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 6.3 - Complex Rational Expressions - Exercise Set - Page 435: 16

Answer

$-\frac{(x-1) }{(x+1) }$.

Work Step by Step

The given expression is $=\frac{\frac{1}{x+1}-1}{\frac{1}{x-1}+1}$ Multiply the numerator and the denominator by $(x-1)(x+1)$. $=\frac{(x-1)(x+1)}{(x-1)(x+1)}\cdot \frac{\frac{1}{x+1}-1}{\frac{1}{x-1}+1}$ Use the distributive property. $=\frac{(x-1)(x+1) \cdot \frac{1}{x+1}-(x-1)(x+1) }{(x-1)(x+1) \cdot \frac{1}{x-1}+(x-1)(x+1) }$ Simplify. $=\frac{(x-1) -(x-1)(x+1) }{(x+1)+(x-1)(x+1) }$ Factor out common terms in the numerator and denominator. $=\frac{(x-1) [1-(x+1)] }{(x+1)[1+(x-1)] }$ $=\frac{(x-1) [1-x-1] }{(x+1)[1+x-1] }$ $=\frac{(x-1) [-x] }{(x+1)[x] }$ Cancel common terms. $=\frac{-(x-1) }{(x+1) }$ $=-\frac{(x-1) }{(x+1) }$.
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