Precalculus (6th Edition) Blitzer

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

Chapter 7 - Section 7.2 - Systems of Linear Equations in Three Variables - Exercise Set - Page 832: 60


The simplified form of the equation is $\frac{5{{x}^{3}}-3{{x}^{2}}+7x-3}{{{\left( {{x}^{2}}+1 \right)}^{2}}}$.

Work Step by Step

Let us consider the expression $\frac{5x-3}{{{x}^{2}}+1}+\frac{2x}{{{\left( {{x}^{2}}+1 \right)}^{2}}}$ And simplify: $\begin{align} & \frac{{{\left( {{x}^{2}}+1 \right)}^{2}}}{{{\left( {{x}^{2}}+1 \right)}^{2}}}\cdot \left( \frac{5x-3}{{{x}^{2}}+1}+\frac{2x}{{{\left( {{x}^{2}}+1 \right)}^{2}}} \right)=\frac{{{\left( {{x}^{2}}+1 \right)}^{2}}}{{{\left( {{x}^{2}}+1 \right)}^{2}}}\cdot \frac{5x-3}{{{x}^{2}}+1}+\frac{{{\left( {{x}^{2}}+1 \right)}^{2}}}{{{\left( {{x}^{2}}+1 \right)}^{2}}}\cdot \frac{2x}{{{\left( {{x}^{2}}+1 \right)}^{2}}} \\ & =\frac{\left( {{x}^{2}}+1 \right)\left( 5x-3 \right)}{{{\left( {{x}^{2}}+1 \right)}^{2}}}+\frac{2x}{{{\left( {{x}^{2}}+1 \right)}^{2}}} \\ & =\frac{{{x}^{2}}\left( 5x-3 \right)+1\left( 5x-3 \right)+2x}{{{\left( {{x}^{2}}+1 \right)}^{2}}} \\ & =\frac{5{{x}^{3}}-3{{x}^{2}}+5x-3+2x}{{{\left( {{x}^{2}}+1 \right)}^{2}}} \end{align}$ Therefore, $\frac{{{\left( {{x}^{2}}+1 \right)}^{2}}}{{{\left( {{x}^{2}}+1 \right)}^{2}}}\cdot \left( \frac{5x-3}{{{x}^{2}}+1}+\frac{2x}{{{\left( {{x}^{2}}+1 \right)}^{2}}} \right)=\frac{5{{x}^{3}}-3{{x}^{2}}+7x-3}{{{\left( {{x}^{2}}+1 \right)}^{2}}}$ Thus, the simplified form of the expression $\frac{5x-3}{{{x}^{2}}+1}+\frac{2x}{{{\left( {{x}^{2}}+1 \right)}^{2}}}$ is $\frac{5{{x}^{3}}-3{{x}^{2}}+7x-3}{{{\left( {{x}^{2}}+1 \right)}^{2}}}$.
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