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 85: 83


The required solution is: $\frac{{{x}^{2}}+5x+8}{\left( x+2 \right)\left( x+1 \right)}$.

Work Step by Step

We have the given algebraic expression: $\left( \frac{2x+3}{x+1}\cdot \frac{{{x}^{2}}+4x-5}{2{{x}^{2}}+x-3} \right)-\frac{2}{x+2}$. 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 $q\ne 0$. Now solve the bracket of the above rational expression: $\begin{align} & \frac{2x+3}{x+1}\cdot \frac{{{x}^{2}}+4x-5}{2{{x}^{2}}+x-3}=\frac{2x+3}{x+1}\cdot \frac{{{x}^{2}}+\left( 5-1 \right)x-5}{2{{x}^{2}}+\left( 3-2 \right)x-3} \\ & =\frac{2x+3}{x+1}\cdot \frac{{{x}^{2}}+5x-1x-5}{2{{x}^{2}}+3x-2x-3} \\ & =\frac{2x+3}{x+1}\cdot \frac{x\left( x+5 \right)-1\left( x+5 \right)}{x\left( 2x+3 \right)-1\left( 2x+3 \right)} \\ & =\frac{2x+3}{x+1}\cdot \frac{\left( x+5 \right)\left( x-1 \right)}{\left( 2x+3 \right)\left( x-1 \right)} \end{align}$. Solving further, $\begin{align} & \frac{2x+3}{x+1}\cdot \frac{{{x}^{2}}+4x-5}{2{{x}^{2}}+x-3}=\frac{\left( 2x+3 \right)}{\left( x+1 \right)}\cdot \frac{\left( x+5 \right)\left( x-1 \right)}{\left( 2x+3 \right)\left( x-1 \right)} \\ & =\frac{x+5}{x+1} \end{align}$. So, the given expression becomes $\left( \frac{2x+3}{x+1}\cdot \frac{{{x}^{2}}+4x-5}{2{{x}^{2}}+x-3} \right)-\frac{2}{x+2}=\frac{x+5}{x+1}-\frac{2}{x+2}$. And simplify the above expression by taking the least common denominator: $\begin{align} & \frac{x+5}{x+1}-\frac{2}{x+2}=\frac{x+5}{\left( x+1 \right)}\times \frac{\left( x+2 \right)}{\left( x+2 \right)}-\frac{2}{\left( x+2 \right)}\times \frac{\left( x+1 \right)}{\left( x+1 \right)} \\ & =\frac{\left( x+5 \right)\left( x+2 \right)}{\left( x+1 \right)\left( x+2 \right)}-\frac{2\left( x+1 \right)}{\left( x+2 \right)\left( x+1 \right)} \\ & =\frac{{{x}^{2}}+5x+2x+10-2x-2}{\left( x+2 \right)\left( x+1 \right)} \\ & =\frac{{{x}^{2}}+5x+8}{\left( x+2 \right)\left( x+1 \right)} \end{align}$ Therefore, the given expression gets reduced to $\left( \frac{2x+3}{x+1}\cdot \frac{{{x}^{2}}+4x-5}{2{{x}^{2}}+x-3} \right)-\frac{2}{x+2}=\frac{{{x}^{2}}+5x+8}{\left( x+2 \right)\left( x+1 \right)}$ Hence, $\left( \frac{2x+3}{x+1}\cdot \frac{{{x}^{2}}+4x-5}{2{{x}^{2}}+x-3} \right)-\frac{2}{x+2}=$ $\frac{{{x}^{2}}+5x+8}{\left( x+2 \right)\left( x+1 \right)}$.
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