## Precalculus (6th Edition)

Published by Pearson

# Chapter R - Review of Basic Concepts - R.5 Rational Expressions - R.5 Exercises: 35

#### Answer

$\color{blue}{\dfrac{2}{9}}$

#### Work Step by Step

Use the rule $\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c}$ to obtain: $=\dfrac{2k+8}{6} \times \dfrac{2}{3k+12}$ Factor the binomials then cancel common factors to obtain: $\require{cancel} =\dfrac{2(k+4)}{3(2)} \times \dfrac{2}{3(k+4)} \\=\dfrac{\cancel{2}\cancel{(k+4)}}{3\cancel{(2)}} \times \dfrac{2}{3\cancel{(k+4)}} \\=\dfrac{1}{3} \times \dfrac{2}{3} \\=\color{blue}{\dfrac{2}{9}}$

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