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Re: If xy > 0, does (x – 1)(y – 1) = 2 ? (1) x + y = –1 (2) x = y [#permalink]
Bunuel wrote:
If xy > 0, does (x – 1)(y – 1) = 2 ?

(1) x + y = –1
(2) x = y


(x – 1)(y – 1) = 2
=> xy - x - y -1 = 0
=> xy = 1 +x +y
So the question basically is whether xy = 1 +x +y

(1) x + y = –1

=> xy = 0 (which will not happen as xy >0)
Sufficient

(2) x = y

=> x^2 = 1 + 2x
=> x^2 - 2x -1 = 0
=> We surely can find 2 values of x that satisfies above equation.
So, it depends on x and y that the condition is met or not.
Insufficient.

So A is the answer
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Re: If xy > 0, does (x – 1)(y – 1) = 2 ? (1) x + y = –1 (2) x = y [#permalink]
Bunuel wrote:
If xy > 0, does (x – 1)(y – 1) = 2 ?

(1) x + y = –1
(2) x = y


D

Not sure if the solution is correct, but here is how I did it:

(1) Solving for the equation:
1 + xy - (x+y) = 2, this comes out to be xy=0 which doesn't confirm with the given situation. Hence,

SUFFICIENT

2) putting x=y in the given, we get x=y= 1+ sqrt 2 or 1-sqrt 2. Both give you 2 as an answer.

SUFFICIENT

Hence, D.


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Re: If xy > 0, does (x – 1)(y – 1) = 2 ? (1) x + y = –1 (2) x = y [#permalink]
@Bunuel- Again different answers here, could you provide the correct answer?
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Re: If xy > 0, does (x 1)(y 1) = 2 ? (1) x + y = 1 (2) x = y [#permalink]
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Re: If xy > 0, does (x 1)(y 1) = 2 ? (1) x + y = 1 (2) x = y [#permalink]
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