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If x^2+y^2=1, is x+y=1 ? (1) xy=0 (2) y=0

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Re: If x^2+y^2=1, is x+y=1 ? (1) xy=0 (2) y=0  [#permalink]

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New post 21 Oct 2014, 08:17
Answer is E.
x^2 + y^2 = (x+y)^2 - 2xy = 1
1) xy = 0 => 2xy = 0 => (x+y)^2 = 1 => |x+y| = 1 => x+y = 1 OR x+y = -1 => Not suff.
2) y = 0 => 2xy = 0 ... (ditto) => Not suff.
1) + 2) Not suff since 2) is like a possibility of 1).
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Re: If x^2+y^2=1, is x+y=1 ? (1) xy=0 (2) y=0  [#permalink]

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New post 20 Nov 2014, 15:31
Please tell me if my assumption is wrong:
Given: x^2+y^2=1
Assume: x^2+y^2+2xy = 1+2xy
(x+y)^2 = 1+2xy
x+y = sqrt(1+2xy)

Stmt1: xy = 0; So, plugging this in: x+y = sqrt(1) = 1 --> hence sufficient?!
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Re: If x^2+y^2=1, is x+y=1 ? (1) xy=0 (2) y=0  [#permalink]

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New post 21 Nov 2014, 04:28
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Re: If x^2+y^2=1, is x+y=1 ? (1) xy=0 (2) y=0  [#permalink]

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New post 17 Oct 2018, 00:54
(A) XY=0. so, either X=0 or Y=0.
If X=0, Y^2=1. Therefore, Y=-1 or +1. In this case, X+Y could be -1 or +1
Insuff
(B) Y=0 , Same as (A)

Answer E
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Re: If x^2+y^2=1, is x+y=1 ? (1) xy=0 (2) y=0   [#permalink] 17 Oct 2018, 00:54

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