## Introductory Algebra for College Students (7th Edition)

Published by Pearson

# Chapter 1 - Section 1.5 - Addition of Real Numbers - Exercise Set - Page 63: 70

#### Answer

$\displaystyle \frac{-15}{x}+\frac{4}{x}.$, simplifies to $\displaystyle \frac{-11}{x}$

#### Work Step by Step

the quotient of $-15$ and a number ... $\rightarrow$... $\displaystyle \frac{-15}{x}$ ...which is increased by ... $\rightarrow \displaystyle \frac{-15}{x}+...$ .... the quotient of $4$ and a number$\quad\rightarrow \displaystyle \frac{-15}{x}+\frac{4}{x}.$ This simplifies, because the demominator is common, to, $...=\displaystyle \frac{ -15+4}{x}$ ... then, adding different signed numbers, we keep the sign of $-15$, because it has the greater absolute value, and subtract their absolute values: $=\displaystyle \frac{-(15-4)}{x}$ $=\displaystyle \frac{-11}{x}$

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