## Intermediate Algebra for College Students (7th Edition)

{$-\dfrac{14}{3},\dfrac{16}{3}$}
Given: $|3y-1|+10=25$ This can also be written as: $|3y-1|=15$ As per definition of absolute value, we have $3y-1=15$ and $3y-1=-15$ Thus, $3y=16 \implies y=\dfrac{16}{3}$ and $3y=-14 \implies y=-\dfrac{14}{3}$ Hence, {$-\dfrac{14}{3},\dfrac{16}{3}$}