College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.6 - Other Types of Equations and Applications - 1.6 Exercises - Page 134: 22


$x=\left\{ \dfrac{4}{3},\dfrac{9}{4} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given rational equation, $ 6=\dfrac{7}{2x-3}+\dfrac{3}{(2x-3)^2} ,$ multiply both sides by the $LCD.$ Then use concepts of quadratic equations to solve for the variable. $\bf{\text{Solution Details:}}$ The $LCD$ of the denominators is the collection of all not common factors with the highest exponent and the common factors with the highest exponent. Hence the $LCD$ of $\left\{ 1,(2x-3),(2x-3)^2 \right\}$ is $ (2x-3)^2 .$ Multiplying both sides of the given equation by the $LCD= (2x-3)^2 $ results to \begin{array}{l}\require{cancel} (2x-3)^2(6)=\left(\dfrac{7}{2x-3}+\dfrac{3}{(2x-3)^2}\right)(2x-3)^2 \\\\ (2x-3)^2(6)=\dfrac{7}{2x-3}\cdot(2x-3)^2+\dfrac{3}{(2x-3)^2}\cdot(2x-3)^2 \\\\ (2x-3)^2(6)=\dfrac{7}{\cancel{2x-3}}\cdot(2x-3)\cancel{^2}+\dfrac{3}{\cancel{(2x-3)^2}}\cdot\cancel{(2x-3)^2} \\\\ (2x-3)^2(6)=(2x-3)(7)+3(1) \\\\ (2x-3)^2(6)=(2x-3)(7)+3 .\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} [(2x)^2-2(2x)(3)+(3)^2](6)=(2x-3)(7)+3 \\\\ (4x^2-12x+9)(6)=(2x-3)(7)+3 .\end{array} Using the Distributive Property and then combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} 24x^2-72x+54=14x-21+3 \\\\ 24x^2+(-72x-14x)+(54+21-3)=0 \\\\ 24x^2-86x+72=0 \\\\ \dfrac{24x^2-86x+72}{2}=\dfrac{0}{2} \\\\ 12x^2-43x+36=0 .\end{array} In the trinomial expression above the value of $ac$ is $ 12(36)=432 $ and the value of $b$ is $ -43 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -16,-27 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 12x^2-16x-27x+36=0 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (12x^2-16x)-(27x-36)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 4x(3x-4)-9(3x-4)=0 .\end{array} Factoring the $GCF= (3x-4) $ of the entire expression above results to \begin{array}{l}\require{cancel} (3x-4)(4x-9)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} 3x-4=0 \\\\\text{OR}\\\\ 4x-9=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 3x-4=0 \\\\ 3x=4 \\\\ x=\dfrac{4}{3} \\\\\text{OR}\\\\ 4x-9=0 \\\\ 4x=9 \\\\ x=\dfrac{9}{4} .\end{array} Upon checking, none of the solutions above will make the denominator of the original equation equal to zero. Hence, $ x=\left\{ \dfrac{4}{3},\dfrac{9}{4} \right\} .$
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