Calculus with Applications (10th Edition)

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

Chapter R - Algebra Reference - R.3 Rational Expressions - R.3 Exercises - Page R-11: 32


$\dfrac{2}{5(k-2)}+\dfrac{3}{4(k-2)}=\dfrac{23}{20(k-2)}$ or $\dfrac{2}{5(k-2)}+\dfrac{3}{4(k-2)}=\dfrac{23}{20k-40}$

Work Step by Step

$\dfrac{2}{5(k-2)}+\dfrac{3}{4(k-2)}$ Evaluate the sum of these two fractions by using the LCD, which is $(5)(4)(k-2)$ in this case: $\dfrac{2}{5(k-2)}+\dfrac{3}{4(k-2)}=\dfrac{(2)(4)+(3)(5)}{(5)(4)(k-2)}=...$ Simplify: $...=\dfrac{8+15}{20(k-2)}=\dfrac{23}{20(k-2)}$
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