Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 1 - Review Exercises: 26

Answer

$-\dfrac{1}{10}$

Work Step by Step

Make the fractions similar using their LCD of 10 as denominator to obtain: $=-(\frac{3\cdot2}{5\cdot2}) - (-\frac{1\cdot5}{2\cdot5}) \\=-\frac{6}{10} - (-\frac{5}{10}) $ RECALL: $a-b = a+(-b)$ Use the rule above to obtain: $=-\dfrac{6}{10} + \dfrac{5}{10} \\=\dfrac{-6+5}{10} \\=\dfrac{-1}{10}$
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