Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Review Exercises - Page 144: 126

Answer

The solution set of the equation is all real numbers except $x=1\text{ and }x=-1$.

Work Step by Step

Consider the equation $\frac{1}{x-1}-\frac{1}{x+1}=\frac{2}{{{x}^{2}}-1}$. The equation is not defined for $x=1\text{ and }x=-1$. First of all, simplify the equation. That is, $\begin{align} & \frac{1}{x-1}-\frac{1}{x+1}=\frac{2}{{{x}^{2}}-1} \\ & \frac{\left( x+1 \right)-\left( x-1 \right)}{\left( x-1 \right)\left( x+1 \right)}=\frac{2}{{{x}^{2}}-1} \\ & \frac{x+1-x+1}{{{x}^{2}}-1}=\frac{2}{{{x}^{2}}-1} \\ & \frac{2}{{{x}^{2}}-1}=\frac{2}{{{x}^{2}}-1} \end{align}$ Now, multiply both sides by ${{x}^{2}}-1$ and get, $\begin{align} & \frac{2}{{{x}^{2}}-1}\cdot \left( {{x}^{2}}-1 \right)=\frac{2}{{{x}^{2}}-1}\cdot \left( {{x}^{2}}-1 \right) \\ & 2=2 \end{align}$ Therefore, all the real numbers except $x=1\text{ and }x=-1$ are solutions for the equation. Hence, the solution set of the equation is all real numbers except $x=1\text{ and }x=-1$.
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