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

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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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New post 14 Nov 2014, 09:53
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A
B
C
D
E

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  45% (medium)

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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) \(x + y = 4\)

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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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New post 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.

Answer E!
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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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New post 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

Answer (C)
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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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New post 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.

Hence answer is C
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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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New post 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

C) should be the answer
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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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New post 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.

Answer: C.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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