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C is the correct answer.
We can simplify the stem. x$y = (x + y) * (x - y ) / (x-y)
We can simplify it further and will end up with x$y = x + y (when x =/=y is not equal to). Therefore, insufficient if x = y.

Statement 2 does not help us much. Insufficient.

Combining the two statements, we get that x+y is 3 and x is not equal to y. It is sufficient as it is right now, but if you would like to solve for x and y, you can. x = 2.5 and y = 0.5;

I am not sure about this OA/OE , can some Math experts validate/invalidate this OE with a better example.

Normally, the question will already contain x != y. In addition, the equation exists and is given: so we cannot have x=y. It is not as we are searching the solution and we are transforming an equation.

Hey fig, don't u think that u are violating the basic rule of Data Sufficiency.

IT is given that x$y = (x^2-y^2)/(x-y).
That's it ...............
We know that the above function is difined if x!=y. But how can we consider that the given statment implies this................. _________________

Hey fig, don't u think that u are violating the basic rule of Data Sufficiency.

IT is given that x$y = (x^2-y^2)/(x-y). That's it ............... We know that the above function is difined if x!=y. But how can we consider that the given statment implies this.................

Well, I would like to say u no

An equation is definied and implies by its existence that x!=y ... In my eyes, it's widely enough

Hey fig, don't u think that u are violating the basic rule of Data Sufficiency.

IT is given that x$y = (x^2-y^2)/(x-y). That's it ............... We know that the above function is difined if x!=y. But how can we consider that the given statment implies this.................

Well, I would like to say u no

An equation is definied and implies by its existence that x!=y ... In my eyes, it's widely enough

Hey fig u mean to say that when x=y the function is not defined.
Now 1 says x+y =3
I will take x=1.5 and y=1.5...............
Then where is the function to calculate with the above values , i mean when they are same the function is not defined.
So i say that 1 is not sufficient _________________

Hey fig, don't u think that u are violating the basic rule of Data Sufficiency.

IT is given that x$y = (x^2-y^2)/(x-y). That's it ............... We know that the above function is difined if x!=y. But how can we consider that the given statment implies this.................

Well, I would like to say u no

An equation is definied and implies by its existence that x!=y ... In my eyes, it's widely enough

Hey fig u mean to say that when x=y the function is not defined. Now 1 says x+y =3 I will take x=1.5 and y=1.5............... Then where is the function to calculate with the above values , i mean when they are same the function is not defined. So i say that 1 is not sufficient

Yes... To me, we have a system of equations :

(1) x$y = (x^2-y^2)/(x-y) (implying by existence of the equation x!=y)
(2) x+y =3

So, in my eyes, (1) implies an equation (3) x!=y that implies the couple (x;y) = (1,5;1,5) a not valid possibility

In the real GMAT, the problem will be clarified by the writings of x!=y.

Hey fig, don't u think that u are violating the basic rule of Data Sufficiency.

IT is given that x$y = (x^2-y^2)/(x-y). That's it ............... We know that the above function is difined if x!=y. But how can we consider that the given statment implies this.................

Well, I would like to say u no

An equation is definied and implies by its existence that x!=y ... In my eyes, it's widely enough

Hey fig u mean to say that when x=y the function is not defined. Now 1 says x+y =3 I will take x=1.5 and y=1.5............... Then where is the function to calculate with the above values , i mean when they are same the function is not defined. So i say that 1 is not sufficient

Yes... To me, we have a system of equations :

(1) x$y = (x^2-y^2)/(x-y) (implying by existence of the equation x!=y) (2) x+y =3

So, in my eyes, (1) implies an equation (3) x!=y that implies the couple (x;y) = (1,5;1,5) a not valid possibility

In the real GMAT, the problem will be clarified by the writings of x!=y.

That's what i am precisely saying........... we definitely need that information.......... _________________

The problem states an equation that needs to avoid certain values of x. Since an equation can exist only for values of x that makes this equation works, impossible values of x are, even if not stated, implicitely banned by the existence of the equation.

In real GMAT, the impossible values of x will be precised next to the equation and not in the statment 1 and 2

Last edited by Fig on 10 Oct 2006, 13:32, edited 2 times in total.

I agree with Fig and disagree with the OE. My thumb of rules for GMAT question is that if something appears in the denominator then it is not zero. However, if you don't know whether it is not zero, you can not divide both sides of an equation by it.

For example, if the stem says x/y=1. We can assume that y is not zero. However, if the stem says x=y. We can not say x/y=1 because y may be zero.

By the same token, if the stem says xy/y=1, we can say that x=1 and y<>0. But if the stem says xy=y, then all we can do is xy-y=0, or y(x-1)=0. In other words there are two solutions to this equation: x=1 and y=anything, or y=0 and x=anything.

What do you all think? _________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

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