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# What is the value of 1/(x^2 - y^2) - 1/(x^2 + 2xy + y^2) ?

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Math Expert
Joined: 02 Sep 2009
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What is the value of 1/(x^2 - y^2) - 1/(x^2 + 2xy + y^2) ?  [#permalink]

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14 Nov 2014, 09:53
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Difficulty:

45% (medium)

Question Stats:

66% (01:44) correct 34% (01:44) wrong based on 98 sessions

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Tough and Tricky questions: Algebra.

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) $$x + y = 4$$

Kudos for a correct solution.

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Joined: 10 Sep 2014
Posts: 98
Re: What is the value of 1/(x^2 - y^2) - 1/(x^2 + 2xy + y^2) ?  [#permalink]

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14 Nov 2014, 12:27
We can start off by factoring the denominators to get

1 / (x+y)(x-y) - 1 / (x+y)(x+y)

Statement 1: not sufficient
Factor to get 2y=(x+y)(x-y)
Plug into what we were given to get 1/2y - 1 / (x+y)(x+y). Not enough info.

Statement 2: not sufficient
Plug 4 in for (x+y) in the equation we were given originally to get 1/4(x-y) - 1/16. Not enough info.

Down to C and E.

Combined we know that 2y must equal 4(x-y) so we can say that 1/4(x-y) - 1/16 = 1/2y - 1/16. We still don't have a value for y and cannot get a value of the equation.

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Joined: 29 Sep 2014
Posts: 15
Re: What is the value of 1/(x^2 - y^2) - 1/(x^2 + 2xy + y^2) ?  [#permalink]

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14 Nov 2014, 13:29
1
First, one should notice that 1/(x^2 - y^2) - 1/(x^2 + 2xy + y^2) = 2*y/[(x+y)^2*(x-y)].

(1) gives 1/(x^2 - y^2) - 1/(x^2 + 2xy + y^2) = 1/(x+y)
INSUFFICIENT

(2) clearly INSUFFICIENT

(1) + (2) gives 1/(x^2 - y^2) - 1/(x^2 + 2xy + y^2) = 1/4

Intern
Joined: 20 Jan 2013
Posts: 33
Re: What is the value of 1/(x^2 - y^2) - 1/(x^2 + 2xy + y^2) ?  [#permalink]

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14 Nov 2014, 18:41
1
Let's Simplify the equation - 1/(x^2 - y^2) - 1/(x^2 + 2xy + y^2)

1/(x+y)(x-y) - 1/(x+y)^2
Take LCM and solve (x+y) - (x-y) /(x-y) (x + y)^2 => 2y / (x-y) (x + y)^2 ------ (1)

Statement 1 : 2y = x^2 - y^2 => 2y = (x+y) (x-y)
Put this value in equation 1) and check,

(x+y) (x-y) / (x-y) (x+y) (x+y) => 1/ x+y .... Not enough info
Not Sufficient. Option A & D are out

Statement 2 : x + y = 4
Put this value in equation 1) and check,
2y / (x-y) (4)^2 => 2y / 16 * (x-y) .... Not enough info
Not Sufficient. Option b is out

Combining both the statement,
We already have 1/ (x+y) from statement 1 as solution of equation and statement 2 provides the value for (x + y) as 4
so we can get 1/4 as final solution from both the statements.

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Re: What is the value of 1/(x^2 - y^2) - 1/(x^2 + 2xy + y^2) ?  [#permalink]

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14 Nov 2014, 19:35
1
1 / x^2-y^2 - 1 / (x+y)^2

=> {x+y -(x-y) } /
=> 2y/{(x^2-y^2)*(x+y)}

From statement 1, 2y = x^2 - y^2
we can say that ,
=> 2y /2y *(x+y) = 1/(x+y). Insufficient

From statement 2, x + y = 4
we can say that ,
=> 2y/ 16*(x-y). Insufficient

Combining both,
we can the value = 1/(x+y) = 1/4

Math Expert
Joined: 02 Sep 2009
Posts: 49303
Re: What is the value of 1/(x^2 - y^2) - 1/(x^2 + 2xy + y^2) ?  [#permalink]

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17 Nov 2014, 12:13
Bunuel wrote:

Tough and Tricky questions: Algebra.

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) $$x + y = 4$$

Kudos for a correct solution.

Official Solution:

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

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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Re: What is the value of 1/(x^2 - y^2) - 1/(x^2 + 2xy + y^2) ?  [#permalink]

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01 Jul 2018, 00:41
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Re: What is the value of 1/(x^2 - y^2) - 1/(x^2 + 2xy + y^2) ? &nbs [#permalink] 01 Jul 2018, 00:41
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