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: 25

Answer

Solution set = $\emptyset$. (no solution)

Work Step by Step

First, we exclude those values of x that yield a zero in any of the denominators. $x\not\in\{-1,+1 \}\qquad (*)$ The last denominator $x^{2}-1$ is factored as $(x-1)(x+1)$ Multiply the equation with the LCD=$(x-1)(x+1)$ $ 1(x+1)+1(x-1)=2\qquad$ ... simplify (distribute) $x+1+x-1=2$ $2x=2 $ $ x=1 \qquad$ ...Checking (*), this is not a valid solution Solution set = $\emptyset$. (no solution)
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