If x does not equal -y is (x-y)/(x+y)>1?

Hey Guys,

I really need help on the thought process with DS not necessary how to solve the problem. I have a hypothetical DS below. The goal of my question isn't to solve the problem but to find what necessary information is need to begin with to solve the problem.


===============================Question 1================================
If x does not equal -y is (x-y)/(x+y)>1?

(1) y<0
(2) x+y>0


Lets take the two scenarios:

a) If x+y>0 which implies (x-y)>(x+y) ---> y>0
b) If x+y<0 which implies (x-y)<(x+y) ----> y<0

Now Statement (1) in the DS meets scenario (b)'s criteria. So does it make it sufficient? If we know y<0 does this imply x+y<0 and therefore (x-y)/(x+y)>1?

When we use multiple "if-then" scenarios does the questionmark on the statement still extend? In other words should statements (a) & (b) correctly be



(x-y)/(x+y)>1 ??
a) If x+y>0 which implies (x-y)>(x+y) ?? ---> y>0 ??
b) If x+y<0 which implies (x-y)<(x+y) ?? ----> y<0 ??


========================================Question 2=========================
To illustrate the point further he is another example:
Is x^2-y^2<0?

(1) |x|<|y|
(2) x+y>0




Is x^2-y^2<0?
can be reworded as x^2<y^2 ??
sqrt(x^2)<sqrt(y^2) ??
|x|<|y| ??

Notice how the question mark extended all the way down as we manipulated the question to get the new question is |x|<|y| ??

In our first statement (1) |x|<|y| matches our question. Hence sufficient.

Is this logic of extending the question mark the same for "if-then" scenarios?