College Algebra (11th Edition)

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

Chapter 1 - Review Exercises - Page 164: 65



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \dfrac{2}{x}-\dfrac{4}{3x}=8+\dfrac{3}{x} ,$ multiply both sides by the $LCD.$ Then use the properties of equality to isolate the variable. Finally, do checking and ensure that the denominator does not become $0.$ $\bf{\text{Solution Details:}}$ The $LCD$ of the denominators, $x,3x,1,$ and $x$ is $3x$ since it is the lowest expression which can be divided exactly by all the given denominators. Multiplying both sides by the $LCD= 3x ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 3x\left( \dfrac{2}{x}-\dfrac{4}{3x} \right) =\left(8+\dfrac{3}{x}\right)3x \\\\ 3(2)-1(4)=8(3x)+3(3) \\\\ 6-4=24x+9 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} -24x=9-6+4 \\\\ -24x=7 \\\\ x=\dfrac{7}{-24} \\\\ x=-\dfrac{7}{24} .\end{array} Upon checking, the solution does not produce an expression with a $0$ in the denominator. Hence, the solution is $ x=-\dfrac{7}{24} .$
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