## College Algebra 7th Edition

$x=\displaystyle \frac{35}{2}$ $x=\displaystyle -\frac{25}{2}$
$\displaystyle 8+5| \frac{1}{3}x-\frac{5}{6}|=33$ $\displaystyle 5|\frac{1}{3}x-\frac{5}{6}|=33-8=25$ $\displaystyle |\frac{1}{3}x-\frac{5}{6}|=25/5=5$ We split the absolute value equation into the positive and negative: $\displaystyle \frac{1}{3}x-\frac{5}{6}=5$ $\displaystyle \frac{1}{3}x=\frac{5*6}{6}+\frac{5}{6}=\frac{35}{6}$ $x=\displaystyle \frac{35*3}{6}=\frac{35}{2}$ $\displaystyle \frac{1}{3}x-\frac{5}{6}=-5$ $\displaystyle \frac{1}{3}x=-\frac{5*6}{6}+\frac{5}{6}=-\frac{25}{6}$ $\displaystyle x=-\frac{25*3}{6}=-\frac{25}{2}$