abelnazareth wrote:

I choose the answer D, but OG says option 2 is not sufficient. Could you please let me know where I am going wrong?

80. If xy > 0, does (x-1)(y-1) = 1?

(1) x + y = xy

(2) x = y

OG Explanation: (2) Substituting y for x in (x-1)(y-1) = 1 gives (y-1)(y-1) = 1 or thus only that y2-2y+1 = 1; this cannot be solved uniquely for y; NOT sufficient.

I was able to solve this as,

y2-2y+1 = 1

y(y-2)=0 --by subtracting 1 from both sides.

Implies y=0 or y-2 = 0

Given xy > 0, which implies neither x or y not equal to 0. Therefore y = 0 is not possible.

Thus y-2 = 0; y = 2. Thus I think option 2 is sufficient.

If \(xy>0\) does \((x-1)(y-1)=1\)?

(1) \(x + y = xy\)

(2) \(x=y\)

\(xy>0\) means that either both \(x\) and \(y\) are positive or both are negative (so neither of unknowns equals to zero: \(x\neq{0}\) and \(y\neq{0}\)).

Question: is \((x-1)(y-1)=1\)? --> is \(xy-x-y+1=1\)? is \(x+y=xy\)?

(1) \(x+y=xy\) --> directly gives us the answer YES. Sufficient.

(2) \(x=y\) -->

question becomes: is \(x+x=x^2\)? --> is \(x(x-2)=0\)? --> is \(x=0\) or \(x=2\)? --> as given that \(x\neq{0}\), then the

question becomes is \(x=2\)? We don't know that (or in other words for \(x=y\) and \(xy>0\) equation \((x-1)(y-1)=1\) holds true for \(x=y=2\), but we don't know whether that's true, thus we can answer whether \((x-1)(y-1)=1\) holds true), hence this statement is not sufficient.

Answer: A.

P.S.

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