## College Algebra (10th Edition)

The solution set is{ $-\frac{27}{2},\frac{27}{2}$}.
$\frac{2}{3}|x|=9$ Isolate the absolute value by making the coefficient of$|x|$ one. $|x|=9\cdot\frac{3}{2}$ $|x|=\frac{27}{2}$ $x=\frac{27}{2}$ or $x=-\frac{27}{2}$ The solution set is{ $-\frac{27}{2},\frac{27}{2}$}.