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# S95-35

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Math Expert
Joined: 02 Sep 2009
Posts: 43853

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16 Sep 2014, 00:50
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Difficulty:

35% (medium)

Question Stats:

65% (01:14) correct 35% (01:37) wrong based on 65 sessions

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What is the value of $$\frac{1}{x^2 - y^2} - \frac{1}{x^2 + 2xy + y^2}$$?

(1) $$2y = x^2 - y^2$$

(2) $$x + y = 4$$
[Reveal] Spoiler: OA

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16 Sep 2014, 00:50
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Official Solution:

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.

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05 Nov 2016, 06:20
any easy solution
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Joined: 10 Jan 2017
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11 Jan 2017, 12:55
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 $$y = 8$$

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14 Apr 2017, 20:46
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.
Re: S95-35   [#permalink] 14 Apr 2017, 20:46
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# S95-35

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