Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter R - Elementary Algebra Review - R.6 Rational Expressions and Equations - R.6 Exercise Set - Page 980: 24


The simplified form of the expression $\frac{-1}{{{x}^{2}}+7x+10}-\frac{3}{{{x}^{2}}+8x+15}$ is $\frac{-4x-9}{\left( x+2 \right)\left( x+3 \right)\left( x+5 \right)}$.

Work Step by Step

$\frac{-1}{{{x}^{2}}+7x+10}-\frac{3}{{{x}^{2}}+8x+15}$ $\begin{align} & {{x}^{2}}+7x+10={{x}^{2}}+5x+2x+10 \\ & =x\left( x+5 \right)+2\left( x+5 \right) \\ & =\left( x+5 \right)\left( x+2 \right) \end{align}$ And $\begin{align} & {{x}^{2}}+8x+15={{x}^{2}}+5x+3x+15 \\ & =x\left( x+5 \right)+3\left( x+5 \right) \\ & =\left( x+5 \right)\left( x+3 \right) \end{align}$ Thus, the expression becomes: $\frac{-1}{{{x}^{2}}+7x+10}-\frac{3}{{{x}^{2}}+8x+15}=\frac{-1}{\left( x+5 \right)\left( x+2 \right)}-\frac{3}{\left( x+5 \right)\left( x+3 \right)}$ $\left( x+5 \right)\left( x+2 \right)$ and $\left( x+5 \right)\left( x+3 \right)$. On solving the LCD is $\left( x+5 \right)\left( x+2 \right)\left( x+3 \right)$. The rational expression is written as a fraction in such a way that the LCD is the denominator of the fraction. Consider, the individual terms of the provided rational terms; $\begin{align} & \frac{-1}{\left( x+5 \right)\left( x+2 \right)}=\frac{-1\left( x+3 \right)}{\left( x+5 \right)\left( x+2 \right)\left( x+3 \right)} \\ & =\frac{-x-3}{\left( x+5 \right)\left( x+2 \right)\left( x+3 \right)} \end{align}$ And, $\begin{align} & \frac{3}{\left( x+5 \right)\left( x+3 \right)}=\frac{3\left( x+2 \right)}{\left( x+5 \right)\left( x+2 \right)\left( x+3 \right)} \\ & =\frac{3x+6}{\left( x+5 \right)\left( x+2 \right)\left( x+3 \right)} \end{align}$ Since the fractions have the common denominator, add the numerators and the LCD is the denominator. $\begin{align} & \frac{-1}{{{x}^{2}}+7x+10}-\frac{3}{{{x}^{2}}+8x+15}=\frac{-x-3}{\left( x+5 \right)\left( x+2 \right)\left( x+3 \right)}-\frac{3x+6}{\left( x+5 \right)\left( x+2 \right)\left( x+3 \right)} \\ & =\frac{-x-3-3x-6}{\left( x+5 \right)\left( x+2 \right)\left( x+3 \right)} \\ & =\frac{-4x-9}{\left( x+5 \right)\left( x+2 \right)\left( x+3 \right)} \end{align}$
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