Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 1 - Introduction to Algebraic Expressions - Study Summary - Practice Exercises: 14

Answer

$\dfrac{25}{6}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To multiply the given expression, $ \dfrac{15}{14}\cdot\dfrac{35}{9} ,$ cancel the common factors between the numerator and the denominator. Then multiply the numerators and multiply the denominators. $\bf{\text{Solution Details:}}$ Cancelling the common factors between the numerator and the denominator, the given expression is equivalent to \begin{array}{l}\require{cancel} \dfrac{15}{14}\cdot\dfrac{35}{9} \\\\= \dfrac{\cancel3(5)}{\cancel7(2)}\cdot\dfrac{\cancel7(5)}{\cancel3(3)} \\\\= \dfrac{5}{2}\cdot\dfrac{5}{3} .\end{array} To multiply fractions, multiply the numerators and multiply the denominators. Hence, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{5}{2}\cdot\dfrac{5}{3} \\\\= \dfrac{5(5)}{2(3)} \\\\= \dfrac{25}{6} .\end{array}
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