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# if xy > 0, does (x - 1)(y - 1) = 1 ? (1) x + y = xy (2) x

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Manager
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if xy > 0, does (x - 1)(y - 1) = 1 ? (1) x + y = xy (2) x [#permalink]

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25 Oct 2006, 08:33
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if xy > 0, does (x - 1)(y - 1) = 1 ?
(1) x + y = xy
(2) x = y
Manager
Joined: 31 Aug 2006
Posts: 211
Followers: 1

Kudos [?]: 2 [0], given: 0

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25 Oct 2006, 08:40
(A)

xy>0 implies x!=0 and y!=0
and x,y are both +ve or x,y are both -ve

Stat(1) - gives only one value , x=y=2 - Hence Suff
Stat(2) - gives two values x=y=1 and x=y=2 - Hence Insuff
Senior Manager
Joined: 01 Sep 2006
Posts: 301
Location: Phoenix, AZ, USA
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25 Oct 2006, 08:41
(1) x+y=xy==> x= y(x-1) or y=x(y-1)
(x-1) = x/y and (y-1)=y/x
(x-11)(y-1)= xy/xy==1

Suffcient

(2) x=y==> (x-1)(y-1)=(x-1)(x-1)=(x^2-1)

assume x=1 (x^2 -1 ) = 0
assume x = -2 ( x^2 -1) = 3

Insuff
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Joined: 01 May 2006
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25 Oct 2006, 12:44
(A) also

(x - 1)(y - 1) = 1?
<=> x*y - x - y + 1 = 1 ? (1)

Stat 1
x + y = xy
<=> x*y - x - y = 0
Thus,
x*y - x - y + 1 = 0 + 1 = 1 => (1) is verified.

SUFF.

Stat 2
x=y
=> x*y - x - y + 1 = x^2 - 2*x + 1 : can be equal to 1 if x = 2 or x = 0 but can be equal to any things else

INSUFF.
25 Oct 2006, 12:44
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