College Algebra (11th Edition)

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

Chapter 1 - Review Exercises - Page 164: 69



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \dfrac{x}{x+2}+\dfrac{1}{x}+3=\dfrac{2}{x^2+2x} ,$ factor all expressions that can be factored. Then multiply both sides by the $LCD.$ Use the concepts of solving quadratic equations to find the values of $x.$ Finally, do checking and ensure that any denominator does not become $0.$ $\bf{\text{Solution Details:}}$ The factored form of the equation above is \begin{array}{l}\require{cancel} \dfrac{x}{x+2}+\dfrac{1}{x}+3=\dfrac{2}{x(x+2)} .\end{array} The $LCD$ of the denominators, $x+2,x,1,$ and $x(x+2)$ is $x(x+2)$ since it is the lowest expression which can be divided exactly by all the given denominators. Multiplying both sides by the $LCD= x(x+2) ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} x(x+2)\left( \dfrac{x}{x+2}+\dfrac{1}{x}+3 \right)=\left(\dfrac{2}{x(x+2)} \right) x(x+2) \\\\ x(x)+(x+2)(1)+x(x+2)(3)=1(2) \\\\ x^2+x+2+3x^2+6x=2 \\\\ (x^2+3x^2)+(x+6x)+(2-2)=0 \\\\ 4x^2+7x=0 .\end{array} Factoring the $GCF$ results to \begin{array}{l}\require{cancel} x(4x+7)=0 .\end{array} Equating each factor to zero (Zero Product Property) results to \begin{array}{l}\require{cancel} x=0 \\\\\text{OR}\\\\ 4x+7=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x=0 \\\\\text{OR}\\\\ 4x+7=0 \\\\ 4x=-7 \\\\ x=-\dfrac{7}{4} .\end{array} If $x=0,$ the part of the given equation, $ \dfrac{1}{x} ,$ becomes $ \dfrac{1}{0} ,$ which is undefined. Hence, only $ x=-\dfrac{7}{4} $ satisfies the given equation.
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