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Re: S95-35 [#permalink]
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Easier perhaps (if I'm correct):

\(\frac{1}{x^2−y^2} - \frac{1}{x^2+2xy+y^2}\)

which is the same than: \(x^2+2xy+y^2 = x^2−y^2\)
hence \((x+y)^2 = x^2−y^2\)
and \((x+y) (x+y) = x^2−y^2\)

Statement (1) \(2y = x^2−y^2\)
Therefore \((x+y)^2 = 2y\)

Statement (2) \(x+y = 4\)
Therefore \(4 * 4 = 2y\)

Which finally gives \(y = 8\)

Answer C, both statements needed.
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Re: S95-35 [#permalink]
marinlst wrote:
Easier perhaps (if I'm correct):

\(\frac{1}{x^2−y^2} - \frac{1}{x^2+2xy+y^2}\)

which is the same than: \(x^2+2xy+y^2 = x^2−y^2\)


marinlst, I do not think that is correct. According to that logic, 1/4 - 1/2 would be the same as 4=2, which is not right.

This is an extremely hard question. The official answer says that you must simplify the original equation to \(\frac{2y}{(x-y)(x+y)^2}\). I was close but I couldn't finish in time. When I combined the two denominators, I ended up with : \(\frac{(x+y)^2-(x+y)(x-y)}{(x+y)^3(x-y)}\). I didn't know what to do from this point, but if I had factored out (x+y) from both the top and bottom I would have eventually gotten \(\frac{(x+y)-(x-y)}{(x+y)^2(x-y)}\) which simplifies to \(\frac{2y}{(x+y)^2(x-y)}\) which would have made it very easy to see the correct answer.

Is completely solving out the equation really the fastest way to the answer? I tried to use guess and check but that didn't work. I guessed with value x=0 and got y=0,-2 for statement 1 and y=4 for statement 2, which doesn't make sense.
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Re: S95-35 [#permalink]
If you have 2 variables and 1 equation (Statement 1 OR 2) you cannot solve it. However, with 2 (different!) equations (Statement 1 AND 2) you can solve it.
Normally this is the case, not sure if always true???
Anyway... using this you can quickly guess the answer.
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Re: S95-35 [#permalink]
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