Precalculus (6th Edition) Blitzer

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

Chapter P - Review Exercises - Page 144: 127


The solution set of the equation is $\left\{ 7 \right\}$.

Work Step by Step

Consider the equation $\frac{4}{x+2}+\frac{2}{x-4}=\frac{30}{{{x}^{2}}-2x-8}$. The equation is not defined for $x=-2\text{ and }x=4$. First of all, simplify the equation. That is, $\begin{align} & \frac{4}{x+2}+\frac{2}{x-4}=\frac{30}{{{x}^{2}}-2x-8} \\ & \frac{4\left( x-4 \right)+2\left( x+2 \right)}{\left( x+2 \right)\left( x-4 \right)}=\frac{30}{{{x}^{2}}-2x-8} \\ & \frac{4x-16+2x+4}{{{x}^{2}}-2x-8}=\frac{30}{{{x}^{2}}-2x-8} \end{align}$ Now, combine the like terms as follows $\frac{6x-12}{{{x}^{2}}-2x-8}=\frac{30}{{{x}^{2}}-2x-8}$ Now, multiply both sides by ${{x}^{2}}-2x-8$ and get, $\begin{align} & \frac{6x-12}{{{x}^{2}}-2x-8}\cdot \left( {{x}^{2}}-2x-8 \right)=\frac{30}{{{x}^{2}}-2x-8}\cdot \left( {{x}^{2}}-2x-8 \right) \\ & 6x-12=30 \\ & 6x=30+12 \\ & 6x=42 \end{align}$ Now, divide both sides by $6$ and get, $\begin{align} & \frac{6x}{6}=\frac{42}{6} \\ & x=7 \end{align}$ Check: Substitute the value of $x=7$ in $\frac{4}{x+2}+\frac{2}{x-4}=\frac{30}{{{x}^{2}}-2x-8}$ to verify the solution. $\begin{align} & \frac{4}{x+2}+\frac{2}{x-4}=\frac{30}{{{x}^{2}}-2x-8} \\ & \frac{4}{7+2}+\frac{2}{7-4}\overset{?}{\mathop{=}}\,\frac{30}{{{7}^{2}}-2\left( 7 \right)-8} \\ & \text{ }\frac{4}{9}+\frac{2}{3}\overset{?}{\mathop{=}}\,\frac{30}{49-14-8} \\ & \text{ }\frac{12+18}{27}\overset{?}{\mathop{=}}\,\frac{30}{27} \\ \end{align}$ Further solve and get, $\frac{30}{27}=\frac{30}{27}$ That is true. Hence, the solution set of the equation is $\left\{ 7 \right\}$.
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