Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 8 - Section 8.3 - Solving Equations by Using Quadratic Methods - Exercise Set - Page 502: 47

Answer

$-3$

Work Step by Step

$\frac{x}{x-1} + \frac{1}{x+1} = \frac{2}{x^{2}-1}$ Using $(a^{2}-b^{2}) = (a+b)(a-b)$ formula, $x^{2}-1 = (x^{2}-1^{2}) = (x+1)(x-1)$ $\frac{x}{x-1} + \frac{1}{x+1} = \frac{2}{(x+1)(x-1)}$ $x=1 $ or $x=-1$ makes the denominator zero. So, $\frac{x}{x-1} + \frac{1}{x+1} = \frac{2}{(x+1)(x-1)} ; $$ x\ne -1, $ or $x \ne 1$ Multiply both sides by $(x+1)(x-1)$ to clear fractions. $(x+1)(x-1)(\frac{x}{x-1} + \frac{1}{x+1} )=(x+1)(x-1)( \frac{2}{(x+1)(x-1)}) ; $$ x\ne -1, $ or $x \ne 1$ $(x+1)x+(x-1) = 2 ; x\ne -1, $ or $x \ne 1$ $x^{2}+x+x-1-2=0 $ $x^{2}+2x-3=0 $ By factoring, $(x+3)(x-1)=0$ $x=-3$ or $x=1$ $x = 1$ makes the denominator zero, so $x=-3$ is the solution.
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