College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.2: 48

Answer

$x=7$

Work Step by Step

$\dfrac{4}{x+5}+\dfrac{2}{x-5}=\dfrac{32}{x^{2}-25}$ Factor the denominator of the fraction on the right side: $\dfrac{4}{x+5}+\dfrac{2}{x-5}=\dfrac{32}{(x-5)(x+5)}$ Multiply the whole equation by $(x-5)(x+5)$: $(x+5)(x-5)\Big[\dfrac{4}{x+5}+\dfrac{2}{x-5}=\dfrac{32}{(x-5)(x+5)}\Big]$ $4(x-5)+2(x+5)=32$ $4x-20+2x+10=32$ Take all terms without $x$ to the right side, simplify, and solve for $x$: $4x+2x=32-10+20$ $6x=42$ $x=\dfrac{42}{6}$ $x=7$
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