## Thinking Mathematically (6th Edition)

Published by Pearson

# Chapter 5 - Number Theory and the Real Number System - Chapter Summary, Review, and Test - Review Exercises - Page 335: 49

#### Answer

$\dfrac{35}{6}$

#### Work Step by Step

Convert each factor to an improper fraction to find: $=\dfrac{3(3)+1}{3} \cdot \dfrac{4(1)+3}{4} \\=\dfrac{10}{3} \cdot \dfrac{7}{4}$ Factor each numerator and denominator, when needed, and then cancel common factors to find: $\\\require{cancel}=\dfrac{2(5)}{3} \cdot \dfrac{7}{2(2)} \\=\dfrac{\cancel{2}(5)}{3} \cdot \dfrac{7}{\cancel{2}(2)} \\=\dfrac{5}{3} \cdot \dfrac{7}{2}$ Multiply the numerators together and the denominators together to find: $=\dfrac{5(7)}{3(2)} \\=\dfrac{35}{6}$

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