# Chapter 1 - Section 1.6 - Other Types of Equations and Applications - 1.6 Exercises: 21

$x=\left\{ \dfrac{3}{4},1 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given rational equation, $2=\dfrac{3}{2x-1}+\dfrac{-1}{(2x-1)^2} ,$ 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 $\left\{ 1,(2x-1),(2x-1)^2 \right\}$ is $(2x-1)^2 .$ Multiplying both sides of the given equation by the $LCD= (2x-1)^2$ results to \begin{array}{l}\require{cancel} (2x-1)^2(2)=\left(\dfrac{3}{2x-1}+\dfrac{-1}{(2x-1)^2}\right)(2x-1)^2 \\\\ (2x-1)^2(2)=\dfrac{3}{2x-1}\cdot(2x-1)^2+\dfrac{-1}{(2x-1)^2}\cdot(2x-1)^2 \\\\ (2x-1)^2(2)=\dfrac{3}{\cancel{2x-1}}\cdot(2x-1)\cancel{^2}+\dfrac{-1}{\cancel{(2x-1)^2}}\cdot\cancel{(2x-1)^2} \\\\ (2x-1)^2(2)=3(2x-1)-1(1) \\\\ (2x-1)^2(2)=(2x-1)(3)-1 .\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)(1)+(1)^2](2)=(2x-1)(3)-1 \\\\ (4x^2-4x+1)(2)=(2x-1)(3)-1 .\end{array} Using the Distributive Property and then combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} 8x^2-8x+2=6x-3-1 \\\\ 8x^2+(-8x-6x)+(2+3+1)=0 \\\\ 8x^2-14x+6=0 \\\\ \dfrac{8x^2-14x+6}{2}=\dfrac{0}{2} \\\\ 4x^2-7x+3=0 .\end{array} In the trinomial expression above the value of $ac$ is $4(3)=12$ and the value of $b$ is $-7 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -3,-4 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 4x^2-3x-4x+3=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} (4x^2-3x)-(4x-3)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(4x-3)-(4x-3)=0 .\end{array} Factoring the $GCF= (4x-3)$ of the entire expression above results to \begin{array}{l}\require{cancel} (4x-3)(x-1)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} 4x-3=0 \\\\\text{OR}\\\\ x-1=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 4x-3=0 \\\\ 4x=3 \\\\ x=\dfrac{3}{4} \\\\\text{OR}\\\\ x-1=0 \\\\ x=1 .\end{array} Upon checking, none of the solutions above will make the denominator of the original equation equal to zero. Hence, $x=\left\{ \dfrac{3}{4},1 \right\} .$

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