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

{$-\dfrac{13}{3},5$}

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, {$-\dfrac{13}{3},5$}