Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.7 - Equations - Exercise Set - Page 106: 24


$a.\quad \left\{\begin{array}{l} x\neq-5\\ x\neq 5 \end{array}\right. $ $ b.\quad$ Solution set = $\{7\}$

Work Step by Step

Factor the denominators: $\displaystyle \frac{4}{x+5}+\frac{2}{x-5}=\frac{32}{(x+5)(x-5)}$ $a.$ The denominators in the equation must not be zero: $\left\{\begin{array}{l} x+5\neq 0\\ x-5\neq 0\\ (x+5)(x-5)\neq 0 \end{array}\right\}\Rightarrow\left\{\begin{array}{l} x\neq-5\\ x\neq 5 \end{array}\right. \qquad(*)$ $b.$ Multiply both sides with the LCD,$\quad (x+5)(x-5)$ $(x+5)(x-5) \displaystyle \left[ \frac{4}{x+5}+\frac{2}{x-5} \right]= (x+5)(x-5) \left[\frac{32}{(x+5)(x-5)} \right]\quad$ ... distribute and simplify $ 4(x-5)+2(x+5)=32\quad$ ... distribute and simplify $4x-20+2x+10=32$ $ 6x-10=32\quad$ ... add $10$ to both sides $6x=42$ $ x=7 \quad$ ... satisfies (*), a valid solution. Solution set = $\{7\}$
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