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Does x = y ? (1) x^2 - y^2 = 0 (2) (x + y)^2 = 0 [#permalink]
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Bunuel wrote:
Does x = y ?

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


Per statement 1, \(x^2=y^2\) ---> \(x= \pm y\) . Thus this statement is not sufficient to answer "is x=y".

Per statement 2, (x+y)^2=0 ---> either x=y=0 , then a "yes" for is x=y but if x=1 and y=-1, then a"no" for "is x=y". This statement is not sufficient.

Combining, still get x=y=0 and x=1 and y=-1 satisfying both the statements (\(x=\pm y\)) giving you an insufficient answer.

E is the correct answer.
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Re: Does x = y ? (1) x^2 - y^2 = 0 (2) (x + y)^2 = 0 [#permalink]
why is A not sufficient? i suppose x²-y² is equal to x²=y² which equals x=y
my understanding is that -ve y² turns +ve when it goes to the side after = sign
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Re: Does x = y ? (1) x^2 - y^2 = 0 (2) (x + y)^2 = 0 [#permalink]
Yce wrote:
why is A not sufficient? i suppose x²-y² is equal to x²=y² which equals x=y
my understanding is that -ve y² turns +ve when it goes to the side after = sign


Plugging in numbers is an easier way to understand.

If x and y = 0 then x^2 - y^2 = 0 and the answer to the question is yes. But if X = 1 and Y = -1 then x^2 - y^2 = 0 but the answer to the question is no.

Since you get a Yes and a No answer, statement 1 is insufficient

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Re: Does x = y ? (1) x^2 - y^2 = 0 (2) (x + y)^2 = 0 [#permalink]
from stmt 1: (x+y)(x-y)=0 --> x+y = 0 or x-y = 0 --> x+y=0 has two cases --> case 1--> x=y, case 2 --> x=-y vise versa, or x=y=0 insuff

from stmt 2: same thing logic from stmt 1 applies here --> x+y=0, x=-y, or x=y=0. insuff

combining stmt 1&2 --> no new info. insuff
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Re: Does x = y I) x^2 - y^2 = 0 II) (x+y)^2 = 0 [#permalink]
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Re: Does x = y I) x^2 - y^2 = 0 II) (x+y)^2 = 0 [#permalink]
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