Bunuel wrote:
If xy + x = z, is |x + y| > z?
(1) x < 0
(2) y > 0
An alternate explanation
x(y+1) - z = 0
First take 1 and 2 both, if x is negative and y is positive, then term x(y+1) is negative, so for x(y+1) - z =0 to be true, z has to be negative so that -z is positive and x(y+1) - z =0. So from this we can deduce that z is also negative--So question stem---> |x+y| is always positive and if z is negative, so |x+y| >z is true.
Now C is correct unless we can make one out of 1 and 2 individually correct or both individually correct. Let's analyze
Take 1) x < 0, if x is negative (and y is positive or negative), nothing can be said about sign of z so we can never know |x+y| >z.
Take 2) y > 0, if y is positive (x can be positive or negative), nothing can be said about sign of z so we can never know |x+y| >z.
So answer is C