Precalculus (6th Edition) Blitzer

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

Chapter P - Section P.6 - Rational Expressions - Exercise Set - Page 86: 88


The required solution is $\frac{1}{y\left( y+2 \right)}$.

Work Step by Step

We have the given algebraic expression $\frac{{{y}^{-1}}-{{\left( y+2 \right)}^{-1}}}{2}$. For an algebraic expression, a rational expression is an expression that can be expressed in the form $\frac{p}{q}$, where, both $p\ \text{and }q$ are polynomials and the denominator $q\ne 0$. And the given expression can also be written as $\frac{{{y}^{-1}}-{{\left( y+2 \right)}^{-1}}}{2}=\frac{\frac{1}{y}-\frac{1}{y+2}}{2}$. We know that a complex rational expression or a complex fraction is an algebraic rational expression in which either the numerator contains a rational expression or the denominator contains a rational expression or both the numerator and denominator contain a rational expression. Here the numerator is $\frac{1}{y}-\frac{1}{y+2}$. Solve the numerator of the given complex rational expression: $\begin{align} & \frac{1}{y}-\frac{1}{y+2}=\frac{1}{y}\times \frac{\left( y+2 \right)}{\left( y+2 \right)}-\frac{1}{\left( y+2 \right)}\times \frac{y}{y} \\ & =\frac{y+2}{y\left( y+2 \right)}-\frac{y}{y\left( y+2 \right)} \\ & =\frac{y+2-y}{y\left( y+2 \right)} \\ & =\frac{2}{y\left( y+2 \right)} \end{align}$. Therefore, the given complex rational expression becomes $\begin{align} & \frac{{{y}^{-1}}-{{\left( y+2 \right)}^{-1}}}{2}=\frac{\frac{2}{y\left( y+2 \right)}}{2} \\ & =\frac{2}{y\left( y+2 \right)}\times \frac{1}{2} \\ & =\frac{1}{y\left( y+2 \right)} \end{align}$. Hence, $\frac{{{y}^{-1}}-{{\left( y+2 \right)}^{-1}}}{2}$ $\frac{1}{y\left( y+2 \right)}$.
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