Answer
tangent $y=2x-4$
normal $y=-\frac{1}{2}x+\frac{7}{2}$
Work Step by Step
Step 1. Given $xy+2x-5y=2$, we have $y+xy'+2-5y'=0$ or $y'=\frac{y+2}{5-x}$ which gives the slope of tangent lines to the curve.
Step 2. At point $(3,2)$, the slope of the tangent line is $m=\frac{2+2}{5-3}=2$ and the equation is $y-2=2(x-3)$ or $y=2x-4$
Step 3. The slope of the normal line is $n=-1/m=-\frac{1}{2}$ and the equation is $y-2=-\frac{1}{2}(x-3)$ or $y=-\frac{1}{2}x+\frac{7}{2}$
