# Chapter 2 - Real Numbers - 2.1 - Rational Numbers: Multiplication and Division - Problem Set 2.1 - Page 48: 43

$\dfrac{8}{21}$

#### 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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