Intermediate Algebra (6th Edition)

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

Chapter 5 - Section 5.8 - Solving Equations by Factoring and Problem Solving - Exercise Set - Page 323: 18

Answer

$y=\left\{ 5,-3 \right\}$

Work Step by Step

Multiplying both sides by the $LCD= 30 $, then the solution to the given equation, $ \dfrac{y^2}{30}=\dfrac{y}{15}+\dfrac{1}{2} $, is \begin{array}{} 1(y^2)=2(y)+15(1) \\\\ y^2=2y+15 \\\\ y^2-2y-15=0 \\\\ (y-5)(y+3)=0 .\end{array} Equating each factor to zero, then $ y=\left\{ 5,-3 \right\} $.
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