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 462: 16


Solution set = $\{-10\}$.

Work Step by Step

First, we exclude those values of x that yield a zero in any of the denominators. $x\not\in\{-1,2 \}\qquad (*)$ Multiply the equation with the LCD=$(x+1)(x-2)$ $ 3x(x-2)+4(x+1)=3(x+1)(x-2)\qquad$ ... simplify (distribute) $3x^{2}-6x+4x+4=3(x^{2}-x-2)$ $ 3x^{2}-2x+4=3x^{2}-3x-6\qquad$ ... add $-4-3x^{2}+3x$ $-2x+3x=-6-4$ $ x=-10 \qquad$ ...Checking (*), this is a valid solution Solution set = $\{-10\}$.
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