Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 6 - Sections 6.1-6.5 - Integrated Review - Expressions and Equations Containing Rational Expressions - Page 380: 10

Answer

$x=-\dfrac{2}{3}$

Work Step by Step

Expressing the terms of the given equation, $ \dfrac{3}{x^2-25}=\dfrac{1}{x+5}+\dfrac{2}{x-5} ,$ in factored form results to \begin{array}{l}\require{cancel} \dfrac{3}{(x+5)(x-5)}=\dfrac{1}{x+5}+\dfrac{2}{x-5} .\end{array} Multiplying both sides by the $LCD= (x+5)(x-5) $, then the solution to the equation above is \begin{array}{l} 1(3)=(x-5)(1)+(x+5)(2) \\\\ 3=x-5+2x+10 \\\\ -x-2x=-5+10-3 \\\\ -3x=2 \\\\ x=\dfrac{2}{-3} \\\\ x=-\dfrac{2}{3} .\end{array}
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