Once we have y<0, you can basically just plug into the equations and see what would work for different versions of y.

The choice y-4<-7 works in some cases, but is not always correct, because rearranged, this choice can be expressed as y<-3, which does not completely overlap with y<0. You can see this if you plug in y=-1. You would get (-1)-4 < -7, which simplifies to -5 < -7, which is not true.

The choice y+1<2 can be rearranged to y<1, which does overlap completely with y<0. Plugging in y=-1 does work, as you get (-1)+1 < 2, which simplifies to 0<2, which is true. Even if you use a more negative number like y=-100, the choice is still true: (-100) + 1 < 2 simplifies to -99 < 2, which is still true.

I agree with your description. However, I feel you don’t consider the corner case when y=0.

y+1 < 2 implies y < 1 and not y< 0. Since y< 1 , y can take values left of 1 , that includes 0. The issue with choosing y=0 is then it invalidates xy< 0. So, I still don’t feel y+1< 2 is correct. On the other hand if it was y+3< 2 or any other combination where, after simplification we get y<0, I would have no doubt in my mind.

We’ll never be able to have y = 0 because we already determined from the given equations that we must have y < 0 in all cases. It isn’t a possibility for it to be zero or greater because it would violate the constraints of the question.

As y is also defined to be an integer, the greatest it could be is y = -1