But (x + y) (x -Y) > 0 means either x + y> 0 , or x - y> 0
Given x + y> 0 does not mean that x - y> 0 In this case we are lucky.

Is there any other approach please?

hi..
(x + y) (x -Y) > 0 means either 'both (x + y) and (x -Y) are greater than 0' or 'both (x + y) and (x -Y) are less than 0' because + * + =+ and -*-=+
so if x=y>0, x-y will also be greater than 0.

Another approach will be logical ..
|x|>|y| can be written as |x-0|>|y-0|, so distance of x from 0, is more than y from 0.
so x+y will depend on the sign of x, if x is negative (x+y) will be negative and if x is positive, (x+y) will be positive..
so we are given x+y>o, which means x>0..
if x>0, |x|=x, so x>|y|..
y will be equal or less than |y|, so x>y

The approach used by @kaziselim is great.
However, if you missed that, here's a different approach:

GIVEN: |x| > |y|
This inequality tells us that the MAGNITUDE of x is greater than the MAGNITUDE of y
In other words, on the number line, the distance from x to 0 is greater than the distance from y to 0

This gives us 4 possible cases:
case i: ----x----y----0------ (x and y are both negative)
case ii: ----x--------0--y---- (x is negative and y is positive)
case iii: ---y--0-------x-- (y is negative and x is positive)
case iv: --0---y----x-- (x and y are both positive)
Notice that, for all 4 cases, I've made the distance from x to 0 greater than the distance from y to 0

Now examine the second piece of given information...
GIVEN: x + y > 0
When we examine the 4 cases above, only cases iii and iv satisfy the condition that x + y > 0
In case iii, x > y
In case iv, x > y
So, it MUST be the case that x > y

Answer: B

Cheers,
Brent

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