Answer :
In quadrant III x-coordinates are less than 0 and y-coordinates are less than 0.
[tex]x<0 \\ y<0[/tex]
You can modify it to get other examples:
[tex]x+1<1 \\ 2y-4<-4 \\ \\ -x>0 \\ y+2<2 \\ \\ \frac{x+y}{2}<\frac{1}{2}y \\ -2y+x>x-y[/tex]
[tex]x<0 \\ y<0[/tex]
You can modify it to get other examples:
[tex]x+1<1 \\ 2y-4<-4 \\ \\ -x>0 \\ y+2<2 \\ \\ \frac{x+y}{2}<\frac{1}{2}y \\ -2y+x>x-y[/tex]
If all solutions are points in Quadrant-3, then the x-coordinates
and y-coordinates of all those points are all negative.
I believe the only possible system of liner inequalities whose
solutions have those characteristics is . . .
x < 0
y < 0 .
