Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 1 - Section 1.2 - Fractions in Algebra - Exercise Set: 120

Answer

$\frac{9}{10}$

Work Step by Step

($\frac {1}{2}$+$\frac {1}{4}$)$\div$($\frac {1}{2}$+$\frac {1}{3}$) ($\frac {1\times2}{2\times2}$+$\frac {1}{4}$)$\div$($\frac {1\times3}{2\times3}$+$\frac {1\times2}{3\times2}$) ($\frac {2}{4}$+$\frac {1}{4}$)$\div$($\frac {3}{6}$+$\frac {2}{6}$) ($\frac {2+1}{4}$)$\div$($\frac {3+2}{6}$) ($\frac {3}{4}$)$\div$($\frac {5}{6}$) $\frac {3}{4}$$\times$$\frac {6}{5}$ $\frac {18}{20}=\frac {18\div2}{20\div2}=\frac{9}{10}$
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