Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 1 - Section 1.3 - Operations on Real Numbers and Order of Operations - Exercise Set - Page 27: 9

Answer

$\frac{4}{3}$

Work Step by Step

$\frac{11}{15}-(-\frac{3}{5})=\frac{11}{15}+\frac{3}{5}=\frac{11}{15}+\frac{3\times3}{5\times3}=\frac{11}{15}+\frac{9}{15}=\frac{20}{15}=\frac{4}{3}$
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