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: 26

Answer

$x=\left\{ -1,\dfrac{2}{3} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given rational equation, $ \dfrac{3x^2}{x-1}+2=\dfrac{x}{x-1} ,$ 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 $\left\{ (x-1),1,(x-1) \right\}$ is $ (x-1) .$ Multiplying both sides of the given equation by the $LCD= (x-1) $ results to \begin{array}{l}\require{cancel} (x-1)\left( \dfrac{3x^2}{x-1}+2\right)=\left(\dfrac{x}{x-1}\right)(x-1) .\end{array} Using the Distributive Property and then combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} (x-1)\cdot\dfrac{3x^2}{x-1}+(x-1)\cdot2=\dfrac{x}{x-1}\cdot(x-1) \\\\ \cancel{(x-1)}\cdot\dfrac{3x^2}{\cancel{(x-1)}}+(x-1)\cdot2=\dfrac{x}{\cancel{(x-1)}}\cdot\cancel{(x-1)} \\\\ 1(3x^2)+(x-1)(2)=x(1) \\\\ 3x^2+2x-2=x \\\\ 3x^2+(2x-x)-2=0 \\\\ 3x^2+x-2=0 .\end{array} In the trinomial expression above the value of $ac$ is $ 3(-2)=-6 $ and the value of $b$ is $ 1 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -2,3 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 3x^2-2x+3x-2=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} (3x^2-2x)+(3x-2)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(3x-2)+(3x-2)=0 .\end{array} Factoring the $GCF= (3x-2) $ of the entire expression above results to \begin{array}{l}\require{cancel} (3x-2)(x+1)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} 3x-2=0 \\\\\text{OR}\\\\ x+1=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 3x-2=0 \\\\ 3x=2 \\\\ x=\dfrac{2}{3} \\\\\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\{ -1,\dfrac{2}{3} \right\} .$
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