Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.5 Rational Expressions - R.5 Exercises: 73

Answer

$\dfrac{-1}{x+1}$

Work Step by Step

Simplify by multiplying the LCD (which is $y$) to both the numerator and the denominator to obtain: $\require{cancel} =\dfrac{x(x+1)\left(\frac{1}{x+1}-\frac{1}{x}\right)}{x(x+1)(\frac{1}{x})} \\=\dfrac{x(x+1)(\frac{1}{x+1})-x(x+1)(\frac{1}{x})}{x(x+1)(\frac{1}{x})} \\=\dfrac{x\cancel{(x+1})(\frac{1}{\cancel{x+1}})-\cancel{x}(x+1)(\frac{1}{\cancel{x}})}{\cancel{x}(x+1)(\frac{1}{\cancel{x}})} \\=\dfrac{x-(x+1)}{x+1} \\=\dfrac{x-x-1}{x+1} \\=\dfrac{-1}{x+1}$
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