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: 56

Answer

$x=3$

Work Step by Step

We may plug the expressions involving $x$ directly into the given equation $y_{1}+y_{2}=y_{3}$, find a common denominator, and solve: $(\frac{2x-1}{x^2+2x-8})+(\frac{2}{x+4})=(\frac{1}{x-2})$ $\frac{2x-1}{(x-2)(x+4)}+\frac{2}{x+4}=\frac{1}{x-2}$ $\frac{2x-1}{(x-2)(x+4)}+(\frac{x-2}{x-2})(\frac{2}{x+4})=(\frac{x+4}{x+4})(\frac{1}{x-2})$ $\frac{2x-1}{(x-2)(x+4)}+\frac{2x-4}{(x-2)(x+4)}=\frac{x+4}{(x-2)(x+4)}$ $\frac{(2x-1)+(2x-4)}{(x-2)(x+4)}=\frac{x+4}{(x-2)(x+4)}$ $\frac{4x-5}{(x-2)(x+4)}=\frac{x+4}{(x-2)(x+4)}$ $4x-5=x+4$ $4x-x=4+5$ $3x=9$ $x=3$
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