Answer: C

Re-writing xy+x=z as x(y+1)=z --(a)

Now looking at the question we can definitely say

|x+y|>z if a) z is -ve as mod of any number is always +ve

OR b) it can be proved that z is positive and > x+y

Taking i) x< 0 our modified equation (a) suggests that sign of z depends on sign of (y+1) so it can be both +ve or -ve. There is no data on the actual numeric value of x ,y or z. so. Not sufficient

Taking ii) y>0 our modified equation (a) suggests that sign of z depends on sign of x so it can be both +ve or -ve. There is no data on the actual numeric value of x ,y or z. so. Not sufficient

Combining i and ii and looking at eq -(a)

x< 0 and y>0 => (y+1)>0. Hence x(y+1) <0 => z <0. From our earlier inference if z<0 then |x+y|>z. Sufficient.

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Thanks

Sindbad

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