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Now xy=x means xy-x=0 or x(1-y)=0 For the equation x(1-y)=0 to hold good, either x has to be 0 or y has to be 1. It can very well be that x=0 and y=1 at the same time. We are not sure. So since it is not clear if x=0 or not, this statement is not sufficient.

Now in statement 2. x+y=y clearly shows x=0.

So it must be B

HTH
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Now xy=x means xy-x=0 or x(1-y)=0 For the equation x(1-y)=0 to hold good, either x has to be 0 or y has to be 1. It can very well be that x=0 and y=1 at the same time. We are not sure. So since it is not clear if x=0 or not, this statement is not sufficient.

This question goes to show that when you have been looking at 700 level questions too long, you try and pick holes in something as simple as x = y - y ....

on the other hand, its good to finally get a question right!

Hi Jimmy86; if (2) where: x + y = x --> y = 0 [no information about x after subtracting x from both sides]

combining (1) & (2) [x*y=x; x+y=x]: x(0) = x for all values of x, x*0 is always a zero. So, we can say x (the result of the multiplication) is a zero. IMO C
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y (x + y) = x and so, x will be zero only if Y is also 0 or else the condition is not satisfied. So, IMO its E.

I agree... its E.

in case statement 2 : x+y = x x=-1 , y=0 gives -1+0 = -1 ; satisfies x=0 , y= 0 gives 0+0 = 0 ; satisfies we see x is not necessarily 0. Insufficient

And combining y(x+y)=x if x=0 ; y=0 then 0*(0+0)=0 ; satisfies if x=1/2 ; y=1/2 then 1/2*(1/2+1/2)=1/2 ; satisfies So x is not necessarily 0.Insufficient

Thus IMO its E

But what if the question states that x is an integer. Then i suppose the only value that of x that satisfies y(x+y)=x is x=0. and answer would be C(not sure, feel i am missing something)
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Life`s battles dont always go, To the stronger or faster man; But sooner or later the man who wins, Is the man who THINKS HE CAN .

KUDOS me if you feel my contribution has helped you.

This question goes to show that when you have been looking at 700 level questions too long, you try and pick holes in something as simple as x = y - y ....

on the other hand, its good to finally get a question right!

Haha yes! I definitely know what you mean. You're so traumatized from every question trying to trick you, that you're reluctant to believe that the answer could be sufficient with something as easy as X=Y-Y. I stared at it for a good 30 seconds trying to pick it apart too. That's hilarious.

(2) if y = x - x, then, it is sufficient to say y = "0".

plug in y (0) into (1) [x*y = x] or x = x*y x = x*0 RHS = x*0 = 0 for any value of a number, 0 times that number = 0 So, RHS is always "0" LHS = x LHS = RHS --> x = RHS x = 0 Sufficient

So, C IMO....except am doing something stupid somewhere. Please correct me.
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Buddy, I dont think u went wrong... I believe I am somewhere wrong..

The first mistake i made was, in substitution... I shud have substituted as (x+y) * y = (x+y)... instead of (X+Y)*Y= X..

But even after this I am getting (X+Y)*Y=(X+Y) ---> Y=1.. and this will finally yield that Both option are not suffiient... And I am not sure how to proceed further...

Can anyone help me pls...

gmatbull wrote:

(2) if y = x - x, then, it is sufficient to say y = "0".

plug in y (0) into (1) [x*y = x] or x = x*y x = x*0 RHS = x*0 = 0 for any value of a number, 0 times that number = 0 So, RHS is always "0" LHS = x LHS = RHS --> x = RHS x = 0 Sufficient

So, C IMO....except am doing something stupid somewhere. Please correct me.

@Gmatbull.. The first mistake i made was, in substitution... I shud have substituted as (x+y) * y = (x+y)... instead of (X+Y)*Y= X..

But even after this I am getting (X+Y)*Y=(X+Y) ---> Y=1.. and this will finally yield that Both option are not suffiient... And I am not sure how to proceed further...

I think you seem to make the expression difficult to yourself. First, realize that (2) has simplified the job: from x+y = x, you know y = 0. (x+y) * y = (x+y) ==> (x+y)0 = x+0 0 = x +0 0 = x you still arrive at C.
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The statements 1 and 2 are contradicting to each other.. Option 1 says that, X * Y = X which tends to give Y = 1,

But option 2 gives, X + Y = X, and So Y = 0. I believe such questions wont appear in GMAT and so less to worry about this question...

gmatbull wrote:

sivai87 wrote:

@Gmatbull.. The first mistake i made was, in substitution... I shud have substituted as (x+y) * y = (x+y)... instead of (X+Y)*Y= X..

But even after this I am getting (X+Y)*Y=(X+Y) ---> Y=1.. and this will finally yield that Both option are not suffiient... And I am not sure how to proceed further...

I think you seem to make the expression difficult to yourself. First, realize that (2) has simplified the job: from x+y = x, you know y = 0. (x+y) * y = (x+y) ==> (x+y)0 = x+0 0 = x +0 0 = x you still arrive at C.

The statements 1 and 2 are contradicting to each other.. Option 1 says that, X * Y = X which tends to give Y = 1,

But option 2 gives, X + Y = X, and So Y = 0. I believe such questions wont appear in GMAT and so less to worry about this question...

Not so. First of all statement (2) is \(x+y=y\) and not \(x+y=x\).

Next, (1) \(xy=x\) --> \(x(y-1)=0\) --> either \(x=0\) (and \(y\) can take any value) OR \(y=1\) (and \(x\) can take any value). So we can not say that \(y=1\) from this statement.

If the question were:

Is \(x = 0\)?

(1) \(xy = x\) (2) \(x+y=x\) (Note that this statement is changed)

Then the answer would be C:

(1) \(xy=x\) --> \(x(y-1)=0\) --> either \(x=0\) (and \(y\) can take any value) OR \(y=1\) (and \(x\) can take any value). Not sufficient. (2) \(x+y=x\) --> \(y=0\). Not sufficient to answer whether \(x=0\).

(1)+(2) As from (2) \(y=0\neq{1}\) then according to (1) \(x=0\). Sufficient.

The statements 1 and 2 are contradicting to each other.. Option 1 says that, X * Y = X which tends to give Y = 1,

But option 2 gives, X + Y = X, and So Y = 0. I believe such questions wont appear in GMAT and so less to worry about this question...

Not so. First of all statement (2) is \(x+y=y\) and not \(x+y=x\).

Next, (1) \(xy=x\) --> \(x(y-1)=0\) --> either \(x=0\) (and \(y\) can take any value) OR \(y=1\) (and \(x\) can take any value). So we can not say that \(y=1\) from this statement.

If the question were:

Is \(x = 0\)?

(1) \(xy = x\) (2) \(x+y=x\) (Note that this statement is changed)

Then the answer would be C:

(1) \(xy=x\) --> \(x(y-1)=0\) --> either \(x=0\) (and \(y\) can take any value) OR \(y=1\) (and \(x\) can take any value). Not sufficient. (2) \(x+y=x\) --> \(y=0\). Not sufficient to answer whether \(x=0\).

(1)+(2) As from (2) \(y=0\neq{1}\) then according to (1) \(x=0\). Sufficient.

So for this revised question answer would be C.

Hope it helps.

I seem to recall that you can't add/subtract variables from both sides of the equation unless you know the sign of the variables. But the question doesn't tell you the sign of x or y so I don't think the problem can be approached this way. I think the only way to answer the question is to plug in numbers for x and y then check to see if the equation holds.

For Option1, which of the below method need to be followed?

Method1 ---- \(xy = x\) ----> \(xy-x = 0\) ----> \(x(y-1) = 0\) ----> \(x = 0, y = 1\) ---- I thought this method is always true in determining a value Method2 ---- \(xy = x\) - Substitute values \(x=0\) & \(y\)can be any value? or \(x\) can be any value and \(y=1?\)