Question 1120283
x-2y> =0
This is -2y>=-x or y <=(1/2)x
The second is -y <=-2x-2 or y >=2x+2
For the first, which goes through the origin, y is BELOW the graph
For the second, at (0, 0) is NOT a solution since 0 is not >=2, so the side AWAY from the origin is what is desired.
What is desired is the sector that lies wholly in quadrant III.
{{{graph(300,300,-10,10,-10,10,(x/2),2x+2)}}}