Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.4 Equations - R.4 Exercises - Page R-16: 30


The solution is $p=6$

Work Step by Step

$\dfrac{5}{p-2}-\dfrac{7}{p+2}=\dfrac{12}{p^{2}-4}$ Factor the denominator of the fraction on the right side of the equation: $\dfrac{5}{p-2}-\dfrac{7}{p+2}=\dfrac{12}{(p-2)(p+2)}$ Multiply the whole equation by $(p-2)(p+2)$ $(p-2)(p+2)\Big(\dfrac{5}{p-2}-\dfrac{7}{p+2}=\dfrac{12}{(p-2)(p+2)}\Big)$ $5(p+2)-7(p-2)=12$ $5p+10-7p+14=12$ Simplify the left side by combining like terms: $-2p+24=12$ Take $24$ to the right side: $-2p=12-24$ $-2p=-12$ Take $2$ to divide the right side: $p=\dfrac{-12}{-2}$ $p=6$
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