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NateTheGreat11
Joined: 26 Dec 2011
Last visit: 04 Feb 2015
Posts: 5
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Given Kudos: 3
Schools: Yale '17 (A)
19 Dec 2013, 18:02
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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) \(x(y-3) = 0\)
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.
1) \(x+y = 3\)
2) \(x(y-3) = 0\)
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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20 Dec 2013, 00:28
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NateTheGreat11
That's not correct. Why cannot y be 1 for (2)? Or 0, or 100?
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