${2x-3y}\leq{-6}$ --> $y\geq{\frac{2}{3}x+2}$. Thi inequality represents ALL points, the area, above the line $y={\frac{2}{3}x+2}$. If you draw this line you'll see that the mentioned area is "above" IV quadrant, does not contains any point of this quadrant.

Else you can notice that if $x$ is positive, $y$ can not be negative to satisfy the inequality $y\geq{\frac{2}{3}x+2}$, so you can not have positive $x$, negative $y$. But IV quadrant consists of such $(x,y)$ points for which $x$ is positive and $y$ negative. Thus answer must be E.

Answer: E.