Answer
$$\frac{1}{2}\ln \left| {2y - 3} \right| = - x + C$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = 3 - 2y \cr
& {\text{Separate the variables}} \cr
& \frac{{dy}}{{3 - 2y}} = dx \cr
& \frac{{dy}}{{2y - 3}} = - dx \cr
& {\text{Integrate both sides}} \cr
& \frac{1}{2}\ln \left| {2y - 3} \right| = - x + C \cr
& \cr
& {\text{Graph}} \cr} $$
