if xy>0 does (x-1)(y-1) >1 1) X+y=xy 2)x=y I know that : DS Archive
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# if xy>0 does (x-1)(y-1) >1 1) X+y=xy 2)x=y I know that

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if xy>0 does (x-1)(y-1) >1 1) X+y=xy 2)x=y I know that [#permalink]

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24 Mar 2005, 16:39
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if xy>0 does (x-1)(y-1) >1

1) X+y=xy
2)x=y
I know that A is sufficient.. i thought B will be sufficient too. as we are substituting x=y.. why is it not sufficient...
!!!!
Manager
Joined: 05 Feb 2005
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Location: San Jose
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24 Mar 2005, 16:51
A.

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

1) X+y=xy
2)x=y
I know that A is sufficient.. i thought B will be sufficient too. as we are substituting x=y.. why is it not sufficient...
!!!!

Putting x=y, LHS = (x-1)^2

Example x = 1, LHS = 0, which is less than 1
x = 3, LHS = 4, which is greater than 1
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24 Mar 2005, 17:26
"A"

state 1....asking xy-(x+y)+1 > 1...or xy-x-y > 0 ?

if x+y = xy....LHS 0...so ans is NO...suff

state 2.....x = y

y^2-2y > 0

so is y(y-2) > 0 ?

if y < 2.....LHS is < 0....if y > 2...LHS is > 0....insuff
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