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 - Section 1.2 - Operations with Real Numbers and Simplifying Algebraic Expressions - Exercise Set - Page 27: 19

Answer

$\dfrac{2}{15}$

Work Step by Step

Make the fractions similar using their LCD of 15 to obtain: $=\dfrac{11}{15} + \left(-\dfrac{9}{15}\right)$ The addends have opposite signs, so subtract the smaller absolute value from the bigger absolute value, and then use the sign of the number with the bigger absolute value to obtain: $=(\frac{11}{15}-\frac{9}{15}) \\=\frac{2}{15}$
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