## Basic College Mathematics (9th Edition)

Published by Pearson

# Chapter 9 - Basic Algebra - 9.4 Order of Operations - 9.4 Exercises - Page 657: 58

#### Answer

=$-\frac{31}{45}$

#### Work Step by Step

$\frac{2}{7}(-\frac{14}{5})-(\frac{4}{3}-\frac{13}{9})\longrightarrow$ To evaluate expression in parentheses, first convert to least common denominator of 9. =$\frac{2}{7}(-\frac{14}{5})-(\frac{4\times3}{3\times3}-\frac{13}{9})$ =$\frac{2}{7}(-\frac{14}{5})-(\frac{12}{9}-\frac{13}{9})$ =$\frac{2}{7}(-\frac{14}{5})-(-\frac{1}{9})\longrightarrow$ Evaluate multiplication. =$-(\frac{28}{35})-(-\frac{1}{9})\longrightarrow$ Reduce using greatest common factor, 7. =$-(\frac{28\div7}{35\div7})-(-\frac{1}{9})\longrightarrow$ Simplify. =$-(\frac{4}{5})-(-\frac{1}{9})\longrightarrow$ Change subtraction of a negative to the additive inverse. =$-(\frac{4}{5})+(\frac{1}{9})\longrightarrow$ To evaluate the addition, first convert to least common denominator of 45. =$-(\frac{4\times9}{5\times9})+(\frac{1\times5}{9\times5})\longrightarrow$ Simplify. =$-(\frac{36}{45})+(\frac{5}{45})\longrightarrow$ Add. =$-\frac{31}{45}$

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