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: 19


$t=\left\{ -\dfrac{1}{2},\dfrac{3}{4} \right\}$

Work Step by Step

Multiplying both sides by the $LCD= 10 $, then the solution to the given equation, $ \dfrac{4t^2}{5}=\dfrac{t}{5}+\dfrac{3}{10} $, is \begin{array}{} 2(4t^2)=2(t)+1(3) \\\\ 8t^2=2t+3 \\\\ 8t^2-2t-3=0 \\\\ (2t+1)(4t-3)=0 .\end{array} Equating each factor to zero, then $ t=\left\{ -\dfrac{1}{2},\dfrac{3}{4} \right\} $.
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