Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 6 - Section 6.2 - Adding and Subtracting Rational Expressions - Exercise Set: 105

Answer

$\dfrac{3}{2x}$

Work Step by Step

RECALL: $x^{-m} = \dfrac{1}{x^{m}}, m \gt 0$ Use the rule above to have: $\\=\dfrac{1}{x^1} + \frac{1}{(2x)^1} \\=\dfrac{1}{x}+\dfrac{1}{2x}$ Make the expressions similar using the LCD which is $2x$ to have: $\\=\dfrac{1(2)}{2(x)} + \dfrac{1}{2x} \\=\dfrac{2}{2x} + \dfrac{1}{2x} \\=\dfrac{3}{2x}$
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