Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 7 - Rational Functions - 7.4 Adding and Subtracting Rational Expressions - 7.4 Exercises: 1

Answer

$\text{LCD: } 84x^2y \\\\\text{Equivalent of 1st Expression: } \dfrac{49}{84x^2y} \\\\\text{Equivalent of 2nd Expression: } \dfrac{9x}{84x^2y}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the $LCD$ of the given expressions, $ \dfrac{7}{12x^2y} $ and $ \dfrac{3}{28xy} ,$ express the denominators in factored form. Then take the highest exponent of each factor. Finally use the $LCD$ to find the required equivalent expression for the given. $\bf{\text{Solution Details:}}$ In factored form, the given expressions are equivalent to \begin{array}{l}\require{cancel} \dfrac{7}{2^2(3)(x^2)(y)} \text{ and } \dfrac{3}{2^2(7)(x)(y)} .\end{array} Taking the highest exponent of each factor, then the $LCD$ is \begin{array}{l}\require{cancel} 2^2(3)(7)(x^2)(y) \\\\= 84x^2y .\end{array} Multiplying the given expression by an expression equal to $1$ such that the denominator becomes the $LCD,$ then the given expressions are equivalent to \begin{array}{l}\require{cancel} \dfrac{7}{12x^2y}\cdot\dfrac{7}{7}= \dfrac{49}{84x^2y} \\\\\text{ and }\\\\ \dfrac{3}{28xy}\cdot\dfrac{3x}{3x}= \dfrac{9x}{84x^2y} .\end{array} Hence, the needed pieces of information are as follows: \begin{array}{l}\require{cancel} \text{LCD: } 84x^2y \\\\\text{Equivalent of 1st Expression: } \dfrac{49}{84x^2y} \\\\\text{Equivalent of 2nd Expression: } \dfrac{9x}{84x^2y} .\end{array}
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