## Precalculus (6th Edition)

$\color{blue}{\dfrac{2}{r+2}}$
Use the rule $\dfrac{a}{b} \div \dfrac{c}{d}= \dfrac{a}{b} \cdot \dfrac{d}{c}$ to obtain: $=\dfrac{6r-18}{9r^2+6r-24} \cdot \dfrac{12r-16}{4r-12}$ Factor each polynomial, then cancel common factors to obtain: $\require{cancel} =\dfrac{6(r-3)}{3(3r^2+2r-8)} \cdot \dfrac{4(3r-4)}{4(r-3)} \\=\dfrac{6(r-3)}{3(3r-4)(r+2)} \cdot \dfrac{4(3r-4)}{4(r-3)} \\=\dfrac{\cancel{6}^2\cancel{(r-3)}}{\cancel{3}\cancel{(3r-4)}(r+2)} \cdot \dfrac{\cancel{4}\cancel{(3r-4)}}{\cancel{4}\cancel{(r-3)}} \\=\color{blue}{\dfrac{2}{r+2}}$