## Algebra 1

$y=1\frac{1}{3}$ OR $y=-2\frac{2}{3}$
$3|3y+2|=18\longrightarrow$ divide each side by 3 $3|3y+2|\div3=18\div3\longrightarrow$ divide $|3y+2|=6\longrightarrow$ definition of absolute value $3y+2=6$ OR $3y+2=-6\longrightarrow$ subtract 2 from all parts $3y+2-2=6-2$ OR $3y+2-2=-6-2\longrightarrow$ subtract $3y=4$ OR $3y=-8\longrightarrow$ divide all parts by 3 $3y\div3=4\div3$ OR $3y\div3=-8\div3\longrightarrow$ divide $y=1\frac{1}{3}$ OR $y=-2\frac{2}{3}$