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S95-35

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S95-35 [#permalink]

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New post 16 Sep 2014, 01:50
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

67% (02:13) correct 33% (01:46) wrong based on 61 sessions

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Re S95-35 [#permalink]

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New post 16 Sep 2014, 01:50
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We must determine the value of the expression \(\frac{1}{x^2 - y^2} - \frac{1}{x^2 + 2xy + y^2}\). When a problem presents an opportunity to factor or find a common denominator, it is usually a good idea to do so.

First, factor the denominator of each fraction. The denominator of the fraction on the left is a difference of squares: \(\frac{1}{x^2 - y^2} = \frac{1}{(x + y)(x - y)}\). The denominator of the fraction on the right is the expanded form of the quadratic expression \((x + y)^{2}\): \(\frac{1}{x^2 + 2xy + y^2} = \frac{1}{(x + y)^2}\).

Thus, the fraction can be rewritten: \(\frac{1}{(x + y)(x - y)} - \frac{1}{(x + y)^2}\). The common denominator of these two fractions is \((x + y)^{2}(x - y)\). Multiply the first fraction by \(\frac{x + y}{x + y}\) and the second fraction by \(\frac{x - y}{x - y}\) and rewrite: \(\frac{x + y}{(x + y)^{2}(x - y)} - \frac{x - y}{(x + y)^{2}(x - y)}\).

Combine the terms by subtracting: \(\frac{x + y - (x - y)}{(x + y)^{2}(x - y)} = \frac{2y}{(x + y)^{2}(x - y)}\).

Statement 1 says that \(2y = x^2 - y^2\), or \(2y = (x + y)(x - y)\). Substitute this into the fraction that we derived above: \(\frac{2y}{(x + y)^{2}(x - y)} = \frac{(x + y)(x - y)}{(x + y)^{2}(x - y)}\). Cancel the factors that appear in both the numerator and the denominator, leaving \(\frac{1}{x + y}\). Without more information about \(x\) or \(y\), we cannot determine the value of this fraction. Statement 1 is NOT sufficient. Eliminate answer choices A and D. The correct answer choice is B, C, or E.

Statement 2 says that \(x + y = 4\). In this case, it will be easier to substitute into the expression \(\frac{1}{(x + y)(x - y)} - \frac{1}{(x + y)^2}\). Doing so gives: \(\frac{1}{4(x - y)} - \frac{1}{(4)^2}\). Without more information about \(x\) and \(y\), however, we cannot determine the value of this expression. Statement 2 is NOT sufficient. Eliminate answer choice B. The correct answer choice is either C or E.

When the statements are taken together, statement 1 allows us to simplify the fraction to \(\frac{1}{x + y}\), and statement 2 tells us that \(x + y = 4\). Substituting, we find: \(\frac{1}{x + y} = \frac{1}{4}\). Together, the statements are sufficient to answer the question.


Answer: C
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Re: S95-35 [#permalink]

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New post 05 Nov 2016, 07:20
any easy solution :shock:

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Re: S95-35 [#permalink]

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New post 11 Jan 2017, 13:55
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]

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New post 14 Apr 2017, 21:46
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] 14 Apr 2017, 21:46
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