## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter P - Section P.6 - Rational Expressions - Exercise Set - Page 84: 23

#### Answer

$= \displaystyle \frac{7}{9},\qquad x\neq-1$

#### Work Step by Step

Division with $\displaystyle \frac{a}{b}$ equals multiplication with $\displaystyle \frac{b}{a}.$ (neither a or b can be 0) $\displaystyle \frac{x+1}{3}\div\frac{3x+3}{7}==\frac{x+1}{3}\cdot\frac{7}{3x+3}$ ... Factor what we can $=\displaystyle \frac{(x+1)}{3}\cdot\frac{7}{3(x+1)}$ ...exclude the values that yield 0 in the denominator: $x\neq-1$ ... cancel common factors $= \displaystyle \frac{7}{9},\qquad x\neq-1$

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