## College Algebra (10th Edition)

The solution set is {$-1,\frac{3}{2}$}.
$|1-4t|+8=13$ Isolate the absolute value $|1-4t|=13-8$ $|1-4t|=5$ $1-4t=5$ or $1-4t=-5$ $-4t=4$ or $-4t=-6$ $t=-1$ or $t=\frac{-6}{-4}=\frac{3}{2}$ The solution set is {$-1,\frac{3}{2}$}