Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 6 - Section 6.6 - Rational Equations - Exercise Set - Page 463: 28


Solution set = $\{3\}$.

Work Step by Step

The LHS denominator $x^{2}+2x-8$ is factored by searching for two factors of $c=-8$ whose sum is $b=2$ ... we find $+4$ and $-2$ $x^{2}+2x-8=(x-2)(x+4)$ First, we exclude those values of x that yield a zero in any of the denominators. $x\not\in\{-4,2 \}\qquad (*)$ Multiply the equation with the LCD=$(x-2)(x+4)$ $(2x-1)+2(x-2)=x+4\qquad$ ... simplify (distribute) $2x-1+2x-4=x+4$ $ 4x-5=x+4\qquad$ ... add $5-x$ to both sides $3x=9$ $ x=3 \qquad$ ...Checking (*), this is a valid solution Solution set = $\{3\}$.
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