Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 4 - Mixed Review Exercises: 14

Answer

$2y^{2}x+\dfrac{3y^{3}}{2x}+\dfrac{5x^{2}}{2}$

Work Step by Step

Dividing the denominator to each of the terms of the numerator, the given expression, $ \dfrac{20y^3x^3+15y^4x+25yx^4}{10yx^2} $ is equivalent to \begin{array}{l}\require{cancel} \dfrac{20y^3x^3}{10yx^2}+\dfrac{15y^4x}{10yx^2}+\dfrac{25yx^4}{10yx^2} .\end{array} Using the law of exponents which states that $\dfrac{a^x}{a^y}=a^{x-y},$ the expression above simplifies to \begin{array}{l}\require{cancel} 2y^{3-1}x^{3-2}+\dfrac{3y^{4-1}x^{1-2}}{2}+\dfrac{5\cancel{y}x^{4-2}}{2\cancel{y}} \\\\= 2y^{2}x^{1}+\dfrac{3y^{3}x^{-1}}{2}+\dfrac{5x^{2}}{2} .\end{array} Using the law of exponents which states that $a^{-x}=\dfrac{1}{a^x}$ or $\dfrac{1}{a^{-x}}=a^x$ the expression above simplifies to \begin{array}{l}\require{cancel} 2y^{2}x^{1}+\dfrac{3y^{3}}{2x^{1}}+\dfrac{5x^{2}}{2} \\\\= 2y^{2}x+\dfrac{3y^{3}}{2x}+\dfrac{5x^{2}}{2} .\end{array}
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