Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter R - Section R.2 - Operations on Real Numbers - R.2 Exercises: 45

Answer

$-\frac{7}{4}$

Work Step by Step

$(-\frac{5}{4}-\frac{2}{3})+\frac{1}{6}=(-\frac{5\times3}{4\times3}-\frac{2\times4}{3\times4})+\frac{1}{6}=(-\frac{15}{12}-\frac{8}{12})+\frac{1}{6}=-(\frac{15}{12}+\frac{8}{12})+\frac{1}{6}=-(\frac{23}{12})+\frac{1}{6}=-\frac{23}{12}+\frac{1}{6}$ $-\frac{23}{12}+\frac{1}{6}=-\frac{23}{12}+\frac{1\times2}{6\times2}=-\frac{23}{12}+\frac{2}{12}=\frac{2}{12}+(-\frac{23}{12})=\frac{2+(-23)}{12}=\frac{-21}{12}=\frac{-21\div3}{12\div3}=-\frac{7}{4}$
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