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: 35

Answer

$\dfrac{3k}{2k^{2}+3k-2}-\dfrac{2k}{2k^{2}-7k+3}=\dfrac{k(k-13)}{(2k-1)(k+2)(k-3)}$

Work Step by Step

$\dfrac{3k}{2k^{2}+3k-2}-\dfrac{2k}{2k^{2}-7k+3}$ Factor the denominators of both rational expressions: $\dfrac{3k}{2k^{2}+3k-2}-\dfrac{2k}{2k^{2}-7k+3}=...$ $...=\dfrac{3k}{(k+2)(2k-1)}-\dfrac{2k}{(2k-1)(k-3)}=...$ Evaluate the subtraction of these two fractions by using the LCD, which is $(2k-1)(k+2)(k-3)$ in this case: $...=\dfrac{3k(k-3)-2k(k+2)}{(2k-1)(k+2)(k-3)}=...$ Simplify: $...=\dfrac{3k^{2}-9k-2k^{2}-4k}{(2k-1)(k+2)(k-3)}=\dfrac{k^{2}-13k}{(2k-1)(k+2)(k-3)}=...$ $...=\dfrac{k(k-13)}{(2k-1)(k+2)(k-3)}$
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