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

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

Example 1.

→ Given a linear equation in disguise, we will try to solve it. To do so, we must first assume that the following equation is true for some number x.

\(\frac{x-1}{2}\) = \(\frac{x+\frac{1}{3}}{4}\)

We want to make this equation look like the linear equations we are used to. For that reason, we will multiply both sides of the equation by 2 and 4, as we normally do with proportions:

2(x+\(\frac{1}{3}\))=4(x-1).

→ Is this a linear equation? How do you know?

→ Yes, this is a linear equation because the expressions on the left and right of the equal sign are linear expressions.

→ Notice that the expressions that contained more than one term were put in parentheses. We do that so we do not make a mistake and forget to use the distributive property.

→ Now that we have a linear equation, we will use the distributive property and solve as usual.

2(x+\(\frac{1}{3}\))=4(x-1)

2x+\(\frac{2}{3}\)=4x-4

2x-2x+\(\frac{2}{3}\)=4x-2x-4

\(\frac{2}{3}\)=2x-4

\(\frac{2}{3}\)+4=2x-4+4

\(\frac{14}{3}\)=2x

\(\frac{1}{2}\)∙\(\frac{14}{3}\)=\(\frac{1}{2}\)∙2x

\(\frac{7}{3}\)=x

→ How can we verify that \(\frac{7}{3}\) is the solution to the equation?

→ We can replace x with \(\frac{7}{3}\) in the original equation.

Since \(\frac{7}{3}\) made the equation true, we know it is a solution to the equation.

Example 2.

→ Can we solve the following equation? Explain.

\(\frac{\frac{1}{5}-x}{7}\) = \(\frac{2 x+9}{3}\)

→ We need to multiply each numerator with the other fraction’s denominator.

→ So,

\(\frac{\frac{1}{5}-x}{7}\) = \(\frac{2 x+9}{3}\)

7(2x+9)=3(\(\frac{1}{5}\)-x).

→ What would be the next step?

→ Use the distributive property.

→ Now we have

7(2x+9)=3(\(\frac{1}{5}\)-x)

14x+63=\(\frac{3}{5}\)-3x

14x+3x+63=\(\frac{3}{5}\)-3x+3x

17x+63=\(\frac{3}{5}\)

17x+63-63=\(\frac{3}{5}\)-63

17x=\(\frac{3}{5}\)–\(\frac{315}{5}\)

17x=-\(\frac{312}{5}\)

\(\frac{1}{17}\) (17x)=(-\(\frac{312}{5}\))\(\frac{1}{17}\)

x=-\(\frac{312}{85}\).

→ Is this a linear equation? How do you know?

→ Yes, this is a linear equation because the left and right side are linear expressions.

Example 3.

Can this equation be solved?

\(\frac{6+x}{7 x+\frac{2}{3}}\)=\(\frac{3}{8}\)

Give students a few minutes to work. Provide support to students as needed.

→ Yes, we can solve the equation because we can multiply each numerator with the other fraction’s denominator and then use the distributive property to begin solving it.

Answer:

\(\frac{6+x}{7 x+\frac{2}{3}}\)=\(\frac{3}{8}\)

(6+x)8=(7x+\(\frac{2}{3}\))3

48+8x=21x+2

48+8x-8x=21x-8x+2

48=13x+2

48-2=13x+2-2

46=13x

\(\frac{46}{13}\)=x

Example 4.

Can this equation be solved?

\(\frac{7}{3 x+9}\)=\(\frac{1}{8}\)

Give students a few minutes to work. Provide support to students as needed.

→ Yes, we can solve the equation because we can multiply each numerator with the other fraction’s denominator and then use the distributive property to begin solving it.

Answer:

\(\frac{7}{3 x+9}\)=\(\frac{1}{8}\)

7(8)=(3x+9)1

56=3x+9

56-9=3x+9-9

47=3x

\(\frac{47}{3}\)=x

Example 5.

In the diagram below, △ABC~ △A’ B’ C’. Using what we know about similar triangles, we can determine the value of x.

→ Begin by writing the ratios that represent the corresponding sides.

Answer:

\(\frac{x-2}{9.5}\) = \(\frac{x+2}{12}\)

It is possible to write several different proportions in this case. If time, discuss this fact with students.

→ Now that we have the ratios, solve for x and find the lengths of \(\overline{A B}\) and \(\overline{A C}\).

Answer:

\(\frac{x-2}{9.5}\) = \(\frac{x+2}{12}\)

(x-2)12=9.5(x+2)

12x-24=9.5x+19

12x-24+24=9.5x+19+24

12x=9.5x+43

12x-9.5x=9.5x-9.5x+43

2.5x=43

x=\(\frac{43}{2.5}\)

x=17.2

|AB|=15.2 cm, and |AC|=19.2 cm.

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

Solve the following equations of rational expressions, if possible.

Question 1.

\(\frac{2 x+1}{9}\)=\(\frac{1-x}{6}\)

Answer:

\(\frac{2 x+1}{9}\)=\(\frac{1-x}{6}\)

9(1-x)=(2x+1)6

9-9x=12x+6

9-9x+9x=12x+9x+6

9=21x+6

9-6=21x+6-6

3=21x

\(\frac{3}{21}\)=\(\frac{21}{2}\)1 x

\(\frac{1}{7}\)=x

Question 2.

\(\frac{5+2 x}{3 x-1}\)=\(\frac{6}{7}\)

Answer:

\(\frac{5+2 x}{3 x-1}\)=\(\frac{6}{7}\)

(5+2x)7=(3x-1)6

35+14x=18x-6

35-35+14x=18x-6-35

14x=18x-41

14x-18x=18x-18x-41

-4x=-41

\(\frac{-4}{-4}\) x=\(\frac{-41}{-4}\)

x=\(\frac{41}{4}\)

Question 3.

\(\frac{x+9}{12}\)=\(\frac{-2 x-\frac{1}{2}}{3}\)

Answer:

\(\frac{x+9}{12}\)=\(\frac{-2 x-\frac{1}{2}}{3}\)

12(-2x-\(\frac{1}{2}\))=(x+9)3

-24x-6=3x+27

-24x+24x-6=3x+24x+27

-6=27x+27

-6-27=27x+27-27

-33=27x

\(\frac{-33}{27}\)=\(\frac{27}{27}\) x

–\(\frac{11}{9}\)=x

Question 8.

\(\frac{8}{3-4 x}\) = \(\frac{5}{2 x+\frac{1}{4}}\)

Answer:

\(\frac{8}{3-4 x}\) = \(\frac{5}{2 x+\frac{1}{4}}\)

8(2x+\(\frac{1}{4}\))=(3-4x)5

16x+2=15-20x

16x+2-2=15-2-20x

16x=13-20x

16x+20x=13-20x+20x

36x=13

\(\frac{36}{36}\) x=\(\frac{13}{36}\)

x=\(\frac{13}{36}\)

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

Students practice solving equations with rational expressions, if a solution is possible.

Solve the following equations of rational expressions, if possible. If an equation cannot be solved, explain why.

Question 1.

\(\frac{5}{6 x-2}\) = \(\frac{-1}{x+1}\)

Answer:

\(\frac{5}{6 x-2}\) = \(\frac{-1}{x+1}\)

5(x+1)=-1(6x-2)

5x+5=-6x+2

5x+5-5=-6x+2-5

5x=-6x-3

5x+6x=-6x+6x-3

11x=-3

x=-\(\frac{3}{11}\)

Question 2.

\(\frac{4-x}{8}\) = \(\frac{7 x-1}{3}\)

Answer:

\(\frac{4-x}{8}\) = \(\frac{7 x-1}{3}\)

8(7x-1)=(4-x)3

56x-8=12-3x

56x-8+8=12+8-3x

56x=20-3x

56x+3x=20-3x+3x

59x=20

\(\frac{59}{59}\) x=\(\frac{20}{59}\)

x=\(\frac{20}{59}\)

Question 3.

\(\frac{3 x}{x+2}\) = \(\frac{5}{9}\)

Answer:

\(\frac{3 x}{x+2}\) = \(\frac{5}{9}\)

9(3x)=(x+2)5

27x=5x+10

27x-5x=5x-5x+10

22x=10

\(\frac{22}{22}\) x=\(\frac{10}{22}\)

x=\(\frac{5}{11}\)

Question 4.

\(\frac{\frac{1}{2} x+6}{3}\) = \(\frac{x-3}{2}\)

Answer:

\(\frac{\frac{1}{2} x+6}{3}\) = \(\frac{x-3}{2}\)

3(x-3)=2(\(\frac{1}{2}\) x+6)

3x-9=x+12

3x-9+9=x+12+9

3x=x+21

3x-x=x-x+21

2x=21

x=\(\frac{21}{2}\)

Question 5.

\(\frac{7-2 x}{6}\) = \(\frac{x-5}{1}\)

Answer:

\(\frac{7-2 x}{6}\) = \(\frac{x-5}{1}\)

6(x-5)=(7-2x)1

6x-30=7-2x

6x-30+30=7+30-2x

6x=37-2x

6x+2x=37-2x+2x

8x=37

\(\frac{8}{8}\) x=\(\frac{37}{8}\)

x=\(\frac{37}{8}\)

Question 6.

\(\frac{2 x+5}{2}\) = \(\frac{3 x-2}{6}\)

Answer:

\(\frac{2 x+5}{2}\) = \(\frac{3 x-2}{6}\)

2(3x-2)=6(2x+5)

6x-4=12x+30

6x-4+4=12x+30+4

6x=12x+34

6x-12x=12x-12x+34

-6x=34

x=-\(\frac{34}{6}\)

x=-\(\frac{17}{3}\)

Question 7.

\(\frac{6 x+1}{3}\) = \(\frac{9-x}{7}\)

Answer:

\(\frac{6 x+1}{3}\) = \(\frac{9-x}{7}\)

(6x+1)7=3(9-x)

42x+7=27-3x

42x+7-7=27-7-3x

42x=20-3x

42x+3x=20-3x+3x

45x=20

\(\frac{45}{45}\)x=\(\frac{20}{45}\)

x=\(\frac{4}{9}\)

Question 8.

\(\frac{\frac{1}{3} x-8}{12}\) = \(\frac{-2-x}{15}\)

Answer:

\(\frac{\frac{1}{3} x-8}{12}\) = \(\frac{-2-x}{15}\)

12(-2-x)=(\(\frac{1}{3}\) x-8)15

-24-12x=5x-120

-24-12x+12x=5x+12x-120

-24=17x-120

-24+120=17x-120+120

96=17x

\(\frac{96}{17}\)=\(\frac{17}{17}\) x

\(\frac{96}{17}\)=x

Question 9.

\(\frac{3-x}{1-x}\)=\(\frac{3}{2}\)

Answer:

\(\frac{3-x}{1-x}\)=\(\frac{3}{2}\)

(3-x)2=(1-x)3

6-2x=3-3x

6-2x+2x=3-3x+2x

6=3-x

6-3=3-3-x

3=-x

-3=x

Question 10.

In the diagram below, △ABC~ △A’ B’ C’. Determine the lengths of \(\overline{A C}\) and \(\overline{B C}\).

Answer:

\(\frac{x+4}{4.5}\) = \(\frac{3 x-2}{9}\)

9(x+4)=4.5(3x-2)

9x+36=13.5x-9

9x+36+9=13.5x-9+9

9x+45=13.5x

9x-9x+45=13.5x-9x

45=4.5x

10=x

|AC|=14 cm, and |BC|=28 cm.

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

Solve the following equations for x.

Question 1.

\(\frac{5 x-8}{3}\) = \(\frac{11 x-9}{5}\)

Answer:

\(\frac{5 x-8}{3}\) = \(\frac{11 x-9}{5}\)

5(5x-8)=3(11x-9)

25x-40=33x-27

25x-25x-40=33x-25x-27

-40=8x-27

-40+27=8x-27+27

-13=8x

–\(\frac{13}{8}\)=x

Question 2.

\(\frac{x+11}{7}\)=\(\frac{2 x+1}{-8}\)

Answer:

\(\frac{x+11}{7}\)=\(\frac{2 x+1}{-8}\)

7(2x+1)=-8(x+11)

14x+7=-8x-88

14x+7-7=-8x-88-7

14x=-8x-95

14x+8x=-8x+8x-95

22x=-95

x=-\(\frac{95}{22}\)

Question 3.

\(\frac{-x-2}{-4}\) = \(\frac{3 x+6}{2}\)

Answer:

\(\frac{-x-2}{-4}\) = \(\frac{3 x+6}{2}\)

-4(3x+6)=2(-x-2)

-12x-24=-2x-4

-12x-24+24=-2x-4+24

-12x=-2x+20

-12x+2x=-2x+2x+20

-10x=20

x=-2