College Algebra (11th Edition)

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

Chapter 1 - Review Exercises - Page 164: 70


no solution

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \dfrac{2}{x+2}+\dfrac{1}{x+4}=\dfrac{4}{x^2+6x+8} ,$ factor all expressions that can be factored. Then multiply both sides by the $LCD$ and isolate the variable. Finally, do checking and ensure that any denominator does not become $0.$ $\bf{\text{Solution Details:}}$ Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the factored form of the equation above is \begin{array}{l}\require{cancel} \dfrac{2}{x+2}+\dfrac{1}{x+4}=\dfrac{4}{(x+2)(x+4)} .\end{array} The $LCD$ of the denominators, $x+2,x+4,$ and $(x+2)(x+4)$ is $(x+2)(x+4)$ since it is the lowest expression which can be divided exactly by all the given denominators. Multiplying both sides by the $LCD= (x+2)(x+4) ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} (x+2)(x+4)\left( \dfrac{2}{x+2}+\dfrac{1}{x+4} \right)=\left( \dfrac{4}{(x+2)(x+4)} \right) (x+2)(x+4) \\\\ (x+4)(2)+(x+2)(1)=1(4) \\\\ 2x+8+x+2=4 \\\\ 2x+x=4-8-2 \\\\ 3x=-6 \\\\ x=-\dfrac{6}{3} \\\\ x=-2 .\end{array} If $x=-2,$ the part of the given equation, $ \dfrac{2}{x+2} ,$ becomes $ \dfrac{2}{0} ,$ which is undefined. Hence, there is $\text{ no solution .}$
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