Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 4 - Section 4.3 - Equations and Inequalities Involving Absolute Value - Exercise Set - Page 283: 17



Work Step by Step

Given: $|6y-2|+4=32$ This can also be written as: $|6y-2|=28$ As per definition of absolute value, we have $6y-2=28$ and $6y-2=-28$ Thus, $6y=30 \implies y=5$ and $6y=-26 \implies y=-\dfrac{13}{3}$ Hence, {$-\dfrac{13}{3},5$}
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