Note, let "!=" read "not equal".

If x != y, is (x-y)/(x+y) > 1?

1) x'>'0
2) y'<'0.






even if (1) and (2) are both true, the sign of (x + y) still isn't known. so at this point you should just dispense with the theory and PICK NUMBERS, with the goal of making the denominator negative one time and positive the next.

x = 1, y = -2: is 3/(-1) '>' 1? no.
x = 2, y = -1: is 3/1 '>' 1? yes.
insufficient
ans = e