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# If x and y are positive, what is the value of x y? (1) (x^2

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If x and y are positive, what is the value of x y? (1) (x^2 [#permalink]

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10 Jun 2008, 12:33
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If x and y are positive, what is the value of x – y?
(1) (x^2 – y^2) / (x + y) = 4
(2) x + y = 7
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MBA Blog: University of Minnesota Carlson School of Management- http://unconventionalapplicant.blogspot.com/

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10 Jun 2008, 12:43
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Statement #1

If you have
$$\frac{(x^2-y^2)}{(x+y)} = 4$$ this the same as

$$\frac{(x+y)(x-y)}{(x+y)} = 4$$ The (x+y) on the top and bottom cancel out leaving only:

$$(x-y)=4$$

Statement #2

x + y = 7, then x = 7 - y. No matter what we do, I don't think it's possible to get a number value for x-y from x+y=7.

We could try to add x to both sides of x = 7 -y to get 2x = 7 + x - y, then 2x - 7 = x-y, but this still is not an actual value.

carlson2010 wrote:
If x and y are positive, what is the value of x – y?
(1) (x^2 – y^2) / (x + y) = 4
(2) x + y = 7

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Last edited by jallenmorris on 10 Jun 2008, 12:48, edited 3 times in total.
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10 Jun 2008, 12:44
jallenmorris wrote:
A

carlson2010 wrote:
If x and y are positive, what is the value of x – y?
(1) (x^2 – y^2) / (x + y) = 4
(2) x + y = 7

You are right. Can you show how? Thanks.
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10 Jun 2008, 12:45
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(1) (x^2 – y^2) / (x + y) = 4
(2) x + y = 7

From statement 1:
(x^2 – y^2) / (x + y) = 4
=>
(x + y) * (x - y) / (x + y) = 4
=>
x - y = 4; SUFFICIENT

Statement 2 is insufficient
counter-example: (x, y) = {(6, 1), (5, 2), ...}
Re: DS algebra   [#permalink] 10 Jun 2008, 12:45
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