Fantastic approach. Thanks!

naaga wrote:

If xy ≠ 0 and x2y2 – xy = 6, which of the following could be y in terms of x?

I. 1/(2x)

II. – 2/x

III. 3/x

A. I only

B. II only

C. I and II

D. I and III

E. II and III

If xy ≠ 0 and x^2*y^2 - xy = 6, which of the following could be y in terms of x? I. 1/(2x)

II. - 2/x

III. 3/x

x^2*y^2-xy=6 --> (xy)^2-xy-6=0 --> (xy-3)(xy+2)=0 --> either xy-3=0 and in this case y=3/x or xy+2=0 and in this case y=-2/x.

Answer: E.

naaga, please format the questions properly.P.S. This is your third question for today with incorrect OA.