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=yxy>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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