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The OA is E. But I think C is the way to go. Pls verify

S1 tells us that x = -y or x = y Insufficient

S2 tells us that x = -y or x = y = 0 Insufficient

1) + 2) x = -y or x = y x = -y or x = y = 0

The root x = -y satisfy both the equations. Now the trap x = -y does not preclude y from being zero. Hence x = -y can mean x = y = 0 in which case the answer is YES or it can mean x = -y in which case the answer is NO. So we cant make a conclusive statement about whether x = y

Re: Simple yet tripping ! [#permalink]
15 May 2011, 01:31

I think this is the trap. x = -y does not preclude y from being zero. Hence x = -y can mean x = y = 0 in which case the answer is YES or it can mean x = -y in which case the answer is NO. So we cant make a conclusive statement whether x = y

jamifahad wrote:

I think this could be the reason..

Basically question is x-y=0 ?

From S1 and S2 x = -y or x = y x = -y or x = y = 0

So either x+y=0 or x-y=0. hence cannot be determined.

Re: Simple yet tripping ! [#permalink]
15 May 2011, 01:39

1

This post received KUDOS

gmat1220 wrote:

Guys The answer sounds unconvincing. Can you verify this?

Does x = y ? (1) x^2 - y^2 = 0 (2) (x + y)^2 = 0

Try to solve it using substitution and it will get clearer:

Q: Does x=y?

Two cases; Case I: x != y; x=-1, y=1 Case II: x = y; x=0, y=0

(1) x^2=y^2 Case I satisfies. Case II satisfies. So, we have two sets both satisfying this statement. However, in one case x=y and in other x!=y. We can't conclude that x=y.

(1) x=-y Case I satisfies. Case II satisfies. So, we have two sets both satisfying this statement. However, in one case x=y and in other x!=y. We can't conclude that x=y.

Combining; The same two cases satisfy. Nothing conclusive. Ans: "E"

If the questions said: "Does x=y if xy!=0". Then "B" would be the answer.

The OA is E. But I think C is the way to go. Pls verify

S1 tells us that x = -y or x = y Insufficient

S2 tells us that x = -y or x = y = 0 Insufficient

1) + 2) x = -y or x = y x = -y or x = y = 0

The root x = -y satisfy both the equations. Now the trap x = -y does not preclude y from being zero. Hence x = -y can mean x = y = 0 in which case the answer is YES or it can mean x = -y in which case the answer is NO. So we cant make a conclusive statement about whether x = y

Statement 1: tells |x|=|y|. this gives us 3 values: a. x=y (xy not equal to 0) b. x=-y c. x=y=0

x and y can take infinite values.

Statement 2: tells us x=-y or x=y=0 So answer can be both "No" and "Yes"

Both statements combined, we get x=-y or x=y=0. No new information. So insufficient.

Answer: E

Admittedly, I almost picked B. But suddenly I remembered I had pledged that I will ALWAYS check for zero in algebra questions. Glad it paid off