College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter R - Section R.7 - Rational Expressions - R.7 Assess Your Understanding: 87

Answer

$\dfrac{19}{9x^2-30x+25}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ \dfrac{(2x+3)\cdot3-(3x-5)\cdot2}{(3x-5)^2} ,$ use the Distributive Property first. Then remove the grouping symbols and combine like terms. Finally, use special products to simplify the denominator. $\bf{\text{Solution Details:}}$ Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{(2x\cdot3+3\cdot3)-(3x\cdot2-5\cdot2)}{(3x-5)^2} \\\\= \dfrac{(6x+9)-(6x-10)}{(3x-5)^2} .\end{array} Removing the grouping symbols and then combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{6x+9-6x+10}{(3x-5)^2} \\\\= \dfrac{19}{(3x-5)^2} .\end{array} Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{19}{(3x)^2-2(3x)(5)+(5)^2} \\\\= \dfrac{19}{9x^2-30x+25} .\end{array}
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