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: 32

Answer

The simplified form of the expression $\frac{1}{x+y}+\frac{2}{{{x}^{2}}+{{y}^{2}}}$ is $\frac{{{x}^{2}}+2x+{{y}^{2}}+2y}{\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)}$.

Work Step by Step

$\frac{1}{x+y}+\frac{2}{{{x}^{2}}+{{y}^{2}}}$ Solve to obtain the LCD of denominator of the fractions that is $\left( x+y \right)$ and $\left( {{x}^{2}}+{{y}^{2}} \right)$. On solving the LCD is $\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)$. The rational expression is written as a fraction in such a way that the LCD is the denominator of the fraction: $\begin{align} & \frac{1}{x+y}=\frac{1\left( {{x}^{2}}+{{y}^{2}} \right)}{\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)} \\ & =\frac{\left( {{x}^{2}}+{{y}^{2}} \right)}{\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)} \end{align}$ And $\frac{2}{{{x}^{2}}+{{y}^{2}}}=\frac{2\left( x+y \right)}{\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)}$ Since the fractions have the common denominator, add the numerators and the LCD is the denominator: $\begin{align} & \frac{1}{x+y}+\frac{2}{{{x}^{2}}+{{y}^{2}}}=\frac{\left( {{x}^{2}}+{{y}^{2}} \right)}{\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)}+\frac{2\left( x+y \right)}{\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)} \\ & =\frac{\left( {{x}^{2}}+{{y}^{2}} \right)+2\left( x+y \right)}{\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)} \end{align}$ Simplify the above expression as: $\begin{align} & \frac{1}{x+y}+\frac{2}{{{x}^{2}}+{{y}^{2}}}=\frac{{{x}^{2}}+{{y}^{2}}+2x+2y}{\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)} \\ & =\frac{{{x}^{2}}+2x+{{y}^{2}}+2y}{\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)} \end{align}$
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