College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.3 - Page 135: 73



Work Step by Step

Multiply all terms by the LCD. Then factor out $f$ from $qf+pf$ and divide both sides by $q+p$. $\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$ for $f$ $pqf(\frac{1}{p}+\frac{1}{q})=\frac{1}{f}(pqf)$ $qf+pf=pq$ $f(q+p)=pq$ $f=\frac{pq}{q+p}$
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