Hi,
From(1): x+y=z => x=z-y
So, |x| = |z-y|
So, |x| = z-y if x>0
& |x| = y-z if x<0
Insufficient

From(2):
x<0. No info about the relation between x,y,z
Insufficient

Both(1)&(2): x<0
So, from(1) we can say |x| = y-z
Sufficient

Hence, C

|x|=y-z??
a) x+y=z
x=z-y
|X| = |z-y| = y-z only when y>=z
= z-y when z>y
insufficient
b)x<0 insufficient
a&b) x<0
x+y=z -> z<y
thus |x| = y-z
Sufficient
IMO C

|x| = y -z?
This gives two equations when x >=0 0 x = y-z => eq 1 and when x< 0 -x = y-z i.e x = z-y => eq 2

1 : X+Y = Z first take +ve x x=1, y=2 and z=3 when we substitute in eq 1 we get False and when we substitute in eq 2 we get True. Since we get both true and false this is insufficient.

2: By itself it doesn't give any relation among x,y and z so Insuff

Now combining both check for only negative values of X So we have to take some values and substitute only in EQ 2 since we know that x < 0

Now we take -1,3, 2 -1 = 2-3

-1 = -1

Take only more example just to be sure

-5 7 2 -5 = 2-7

-5 = 5

So Sufficient answer is C