# Chapter 4 - Proportions, Percents, and Solving Inequalities - 4.4 - Inequalities - Problem Set 4.4 - Page 169: 7

True

#### Work Step by Step

$\frac{3}{4}$ + $\frac{2}{3}$ $\div$ $\frac{1}{5}$ $\gt$ $\frac{2}{3}$ + $\frac{1}{2}$ $\div$ $\frac{3}{4}$ To check whether this inequality is true or false, we simplify it. $\frac{3}{4}$ + $\frac{2}{3}$ $\times$ $\frac{5}{1}$ $\gt$ $\frac{2}{3}$ + $\frac{1}{2}$ $\times$ $\frac{4}{3}$ $\frac{3}{4}$ + $\frac{10}{3}$ $\gt$ $\frac{2}{3}$ + $\frac{4}{6}$ We need to make the denominators the same in order to simplify this inequality. $\frac{3}{4}$ $\times$ $\frac{3}{3}$ + $\frac{10}{3}$ $\times$ $\frac{4}{4}$ $\gt$ $\frac{2}{3}$ $\times$ $\frac{4}{4}$ + $\frac{4}{6}$ $\times$ $\frac{2}{2}$ $\frac{9}{12}$ + $\frac{40}{12}$ $\gt$ $\frac{8}{12}$ + $\frac{8}{12}$ $\frac{49}{12}$ $\gt$ $\frac{16}{12}$ Multiply both sides by 12. 49 $\gt$ 16 Because 49 is indeed greater than 16, this inequality is true.

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