# Eureka Math Algebra 2 Module 1 Lesson 24 Answer Key

## Engage NY Eureka Math Algebra 2 Module 1 Lesson 24 Answer Key

### Eureka Math Algebra 2 Module 1 Lesson 24 Example Answer Key

Example 1.
Make a conjecture about the product $$\frac{x^{3}}{4 y} \cdot \frac{y^{2}}{x}$$. What will it be? Explain your conjecture, and give evidence that it is correct.
→ We begin by multiplying the numerators and denominators.
$$\frac{x^{3}}{4 y} \cdot \frac{y^{2}}{x}=\frac{x^{3} y^{2}}{4 y x}$$

→ Identify the greatest common factor (GCF) of the numerator and denominator. The GCF of x3y2 and 4xy is xy.
$$\frac{x^{3}}{4 y} \cdot \frac{y^{2}}{x}=\frac{(x y) x^{2} y}{4(x y)}$$

→ Finally, we divide the common factor xy from the numerator and denominator to find the reduced form of the product:
$$\frac{x^{3}}{4 y} \cdot \frac{y^{2}}{x}=\frac{x^{2} y}{4}$$

Note that the phrases “cancel xy” or “cancel the common factor” are intentionally avoided in this lesson. The goal is to highlight that it is division that allows these expressions to be simplified. Ambiguous words like “cancel” can lead students to simplify $$\frac{\sin (x)}{x}$$ to sin—they “canceled” the x!

It is important to understand why the numerator and denominator may be divided by x. The rule $$\frac{n a}{n b}=\frac{a}{b}$$ works for rational expressions as well. Performing a simplification such as $$\frac{x}{x^{3} y}=\frac{1}{x^{2} y}$$ requires doing the following steps: Example 2.
Find the following product: $$\left(\frac{3 x-6}{2 x+6}\right) \cdot\left(\frac{5 x+15}{4 x+8}\right)$$
First, factor the numerator and denominator of each rational expression.
→ Identify any common factors in the numerator and denominator. The GCF of the numerator and denominator is x + 3
Then, divide the common factor (x + 3) from the numerator and denominator, and obtain the reduced form of the product. Example 3.
Find the quotient and reduce to lowest terms: $$\frac{x^{2}-4}{3 x} \div \frac{x-2}{2 x}$$
→ First, we change the division of $$\frac{x^{2}-4}{3 x}$$ by $$\frac{x-2}{2 x}$$ into multiplication of $$\frac{x^{2}-4}{3 x}$$ by the multiplicative inverse of $$\frac{x-2}{2 x}$$. → Then, we perform multiplication as in the previous examples and exercises. That ¡s, we factor the numerator and denominator and divide any common factors present in both the numerator and denominator. ### Eureka Math Algebra 2 Module 1 Lesson 24 Exercise Answer Key

Exercise 1.
Summarize what you have learned so far with your neighbor

Exercise 2.
Find the following product and reduce to lowest terms: $$\left(\frac{2 x+6}{x^{2}+x-6}\right) \cdot\left(\frac{x^{2}-4}{2 x}\right)$$. The factors 2, x + 3, and x – 2 can be divided from the numerator and the denominator In order to reduce the rational expression to lowest terms. Exercise 3.
Find the following product and reduce to lowest terms: $$\left(\frac{4 n-12}{3 m+6}\right)^{-2} \cdot\left(\frac{n^{2}-2 n-3}{m^{2}+4 m+4}\right)$$. Exercise 4.
Find the quotient and reduce to lowest terms: $$\frac{x^{2}-5 x+6}{x+4}$$ ÷ $$\frac{x^{2}-9}{x^{2}+5 x+4}$$. Exercise 5.
Simplify the rational expression.  ### Eureka Math Algebra 2 Module 1 Lesson 24 Problem Set Answer Key

Question 1.
Perform the following operations:
a. Multiply $$\frac{1}{3}$$(x – 2) by 9.
3x – 6

b. Divide $$\frac{1}{4}$$ (x – 8) by $$\frac{1}{12}$$.
3x – 24

c. Multiply $$\frac{1}{4}$$ ($$\frac{1}{3}$$x + 2) by 12.
x + 6

d. Divide $$\frac{1}{3}\left(\frac{2}{5} x-\frac{1}{5}\right)$$ by $$\frac{1}{15}$$.
2x – 1

e. Multiply $$\frac{2}{3}\left(2 x+\frac{2}{3}\right)$$ by $$\frac{9}{4}$$.
3x + 1

f. Multiply 0.03 (4 – x) by 100.
12 – 3x

Question 2.
Write each rational expression as an equivalent rational expression in lowest terms.
a. $$\left(\frac{a^{3} b^{2}}{c^{2} d^{2}} \cdot \frac{c}{a b}\right) \div \frac{a}{c^{2} d^{3}}$$
abcd

b. $$\frac{a^{2}+6 a+9}{a^{2}-9} \cdot \frac{3 a-9}{a+3}$$
3

c. $$\frac{6 x}{4 x-16} \div \frac{4 x}{x^{2}-16}$$
$$\frac{3(x+4)}{8}$$

d. $$\frac{3 x^{2}-6 x}{3 x+1} \cdot \frac{x+3 x^{2}}{x^{2}-4 x+4}$$
$$\frac{3 x^{2}}{x-2}$$

e. $$\frac{2 x^{2}-10 x+12}{x^{2}-4} \cdot \frac{2+x}{3-x}$$
– 2

f. $$\frac{a-2 b}{a+2 b}$$ ÷ (4b2 – a2)
$$-\frac{1}{(a+2 b)^{2}}$$

g. $$\frac{d+c}{c^{2}+d^{2}} \div \frac{c^{2}-d^{2}}{d^{2}-d c}$$
$$-\frac{d}{c^{2}+d^{2}}$$

h. $$\frac{12 a^{2}-7 a b+b^{2}}{9 a^{2}-b^{2}} \div \frac{16 a^{2}-b^{2}}{3 a b+b^{2}}$$
$$\frac{b}{4 a+b}$$

i. $$\left(\frac{x-3}{x^{2}-4}\right)^{-1} \cdot\left(\frac{x^{2}-x-6}{x-2}\right)$$
(x + 2)2

j. $$\left(\frac{x-2}{x^{2}+1}\right)^{-3} \div\left(\frac{x^{2}-4 x+4}{x^{2}-2 x-3}\right)$$
$$\frac{(x-3)(x+1)\left(x^{2}+1\right)^{3}}{(x-2)^{5}}$$

k. $$\frac{6 x^{2}-11 x-10}{6 x^{2}-5 x-6} \cdot \frac{6-4 x}{25-20 x+4 x^{2}}$$
$$-\frac{2}{2 x-5}, \text { or } \frac{2}{5-2 x}$$

l. $$\frac{3 x^{3}-3 a^{2} x}{x^{2}-2 a x+a^{2}} \cdot \frac{a-x}{a^{3} x+a^{2} x^{2}}$$
$$-\frac{3}{a^{2}}$$

Question 3.
Write each rational expression as an equivalent rational expression in lowest terms.
a. $$\frac{2}{5 a^{2} b}$$

b. $$\frac{x-6}{(x+2)\left(x^{2}-1\right)}$$

c. $$\frac{1}{x-2}$$

Question 4.
Suppose that x = $$\frac{t^{2}+3 t-4}{3 t^{2}-3}$$ and y = $$\frac{t^{2}+2 t-8}{2 t^{2}-2 t-4}$$ , ,for t ≠ 1, t ≠ – 1, t ≠ 2, and t ≠ – 4. Show that the value of x2y-2 does not depend on the value of t. Since x2y-2 = $$\frac{4}{9}$$ the value of x2y-2 does not depend on t.

Question 5.
Determine which of the following numbers is larger without using a calculator, $$\frac{15^{16}}{16^{15}}$$ or $$\frac{20^{24}}{24^{20}}$$. (Hint: We can compare two positive quantities a and b by computing the quotient $$\frac{a}{b}$$. If $$\frac{a}{b}$$ > 1, then a > b. Likewise, if 0 < $$\frac{a}{b}$$ < 1,then a < b.) Extension:

Question 6.
One of two numbers can be represented by the rational expression $$\frac{x-2}{x}$$, where x ≠ 0 and x ≠ 2.
a. Find a representation of the second number if the product of the two numbers is 1.
Let the second number be y. Then $$\left(\frac{x-2}{x}\right)$$ ∙ y = 1, so we have
y = 1 ÷ $$\left(\frac{x-2}{x}\right)$$
= 1 ∙ $$\left(\frac{x}{x-2}\right)$$
= $$\frac{x}{x-2}$$

b. Find a representation of the second number if the product of the two numbers is 0.
Let the second number be z. Then $$\left(\frac{x-2}{x}\right)$$ ∙ z = 0, so we have
z = 0 ÷ $$\left(\frac{x-2}{x}\right)$$
= 0 ∙ $$\left(\frac{x}{x-2}\right)$$
= 0.

### Eureka Math Algebra 2 Module 1 Lesson 24 Exit Ticket Answer Key

Perform the indicated operations, and reduce to lowest terms.

Question 1.    