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GMATClub Tests M28-03

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GMATClub Tests M28-03 [#permalink] New post 19 Dec 2013, 17:02
If x \neq y and y = \frac{x^2-y^2}{x-y}, then what is the value of y?

1) x+y = 3

2) x(y-3) = 0

[Reveal] Spoiler:
A



My question is why not D? below is my logic

Through manipulation we get:
y = (x+y)
Thus, x = 0 and y = (x+y)

Statement 1 gives the very value of x+y. Sufficient.
Statement 2 prompts two scenarios:
x(y-3) = 0

x= 0
OR
(y-3) = 0

Scenario A
X = 0

OR

Scenario B
Y = 3

From Scenario A we find that if X = 0, y can take any value

From Scenarios B we find that if y = 3, then x can take any value; BUT since we know from the question stem that X has to be equal to zero then if Y = 3, X = 0.
So now looking back at Scenario A, if X = 0, then y *cannot* take any value because it must be equal to 3. Hence D.
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Re: GMATClub Tests M28-03 [#permalink] New post 19 Dec 2013, 23:28
Expert's post
NateTheGreat11 wrote:
If x \neq y and y = \frac{x^2-y^2}{x-y}, then what is the value of y?

1) x+y = 3

2) x(y-3) = 0

[Reveal] Spoiler:
A



My question is why not D? below is my logic

Through manipulation we get:
y = (x+y)
Thus, x = 0 and y = (x+y)

Statement 1 gives the very value of x+y. Sufficient.
Statement 2 prompts two scenarios:
x(y-3) = 0

x= 0
OR
(y-3) = 0

Scenario A
X = 0

OR

Scenario B
Y = 3

From Scenario A we find that if X = 0, y can take any value

From Scenarios B we find that if y = 3, then x can take any value; BUT since we know from the question stem that X has to be equal to zero then if Y = 3, X = 0.
So now looking back at Scenario A, if X = 0, then y *cannot* take any value because it must be equal to 3. Hence D.


That's not correct. Why cannot y be 1 for (2)? Or 0, or 100?
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Re: GMATClub Tests M28-03   [#permalink] 19 Dec 2013, 23:28
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