## College Algebra (10th Edition)

The solution set is {$-\frac{5}{2}$}
$\frac{1}{2}-\frac{1}{3}p=\frac{4}{3}$ The LCD is 6 $6(\frac{1}{2}-\frac{1}{3}p)=\frac{4}{3}6$ Multiply both sides by 6. $6\cdot(\frac{1}{2})-6\cdot(\frac{1}{3}p)=\frac{4}{3}\cdot(6)$ Distribute and cancel out common factors. $3-2p=8$ Solve for p. $3-3-2p=8-3$ $-2p=5$ $\frac{-2}{-2}p=\frac{5}{-2}$ $p=-\frac{5}{2}$ The solution set is {$-\frac{5}{2}$}