Verified Answer :

naǫ

In quadrant III x-coordinates are less than 0 and y-coordinates are less than 0.

\(x<0 \\ y<0\)

You can modify it to get other examples:
\(x+1<1 \\ 2y-4<-4 \\ \\ -x>0 \\ y+2<2 \\ \\ \frac{x+y}{2}<\frac{1}{2}y \\ -2y+x>x-y\)

AL2006


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 .