\({2x-3y}\leq{-6}\) --> \(y\geq{\frac{2}{3}x+2}\). 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.