## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 1 - Section 1.1 - Real Numbers - 1.1 Exercises - Page 10: 32

#### Answer

a.) $\frac{8}{3}$ b.) 3

#### Work Step by Step

a.) $\frac{2}{\frac{2}{3}}$ - $\frac{\frac{2}{3}}{2}$ Rewrite the problem to multiply by switching the denominator by its reciprocal $\frac{2}{1}$$\times$$\frac{3}{2}$ - $\frac{2}{3}$$\times$$\frac{1}{2}$ Simplify 3 - $\frac{1}{3}$ Least Common Denominator of 1 and 3 = 3 $\frac{9}{3}$ - $\frac{1}{3}$ = $\frac{8}{3}$ b.) $\frac{\frac{2}{5} + \frac{1}{2}}{\frac{1}{10} + \frac{3}{15}}$ Find the Least Common Denominator for both the numerator and denominator Numerator's LCM of 5 and 2 = 10 Denominator's LCM of 10 and 15 = 30 $\frac{\frac{4}{10} + \frac{5}{10}}{\frac{3}{30} + \frac{6}{30}}$ Simplify $\frac{\frac{9}{10}}{\frac{9}{30}}$ Multiply by denominator's reciprocal $\frac{9}{10}$$\times$$\frac{30}{9}$ = $\frac{270}{90}$ = 3

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