Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 1 - Section 1.5 - Addition of Real Numbers - Exercise Set - Page 64: 92


False. Change to $\displaystyle \frac{3}{4}+(-\frac{3}{5})=\frac{3}{20}$

Work Step by Step

Writing the fractions with common denominator of 20, $\displaystyle \frac{3}{4}=\frac{3\times 5}{4\times 5}=\frac{15}{20},\qquad \frac{3}{5}=\frac{3\times 4}{5\times 4}=\frac{12}{20}$ $\displaystyle \frac{3}{4}+(-\frac{3}{5})=\frac{15}{20}+(-\frac{12}{20})$ ... adding numbers with different signs, the sum has the sign of the number with the greater absolute value. Here, it is positive, because $\displaystyle \frac{15}{20}$ has greater value. So, the statement is false. The true statement is obtained by: $\displaystyle \frac{15}{20}+(-\frac{12}{20})$= $)=\displaystyle \frac{15}{20}+\frac{-12}{20}=\frac{15-12}{20}=\frac{3}{20}$ That is, $\displaystyle \frac{3}{4}+(-\frac{3}{5})=\frac{3}{20}$
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