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 - 1.3 Fraction Notation - 1.3 Exercise Set - Page 28: 100

Answer

$\dfrac{65m}{49p}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the complex fraction, $ \dfrac{3\frac{5}{7}mn}{2\frac{4}{5}np} ,$ change the mixed number to an improper fraction. In the resulting complex fraction, multiply the numerator by the reciprocal of the denominator. Then cancel any common factors between the numerator and the denominator. $\bf{\text{Solution Details:}}$ The mixed number, $a\frac{b}{c},$ is equivalent to $\dfrac{ac+b}{c}.$ Hence, the given expression is equivalent to \begin{array}{l}\require{cancel} \dfrac{3\frac{5}{7}mn}{2\frac{4}{5}np} \\\\= \dfrac{\frac{26}{7}mn}{\frac{14}{5}np} \\\\= \dfrac{\dfrac{26mn}{7}}{\dfrac{14np}{5}} .\end{array} To divide the fractions above, get the reciprocal of the divisor and change the operator to multiplication. Hence, the expression above simplifies to \begin{array}{l}\require{cancel} \dfrac{\dfrac{26mn}{7}}{\dfrac{14np}{5}} \\\\= \dfrac{26mn}{7}\div\dfrac{14np}{5} \\\\= \dfrac{26mn}{7}\cdot\dfrac{5}{14np} \\\\= \dfrac{\cancel{2}(13)m\cancel{n}}{7}\cdot\dfrac{5}{\cancel{2}(7)\cancel{n}p} \\\\= \dfrac{65m}{49p} .\end{array}
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