Elementary Algebra

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

Chapter 2 - Real Numbers - 2.1 - Rational Numbers: Multiplication and Division - Problem Set 2.1: 43

Answer

$\dfrac{8}{21}$

Work Step by Step

First, factor each expression to obtain: $=\dfrac{6a}{2b(7)} \cdot \dfrac{2b(8)}{6a(3)}$ Cancel the common factors to obtain: $\require{cancel} =\dfrac{\cancel{6a}}{\cancel{2b}(7)} \cdot \dfrac{\cancel{2b}(8)}{\cancel{6a}(3)} \\=\dfrac{1}{7} \cdot \dfrac{8}{3}$ Multiply the numerators together and the denominators together to obtain: $=\dfrac{8}{21}$
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