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
Exponent problem
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- DanaJ
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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.
(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.
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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)
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)
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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.