IMO the answer to this is E as from both of the statements it cannot be inferred that y >0 or y <0.

please elaborate more, how you solved the stem 2.

2) -Y < X => X> -y
if X = 2 and Y =3
2>-3 therefore Y > 0

please clarify!

(1) 7x - 2y > 0

==>
x > 2/7*y (*)

these values satisfy (*):
x = 1, y = 7 ==> y > 0
x = 1, y = -1/2 ==> y < 0

INSUFF



(2) -y < x

==>
x > -y (**)

these values satisfy (**) :
x = 1, y = 7 ==> y > 0
x = 1, y = -1/2 ==> y < 0

INSUFF

Note that these values satisfy both (*) and (**), but does not tell us distinctly if y < 0.

Ans: E

(1)
y can be <0 or >0.

(2)
same case as (1). Try plugging positive and negative numbers

combined
tells us
8x>y
Insuff

Answer E

If x and y are integers and x > 0, is y > 0?
(1) 7x - 2y > 0
(2) -y < x

x>0, x & y intergers
possible values: x : + only; y: +, -, 0

1) x> 2/7 y => y has to be a multiple of 7 for x to be an int => y: 7, -7, 0; x: +
x y x>0 7x-2y>0 y>0
3 7 y y y
3 -7 y y n
1 0 y y n

Insufficient

2) x: +, y: +, -, 0 => to be scenario values same in (1) & (2), take y: 7, -7,0
x y x>0 x+y>0 y>0
1 7 y y y
8 -7 y y n
2 0 y y n

Insufficient

1& 2) x: +, y: 7, -7, 0

x y x>0 7x-2y>0 x+y>0 y>0
3 7 y y y
3 -7 y n n
1 0 y y y
Insufficient (answer E)

Hope it helps. Need one clarification: 7x-2y>0 & x+y>0 => 8x>y (is it correct???)

pnk wrote: Need one clarification: 7x-2y>0 & x+y>0 => 8x>y (is it correct???)

Yes. Inequalities with the same sign (and only the same sign) can be added (and only added).