Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 7 - Algebraic Fractions - Chapters 1-7 Cumulative Review Problem Set - Page 324: 12



Work Step by Step

First, we simplify the expression within the parenthesis: $(\frac{1}{3}+\frac{1}{4})^{-1}$ =$(\frac{1(4)+1(3)}{12})^{-1}$ =$(\frac{4+3}{12})^{-1}$ =$(\frac{7}{12})^{-1}$ Since the whole fraction is raised to the negative 1 power, we need to take a reciprocal of the fraction. Therefore, $(\frac{7}{12})^{-1}=\frac{7^{-1}}{12^{-1}}=\frac{12}{7}$
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