## 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.
$$\frac{6+x}{7 x+\frac{2}{3}}$$=$$\frac{3}{8}$$
(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.
$$\frac{7}{3 x+9}$$=$$\frac{1}{8}$$
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.
$$\frac{x-2}{9.5}$$ = $$\frac{x+2}{12}$$

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 $$\overline{A C}$$.
$$\frac{x-2}{9.5}$$ = $$\frac{x+2}{12}$$
(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.
$$\frac{2 x+1}{9}$$=$$\frac{1-x}{6}$$
$$\frac{2 x+1}{9}$$=$$\frac{1-x}{6}$$
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.
$$\frac{5+2 x}{3 x-1}$$=$$\frac{6}{7}$$
$$\frac{5+2 x}{3 x-1}$$=$$\frac{6}{7}$$
(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.
$$\frac{x+9}{12}$$=$$\frac{-2 x-\frac{1}{2}}{3}$$
$$\frac{x+9}{12}$$=$$\frac{-2 x-\frac{1}{2}}{3}$$
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.
$$\frac{8}{3-4 x}$$ = $$\frac{5}{2 x+\frac{1}{4}}$$
$$\frac{8}{3-4 x}$$ = $$\frac{5}{2 x+\frac{1}{4}}$$
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.
$$\frac{5}{6 x-2}$$ = $$\frac{-1}{x+1}$$
$$\frac{5}{6 x-2}$$ = $$\frac{-1}{x+1}$$
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.
$$\frac{4-x}{8}$$ = $$\frac{7 x-1}{3}$$
$$\frac{4-x}{8}$$ = $$\frac{7 x-1}{3}$$
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.
$$\frac{3 x}{x+2}$$ = $$\frac{5}{9}$$
$$\frac{3 x}{x+2}$$ = $$\frac{5}{9}$$
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.
$$\frac{\frac{1}{2} x+6}{3}$$ = $$\frac{x-3}{2}$$
$$\frac{\frac{1}{2} x+6}{3}$$ = $$\frac{x-3}{2}$$
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.
$$\frac{7-2 x}{6}$$ = $$\frac{x-5}{1}$$
$$\frac{7-2 x}{6}$$ = $$\frac{x-5}{1}$$
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.
$$\frac{2 x+5}{2}$$ = $$\frac{3 x-2}{6}$$
$$\frac{2 x+5}{2}$$ = $$\frac{3 x-2}{6}$$
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.
$$\frac{6 x+1}{3}$$ = $$\frac{9-x}{7}$$
$$\frac{6 x+1}{3}$$ = $$\frac{9-x}{7}$$
(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.
$$\frac{\frac{1}{3} x-8}{12}$$ = $$\frac{-2-x}{15}$$
$$\frac{\frac{1}{3} x-8}{12}$$ = $$\frac{-2-x}{15}$$
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.
$$\frac{3-x}{1-x}$$=$$\frac{3}{2}$$
$$\frac{3-x}{1-x}$$=$$\frac{3}{2}$$
(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 $$\overline{B C}$$.

$$\frac{x+4}{4.5}$$ = $$\frac{3 x-2}{9}$$
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.
$$\frac{5 x-8}{3}$$ = $$\frac{11 x-9}{5}$$
$$\frac{5 x-8}{3}$$ = $$\frac{11 x-9}{5}$$
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.
$$\frac{x+11}{7}$$=$$\frac{2 x+1}{-8}$$
$$\frac{x+11}{7}$$=$$\frac{2 x+1}{-8}$$
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.
$$\frac{-x-2}{-4}$$ = $$\frac{3 x+6}{2}$$
$$\frac{-x-2}{-4}$$ = $$\frac{3 x+6}{2}$$