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: 74

Answer

$\dfrac{-3}{y+3}$

Work Step by Step

Simplify by multiplying the LCD (which is $y(y+3)$) to both the numerator and the denominator to obtain: $\require{cancel} =\dfrac{y(y+3)\left(\frac{1}{y+3}-\frac{1}{y}\right)}{y(y+3)(\frac{1}{x})} \\=\dfrac{y(y+3)(\frac{1}{y+3})-y(y+3)(\frac{1}{y})}{y(y+3)(\frac{1}{y})} \\=\dfrac{y\cancel{(y+3)}(\frac{1}{\cancel{y+3}})-\cancel{y}(y+3)(\frac{1}{\cancel{y}})}{\cancel{y}(y+3)(\frac{1}{\cancel{y}})} \\=\dfrac{y-(y+3)}{y+3} \\=\dfrac{y-y-3}{y+3} \\=\dfrac{-3}{y+3}$
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