## Algebra 2 (1st Edition)

$|8r+1|=3r$ implies $8r+1=3r\\1=-5r\\r=-0.2$ or $8r+1=-3r\\1=-11r\\r=\frac{-1}{11}$. Here $r=-0.2$ and $r=\frac{-1}{11}$ are extraneous solutions. Thus we have no solutions.