Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Appendix A - Review - A.6 Solving Equations - A.6 Assess Your Understanding - Page A52: 117

Answer

$R=\dfrac{R_1R_2}{R_1+R_2}$

Work Step by Step

The LCD on the right hand side is $R_1R_2$. Make the expressions on the right hand side similar using their LCD. $\dfrac{1}{R}=\dfrac{R_2}{R_1R_2}+\dfrac{R_1}{R_1R_2}$ Add numerators because denominators are same. $\dfrac{1}{R}=\dfrac{R_2+R_1}{R_1R_2}$ Multiply both sides by $\frac{R\cdot R_1\cdot R_2}{R_1+R_2}$. $\dfrac{RR_1R_2}{R_1+R_2}\cdot \dfrac{1}{R}=\dfrac{RR_1R_2}{R_1+R_2}\cdot \dfrac{R_2+R_1}{R_1R_2}$ Simplify by cancelling common factors. $\require{cancel} \begin{align*} \dfrac{\cancel{R}R_1R_2}{R_1+R_2}\cdot \dfrac{1}{\cancel{R}}&=\dfrac{R\cancel{R_1R_2}}{\cancel{R_1+R_2}}\cdot \dfrac{\cancel{R_2+R_1}}{\cancel{R_1R_2}}\\ \\\dfrac{R_1R_2}{R_1+R_2}&=R \end{align*}$
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