Intermediate Algebra (6th Edition)

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

Chapter 6 - Section 6.3 - Simplifying Complex Fractions - Exercise Set - Page 360: 3

Answer

$\dfrac{7}{13}$

Work Step by Step

Simplify the numerator and the denominator to have; $=\dfrac{\frac{5}{5} + \frac{2}{5}}{\frac{10}{5} + \frac{3}{5}} \\=\dfrac{\frac{7}{5}}{\frac{13}{5}}$ RECALL: $\dfrac{\frac{a}{b}}{\frac{c}{d}}=\dfrac{a}{b} \cdot \dfrac{d}{c}$ Use the rule above to have: $\\=\dfrac{7}{5} \cdot \dfrac{5}{13}$ Cancel the common factors to have: $\require{cancel} \\=\dfrac{7}{\cancel{5}} \cdot \dfrac{\cancel{5}}{13} \\=\dfrac{7}{13}$
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