College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter R - Section R.1 - Real Numbers - R.1 Assess Your Understanding - Page 16: 87

Answer

$\dfrac{15}{8}$

Work Step by Step

Perform multiplication first to obtain: $\require{cancel} =\dfrac{2(3)}{4}+\dfrac{3}{8} \\=\dfrac{6}{4}+\dfrac{3}{8}$ Simplify the first fraction by canceling the common factor $2$ to obtain: $\require{cancel} =\dfrac{\cancel{6}3}{\cancel{4}2} + \dfrac{3}{8} \\=\dfrac{3}{2}+\dfrac{3}{8}$ Make the fractions similar using their LCD of $8$ to obtain: $=\dfrac{3(4)}{2(4)}+\dfrac{3}{8} \\=\dfrac{12}{8} + \dfrac{3}{8} \\=\dfrac{12+3}{8} \\=\dfrac{15}{8}$
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