x-4=12
8x-4=12
8x-4+4=12+4
8x=16
$$\frac{8}{8}$$x=$$\frac{16}{8}$$
x=2
The left side of the equation is 7(2)-4+2=14-4+2=12. The right side is also 12. Since the left side equals the right side, x=2 is the solution to the equation.

Question 8.
Use the properties of equality to solve the equation 3x+2-x=11x+9. State which property justifies your first step and why you chose it. Check your solution.
I used the commutative and distributive properties on the left side of the equal sign to simplify the expression to fewer terms.
3x+2-x=11x+9
3x-x+2=11x+9
(3-1)x+2=11x+9
2x+2=11x+9
2x-2x+2=11x-2x+9
(2-2)x+2=(11-2)x+9
2=9x+9
2-9=9x+9-9
-7=9x
–$$\frac{7}{9}$$=$$\frac{9}{9}$$ x
–$$\frac{7}{9}$$=x
The left side of the equation is 3($$\frac{-7}{9}$$)+2-$$\frac{-7}{9}$$=-$$\frac{21}{9}$$+$$\frac{18}{9}$$+$$\frac{7}{9}$$=$$\frac{4}{9}$$. The right side is 11(-$$\frac{7}{9}$$)+9=$$\frac{-77}{9}$$+$$\frac{81}{9}$$=$$\frac{4}{9}$$. Since the left side equals the right side, x=-$$\frac{7}{9}$$ is the solution to the equation.

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

Students solve equations using properties of equality.

For each problem, show your work, and check that your solution is correct.

Question 1.
Solve the linear equation x+4+3x=72. State the property that justifies your first step and why you chose it.
I used the commutative and distributive properties on the left side of the equal sign to simplify the expression to fewer terms.
x+4+3x=72
x+3x+4=72
(1+3)x+4=72
4x+4=72
4x+4-4=72-4
4x=68
$$\frac{4}{4}$$ x=$$\frac{68}{4}$$
x=17
The left side is equal to 17+4+3(17)=21+51=72, which is equal to the right side. Therefore, x=17 is a solution to the equation x+4+3x=72.

Question 2.
Solve the linear equation x+3+x-8+x=55. State the property that justifies your first step and why you chose it.
I used the commutative and distributive properties on the left side of the equal sign to simplify the expression to fewer terms.
x+3+x-8+x=55
x+x+x+3-8=55
(1+1+1)x+3-8=55
3x-5=55
3x-5+5=55+5
3x=60
$$\frac{3}{3}$$ x=$$\frac{60}{3}$$
x=20
The left side is equal to 20+3+20-8+20=43-8+20=35+20=55, which is equal to the right side. Therefore, x=20 is a solution to x+3+x-8+x=55.

Question 3.
Solve the linear equation $$\frac{1}{2}$$ x+10=$$\frac{1}{4}$$ x+54. State the property that justifies your first step and why you chose it.
I chose to use the subtraction property of equality to get all of the constants on one side of the equal sign.
$$\frac{1}{2}$$ x+10=$$\frac{1}{4}$$ x+54
$$\frac{1}{2}$$ x+10-10=$$\frac{1}{4}$$ x+54-10
$$\frac{1}{2}$$ x=$$\frac{1}{4}$$ x+44
$$\frac{1}{2}$$ x-$$\frac{1}{4}$$ x=$$\frac{1}{4}$$ x-$$\frac{1}{4}$$ x+44
$$\frac{1}{4}$$ x=44
4âˆ™$$\frac{1}{4}$$ x=4âˆ™44
x=176
The left side of the equation is $$\frac{1}{2}$$ (176)+10=88+10=98. The right side of the equation is
$$\frac{1}{4}$$ (176)+54=44+54=98. Since both sides equal 98, x=176 is a solution to the equation
$$\frac{1}{2}$$ x+10=$$\frac{1}{4}$$ x+54.

Question 4.
Solve the linear equation $$\frac{1}{4}$$ x+18=x. State the property that justifies your first step and why you chose it.
I chose to use the subtraction property of equality to get all terms with an x on one side of the equal sign.
$$\frac{1}{4}$$ x+18=x
$$\frac{1}{4}$$ x-$$\frac{1}{4}$$ x+18=x-$$\frac{1}{4}$$ x
18=$$\frac{3}{4}$$ x
$$\frac{4}{3}$$âˆ™18=$$\frac{4}{3}$$âˆ™$$\frac{3}{4}$$ x
24=x
The left side of the equation is $$\frac{1}{4}$$ (24)+18=6+18=24, which is what the right side is equal to. Therefore, x=24 is a solution to $$\frac{1}{4}$$ x+18=x.

Question 5.
Solve the linear equation 17-x=$$\frac{1}{3}$$âˆ™15+6. State the property that justifies your first step and why you chose it.
The right side of the equation can be simplified to 11. Then, the equation is
17-x=11,
and x=6. Both sides of the equation equal 11; therefore, x=6 is a solution to the equation 17-x=$$\frac{1}{3}$$âˆ™15+6. I was able to solve the equation mentally without using the properties of equality.

Question 6.
Solve the linear equation $$\frac{x+x+2}{4}$$=189.5. State the property that justifies your first step and why you chose it.
I chose to use the multiplication property of equality to get all terms with an x on one side of the equal sign.
$$\frac{x+x+2}{4}$$=189.5
x+x+2=4(189.5)
2x+2=758
2x+2-2=758-2
2x=756
$$\frac{2}{2}$$ x=$$\frac{756}{2}$$
x=378
The left side of the equation is $$\frac{378+378+2}{4}$$=$$\frac{758}{4}$$=189.5, which is equal to the right side of the equation. Therefore, x=378 is a solution to $$\frac{x+x+2}{4}$$=189.5.

Question 7.
Alysha solved the linear equation 2x-3-8x=14+2x-1. Her work is shown below. When she checked her answer, the left side of the equation did not equal the right side. Find and explain Alyshaâ€™s error, and then solve the equation correctly.
2x-3-8x=14+2x-1
-6x-3=13+2x
-6x-3+3=13+3+2x
-6x=16+2x
-6x+2x=16
-4x=16
$$\frac{-4}{-4}$$ x=$$\frac{16}{-4}$$
x=-4
$$\frac{-8}{-8}$$ x=$$\frac{16}{-8}$$