# Eureka Math Algebra 2 Module 1 Lesson 26 Answer Key

## Engage NY Eureka Math Algebra 2 Module 1 Lesson 26 Answer Key

### Eureka Math Algebra 2 Module 1 Lesson 26 Example Answer Key

Example 1.
Solve the following equation: $$\frac{x+3}{12}=\frac{5}{6}$$.
Equating Numerators Method: Obtain expressions on both sides with the same denominator and equate numerators. Thus, x + 3 = 10, and x = 7; therefore, 7 is the solution to our original equation.

Clearing Fractions Method: Multiply both sides by a common multiple of the denominators to clear the fractions, and then solve the resulting equation.
$$12 \cdot\left(\frac{x+3}{12}\right)=12 \cdot\left(\frac{5}{6}\right)$$
x + 3 = 10
We can see, once again, that the solution is 7.

### Eureka Math Algebra 2 Module 1 Lesson 26 Exercise Answer Key

Solve the following equations for x, and give evidence that your solutions are correct.

Exercise 1.
$$\frac{x}{2}+\frac{1}{3}=\frac{5}{6}$$ Exercise 2.
$$\frac{2 x}{9}+\frac{5}{9}=\frac{8}{9}$$ Exercise 3.
Solve the following equation: $$\frac{3}{x}=\frac{8}{x-2}$$.
Method 1: Convert both expressions to equivalent expressions with a common denominator. The common denominator is x(x – 2), so we use the identity property of multiplication to multiply the left side by $$\frac{x-2}{x-2}$$ and the right side by $$\frac{x}{x}$$. This does not change the value of the expression on either side of the equation. Since the denominators are equal, we can see that the numerators must be equal; thus, 3x – 6 = 8x. Solving for x gives a solution of –$$\frac{6}{5}$$. At the outset of this example, we noted that x cannot take on the value of 0 or 2, but there is nothing preventing x from taking on the value –$$\frac{6}{5}$$. Thus, we have found a solution. We can check our work. Substituting –$$\frac{6}{5}$$ into $$\frac{3}{x}$$ gives us $$\frac{3}{(-6 / 5)}$$ = –$$\frac{5}{2}$$ and substituting –$$\frac{6}{5}$$ into $$\frac{8}{x-2}$$ gives us $$\frac{8}{(-6 / 5)-2}$$ = –$$\frac{5}{2}$$. Thus, when x = –$$\frac{6}{5}$$, we have $$\frac{3}{x}$$ = $$\frac{8}{x – 2}$$; therefore, –$$\frac{6}{5}$$ is indeed a solution.

Method 2: Multiply both sides of the equation by the common denominator x(x – 2), and solve the resulting equation.
x(x – 2) $$\left(\frac{3}{x}\right)$$ = x(x – 2) $$\left(\frac{8}{x-2}\right)$$
3(x – 2) = 8x
3x – 6 = 8x
From this point, we follow the same steps as we did in Method 1, and we get the same solution: –$$\frac{6}{5}$$

Exercise 4.
Solve the following equation for a: $$\frac{1}{a+2}+\frac{1}{a-2}=\frac{4}{a^{2}-4}$$
First, we notice that we must have a ≠ 2 and a ≠ – 2. Then, we apply Method 1: Since the denominators are equal, we know that the numerators are equal; thus, we have 2a = 4, which means that a = 2. Thus, the only solution to this equation is 2. However, a is not allowed to be 2 because if a = 2, then $$\frac{1}{a-2}$$ is not defined. This means that the original equation, $$\frac{1}{a+2}+\frac{1}{a-2}=\frac{4}{a^{2}-4}$$, has no solution.

Exercise 5.
Solve the following equation. Remember to check for extraneous solutions.
$$\frac{4}{3 x}+\frac{5}{4}=\frac{3}{x}$$
First, note that we must have x ≠ 0.
Equating numerators: $$\frac{16}{12 x}+\frac{15 x}{12 x}=\frac{36}{12 x}$$
Then, we have 16 + 15x = 36, and the solution is x = $$\frac{4}{3}$$.
Clearing fractions: $$12 x\left(\frac{4}{3 x}+\frac{5}{4}\right)=12 x\left(\frac{3}{x}\right)$$
Then, we have 16 + 15x = 36, and the solution is x = $$\frac{4}{3}$$
The solution $$\frac{4}{3}$$ is valid since the only excluded value is 0.

Exercise 6.
Solve the following equation. Remember to check for extraneous solutions.
$$\frac{7}{b+3}+\frac{5}{b-3}=\frac{10 b-2}{b^{2}-9}$$
First, note that we must have x ≠ 3 and x ≠ – 3.
Equating numerators: $$\frac{7(b-3)}{(b-3)(b+3)}$$ + $$\frac{5(b+3)}{(b-3)(b+3)}$$ = $$\frac{10 b-2}{(b-3)(b+3)}$$
Matching numerators, we hove 7b – 21 + 5b + 15 = 10b – 2, which leads to 2b = 4; therefore, b = 2.
Clearing fractions: (b – 3)(b + 3) $$\left(\frac{7}{b+3}+\frac{5}{b-3}\right)$$ = (b – 3)(b + 3) $$\left(\frac{10 b-2}{b^{2}-9}\right)$$
We have 7(b – 3) + 5(b + 3) = 10b – 2, which leads to 2b = 4; therefore, b = 2.
The solution 2 is valid since the only excluded values are 3 and – 3.

Exercise 7.
Solve the following equation. Remember to check for extraneous solutions.
$$\frac{1}{x-6}+\frac{x}{x-2}=\frac{4}{x^{2}-8 x+12}$$
First, note that we must have x ≠ 6 and x ≠ 2.
Equating numerators:
$$\frac{x-2}{(x-6)(x-2)}$$ + $$\frac{x^{2}-6 x}{(x-6)(x-2)}$$ = $$\frac{4}{(x-6)(x-2)}$$
x2 – 5x – 2 = 4
x2 – 5x – 6 = 0
(x -6) (x + 1) = 0
The solutions are 6 and – 1.
Clearing fractions:
$$\left(\frac{1}{x-6}+\frac{x}{x-2}\right)$$ (x – 6) (x – 2) = $$\left(\frac{4}{(x-6)(x-2)}\right)$$ (x – 6) (x – 2)
(x – 2) + x(x – 6) = 4
– 6x + x – 2 = 4
x2 – 5x -6 = 0
(x – 6) (x + 1) = 0
The solutions are 6 and – 1.
Because x is not allowed to be 6 in order to avoid division by zero, the solution 6 is extraneous; thus, – 1 is the only solution to the given rational equation.

### Eureka Math Algebra 2 Module 1 Lesson 26 Problem Set Answer Key

Question 1.
Solve the following equations, and check for extraneous solutions.
a. $$\frac{x-8}{x-4}$$ = 2
0

b. $$\frac{4 x-8}{x-2}$$ = 4
All real numbers except 2

c. $$\frac{x-4}{x-3}$$ = 1
No solution

d. $$\frac{4 x-8}{x-2}$$ = 3
No solution

e. $$\frac{1}{2 a}-\frac{2}{2 a-3}$$ = 0


f. $$\frac{3}{2 x+1}=\frac{5}{4 x+3}$$
– 2

g. $$\frac{4}{x-5}-\frac{2}{5+x}=\frac{2}{x}$$
$$-\frac{5}{3}$$

h. $$\frac{y+2}{3 y-2}+\frac{y}{y-1}=\frac{2}{3}$$
$$\frac{5}{6}$$, – 2

i. $$\frac{3}{x+1}-\frac{2}{1-x}$$ = 1
0, 5

j. $$\frac{4}{x-1}+\frac{3}{x}$$ – 3 = 0
$$\frac{1}{3}$$, 3

k. $$\frac{x+1}{x+3}-\frac{x-5}{x+2}=\frac{17}{6}$$
0, $$-\frac{55}{17}$$

l. $$\frac{x+7}{4}-\frac{x+1}{2}=\frac{5-x}{3 x-14}$$
5, 6

m. $$\frac{b^{2}-b-6}{b^{2}}-\frac{2 b+12}{b}=\frac{b-39}{2 b}$$
3, $$\frac{4}{3}$$

n. $$\frac{1}{p(p-4)}+1=\frac{p-6}{p}$$
$$\frac{23}{6}$$

o. $$\frac{1}{a+3}=\frac{h+4}{h-2}+\frac{6}{h-2}$$
– 8, – 4

p. $$\frac{m+5}{m^{2}+m}=\frac{1}{m^{2}+m}-\frac{m-6}{m+1}$$
4, 1

Question 2.
Create and solve a rational equation that has 0 as an extraneous solution.
One such equation is $$\frac{1}{x-1}+\frac{1}{x}=\frac{1}{x-x^{2}}$$.

Question 3.
Create and solve a rational equation that has 2 as an extraneous solution.
One such equation is $$\frac{1}{x-2}+\frac{1}{x+2}=\frac{4}{x^{2}-4}$$.

Extension:

Question 4.
Two lengths a and b, where a > b, are in golden ratio if the ratio of a + b is to a is the same as a is to b. Symbolically, this is expressed as $$\frac{a}{b}=\frac{a+b}{a}$$. We denote this common ratio by the Greek letter phi (pronounced “fee”) with symbol φ, so that If a and b are in common ratio, then φ = $$\frac{a}{b}=\frac{a+b}{a}$$. By setting b = 1, we find that φ = a and φ is the positive number that satisfies the equation φ = $$\frac{\varphi+1}{\varphi}$$ Solve this equation to find the numerical value for φ. We can apply either method from the previous lesson to solve this equation.
φ = $$\frac{\varphi+1}{\varphi}$$
φ2 = φ + 1
φ2 = φ + 1
φ2 – φ – 1 = 0
Applying the quadratic formula, we have two solutions:
φ = $$\frac{1+\sqrt{5}}{2}$$ or φ = $$\frac{1-\sqrt{5}}{2}$$
Since φ is a positive number, and $$\frac{1-\sqrt{5}}{2}$$ < 0, we have φ = $$\frac{1+\sqrt{5}}{2}$$

Question 5.
Remember that if we use x to represent an integer, then the next integer can be represented by x + 1.
a. Does there exist a pair of consecutive integers whose reciprocals sum to $$\frac{5}{6}$$? Explain how you know.
Yes, 2 and 3 because $$\frac{1}{2}+\frac{1}{3}=\frac{3}{6}+\frac{2}{6}=\frac{5}{6}$$

b. Does there exist a pair of consecutive integers whose reciprocals sum to $$\frac{3}{4}$$? Explain how you know.
If x represents the first integer, then x + 1 represents the next integer. Suppose $$\frac{1}{x}+\frac{1}{x+1}=\frac{3}{4}$$. Then,
$$\frac{1}{x}+\frac{1}{x+1}=\frac{3}{4}$$
$$\frac{4(x+1)+4 x}{4 x(x+1)}=\frac{3 x(x+1)}{4 x(x+1)}$$
8x + 4 = 3x2 + 3x
3x2 – 5x – 4 = 0.
The solutions to this quadratic equation are $$\frac{5+\sqrt{73}}{6}$$ and $$\frac{5-\sqrt{73}}{6}$$, so there are no integers that solve this equation. Thus, there are no pairs of consecutive integers whose reciprocals sum to $$\frac{3}{4}$$.

c. Does there exist a pair of consecutive even integers whose reciprocals sum to $$\frac{3}{4}$$? Explain how you know.
If x represents the first integer, then x + 2 represents the next even integer. Suppose $$\frac{1}{x}+\frac{1}{x+2}=\frac{3}{4}$$. Then,
$$\frac{1}{x}+\frac{1}{x+2}=\frac{3}{4}$$
$$\frac{4(x+2)+4 x}{4 x(x+2)}=\frac{3 x(x+2)}{4 x(x+2)}$$
8x + 8 = 3x2 + 6x
3x2 – 2x – 8 = 0.
The solutions to this quadratic equation are –$$\frac{4}{3}$$ and 2; therefore, the only even integer x that solves the equation is 2. Then, 2 and 4 are consecutive even integers whose reciprocals sum to $$\frac{3}{4}$$.

d. Does there exist a pair of consecutive even integers whose reciprocals sum to $$\frac{5}{6}$$? Explain how you know.
If x represents the first integer, then x + 2 represents the next even integer. Suppose $$\frac{1}{x}+\frac{1}{x+2}=\frac{5}{6}$$. Then,
$$\frac{1}{x}+\frac{1}{x+2}=\frac{5}{6}$$
$$\frac{6(x+2)+6 x}{6 x(x+2)}=\frac{5 x(x+2)}{6 x(x+2)}$$
12x + 12 = 5x2 + 10x
5x2 – 2x – 12 = 0.
The solutions to this quadratic equation are $$\frac{1+\sqrt{61}}{5}$$ and $$\frac{1+\sqrt{61}}{5}$$, so there are no integers that solve this equation. Thus, there are no pairs of consecutive even integers whose reciprocals sum to $$\frac{5}{6}$$.

### Eureka Math Algebra 2 Module 1 Lesson 26 Exit Ticket Answer Key

Find all solutions to the following equation. If there are any extraneous solutions, identify them and explain why they are extraneous.
$$\frac{7}{b+3}+\frac{5}{b-3}=\frac{10 b}{b^{2}-9}$$
Using the equating numerators method: $$\frac{7(b-3)}{(b-3)(b+3)}+\frac{5(b+3)}{(b-3)(b+3)}=\frac{10 b}{(b-3)(b+3)}$$
However, since the excluded values are 3 and – 3, the solution 3 is an extraneous solution, and there is no solution to $$\frac{7}{b+3}+\frac{5}{b-3}=\frac{10 b}{b^{2}-9}$$.