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xy >0 does (x-1)(y-1) =1 x+y=xy x=y spoiler below the

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Senior Manager
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xy >0 does (x-1)(y-1) =1 x+y=xy x=y spoiler below the [#permalink] New post 07 Aug 2007, 13:40
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E

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xy >0 does (x-1)(y-1) =1

x+y=xy

x=y


spoiler below the line:













________________________________________________

so i thought that the answer was D. if x=y then we have y^2-2y+1=1

which simplifies to y(y-2)=0 so y = 0 or y = 2.

since x=y we have:

(0-1)(0-1)=1 and (2-1)(2-1)=1.

what did i miss here?
Senior Manager
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Re: GREEN OG DS #80 [#permalink] New post 07 Aug 2007, 13:58
anonymousegmat wrote:
xy >0 does (x-1)(y-1) =1

x+y=xy

x=y



The question can be rewritten as: "xy - x - y = 0?"

(1) Suff.

(2) Insuff.

A.
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Re: GREEN OG DS #80 [#permalink] New post 07 Aug 2007, 14:00
anonymousegmat wrote:

________________________________________________

so i thought that the answer was D. if x=y then we have y^2-2y+1=1

which simplifies to y(y-2)=0 so y = 0 or y = 2.

since x=y we have:

(0-1)(0-1)=1 and (2-1)(2-1)=1.

what did i miss here?


You can not solve it the like this. From your equation, you solve it like you already assume (x-1)(y-1)=1, and of course the answer you get will be the one fit the criteria. You need to look at the ones doesn't fit as well. What about x=y=3? It does fit the description that xy>0, but in case of x=y=3 (x-1)(y-1) does not equal to 1. Therefore, insufficient.
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Re: GREEN OG DS #80 [#permalink] New post 07 Aug 2007, 15:40
eileen1017 wrote:
anonymousegmat wrote:

________________________________________________

so i thought that the answer was D. if x=y then we have y^2-2y+1=1

which simplifies to y(y-2)=0 so y = 0 or y = 2.

since x=y we have:

(0-1)(0-1)=1 and (2-1)(2-1)=1.

what did i miss here?


You can not solve it the like this. From your equation, you solve it like you already assume (x-1)(y-1)=1, and of course the answer you get will be the one fit the criteria. You need to look at the ones doesn't fit as well. What about x=y=3? It does fit the description that xy>0, but in case of x=y=3 (x-1)(y-1) does not equal to 1. Therefore, insufficient.



you would think that after i TYPED the question I would have noticed that... lol
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 [#permalink] New post 07 Aug 2007, 17:13
St1:
x+y = xy
if x = 2, y = 2, then (x-1)(y-1) = 1
Only one pair works. Sufficient.

St2:
x=y.
if x = 1, y = 1, then (x-1)(y-1) = 0
if x = 2, y = 2, then (x-1)(y-1) = 1
Insufficient.

Ans A
  [#permalink] 07 Aug 2007, 17:13
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