tough ds

[runzun]

Is x>0
1)x-y>-8

2)x+y=8

[Anurag@Gurome]

Is x>0
1)x-y>-8
2)x+y=8

(1) x - y > -8
If x = 2, y = 6 then x - y = 2 - 6 = -4 > -8, here x > 0.
If x = -2, y = 4, then x - y = -2 - 4 = -6 > -8, here x < 0.
From the above examples, it can be observed that x may or may not be > 0.
Therefore, statement 1 alone is NOT sufficient.

(2) x + y = 8
If x = 5, y = 3, then x + y =8; here x > 0.
If x = -2, y = 10, then x + y = 8; here x < 0.
Again from the above examples, it can be observed that x may or may not be > 0.
Therefore, statement 2 alone is NOT sufficient.

Now combine statements 1 and 2, x + y = 8 so that x - y > -8
Let us assume x to be any value < 0, say x = -1/4, and y = 33/4, then x + y = 8 and x - y = -17/2 or -8.5, which is < -8. From this it can be seen that if we take x < 0 then x - y < -8, which is a contradiction. This certainly means that x > 0.

The correct answer is C.

[gmatboost]

Picking number is a reliable strategy on DS inequalities, but I believe that if you can figure out the algebra, it is usually faster. In this case:

Statement 1 can be rearranged to x > y - 8
Is x > 0? We have no idea what y is, Insufficient

Statement 2 can be rearranged to x = 8 - y
Is x > 0? Again, we have no idea what y is, Insufficient

Combined: Since we are asked about x, let's rearrange Statement 2: y = 8 - x
This allows us to plug that into x > y - 8 (from St 1) and get an inequality with only x
x > (8 - x) - 8
x > -x
2x > 0
x > 0, Sufficient

[HSPA]

Addition of A and B shall be an easy technique here.

x+y = 8
x-y = -7.9....1,2,3...

when we add, least value of x is (0.1)/2 which is > 0 [Condition: point is limited to tenth's place]