Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 5 - Number Theory and the Real Number System - Chapter Summary, Review, and Test - Review Exercises - Page 335: 58

Answer

$-\dfrac{20}{3}$

Work Step by Step

Simplify the expressions inside the parentheses to using their LCD obtain: \begin{align*} &=\left(\frac{3}{6}+\frac{2}{6}\right)\div \left(\frac{2}{8}-\frac{3}{8}\right)\\\\ &=\left(\frac{5}{6}\right)\div \left(-\frac{1}{8}\right)\\\\ \end{align*} Use the rule $\dfrac{a}{b} \div \dfrac{c}{d}=\dfrac{a}{b} \times \dfrac{d}{c}$ to obtain: \begin{align*} \require{cancel} &=\left(\frac{5}{6}\right) \times \left(-\frac{8}{1}\right)\\\\ &=\left(\frac{5}{\cancel{6}3}\right) \times \left(-\frac{\cancel{8}4}{1}\right)\\\\ &=\frac{-20}{3} \end{align*}
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