Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 6 - Review - Page 405: 70

Answer

$x=\left\{\dfrac{3}{2}, 5 \right\}$

Work Step by Step

Multiplying both sides by the $LCD= x^2 $, the value of the variable that satisfies the given equation, $ 2+\dfrac{15}{x^2}=\dfrac{13}{x} ,$ is \begin{array}{l}\require{cancel} x^2(2)+1(15)=x(13) \\\\ 2x^2+15=13x \\\\ 2x^2-13x+15=0 \\\\ (2x-3)(x-5)=0 \\\\ x=\left\{\dfrac{3}{2}, 5 \right\} .\end{array}
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