amitansu wrote:The q is : If xy>0, does (x-1)(y-1)=1 ?

1) x+y=xy
2) x=y

Stem 1 is sufficient and i agree with that.

How about stem 2 ?

Amit

xy-(x+y)+1=1

Stm1. x+y=xy so, the given eqn holds correct Sufficient

Stm2. x=y, The eqn will be
x^2-2x+1=1--> (x-1)^2=1
or
y^2-2y+1=1--> (y-1)^2=1

Insufficient to satisfy the equation

aspire750 wrote:

amitansu wrote:The q is : If xy>0, does (x-1)(y-1

xy-(x+y)+1=1

Stm1. x+y=xy so, the given eqn holds correct Sufficient

Stm2. x=y, The eqn will be
x^2-2x+1=1--> (x-1)^2=1
or
y^2-2y+1=1--> (y-1)^2=1

Insufficient to satisfy the equation

Ok.

from 2 : x-1=+ or - 1
and y-1=+ or -1

so, x is either 2 or 0 and y is also either 2 or 0
since it is given xy>0 , x and y have to be either
+ve or -ve at the same time.

So, x and y could be 2 each since they can't be 0 to be > 0.

Amit

what is the ans?

for 2:

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

if x=y

=> x^2 -x -x = 0

=> x^2 = 2x

=> x=2 i.e. y=2 ( as x=y)

therfore (x-1)(y-1) = 1....................sufficient

It doesn't ask if you can solve the problem, it says:

is (x-1)(y-1)=1

1. is obvios

2. Well, x=y.

But if x=y=8, then 7*7=49<>1, so 2 is not sufficient to prove that (x-1)(y-1)=1. It can be solved to x=2, but that's not what we are asked now, is it?

varmaskarma wrote:what is the ans?

for 2:

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

if x=y

=> x^2 -x -x = 0

=> x^2 = 2x

=> x=2 i.e. y=2 ( as x=y)

therfore (x-1)(y-1) = 1....................sufficient

The ans is : step 2 is insufficient .
OG solution says it can't find a unique value of x aor y.
I couldn't convince myself for that...

BUt yes, as santa_dem says, the value of x and y could be anything be it's 2 or 8 or whatever....the ans could be yes and no also.so not sufficient.To me this is more convincing.

Thank you all.

Amit

OG is correct see below
If x=y then stat 2
IF (x-1)^2= X^2-2X+1= 1
X^2-2X=0
x=2 OR x=0
buT if x=2 then xy=0 contradictory from info given