Exponent problem

[relaxin99]

0 < x < y; what is the value of (x+y)^2/(x-y)^2

1) x^2 + y^2 = 3xy
2) xy = 3

With the answer, can you all please explain how you came up with the answer thanks

[DanaJ]

Let's work on the initial equation first:
(x+y)^2 = x^2 + y^2 + 2xy
(x-y)^2 = x^2 + y^2 - 2xy.

Let's use stmt 1 first: since x^2 + y^2 = 3xy, you can replace that in the initial equation and you get that it is 5xy/xy = 5. So 1 is sufficient.

2 will not be sufficient since we do not know anything about x^2 + y^2.

So I'd go with A on this one.

[relaxin99]

This comes absolutely efortlessly to you...great, great

[cramya]

Good solution Dana!

For B since x and y are not given to be integers many different values possible satisfying 0<x<y & xy=3 (x=1 y =3 x=1/3 y=9 etc..) leading to different values for (x+y)^2/(x-y)^2 so INSUFF

Choose A (xy cancels out leading in to one definite answer)

[DanaJ]

Thank you guys for your kind words..

[NicoFalcon]

I try to follow this advice throughly. Always when you see an equation FACTORIZE ,if it`s unfactorized, and UNFACTORIZE, if it is factorized. Always try to simplify whenever you can, and things will usually start to be clear.