If xy ≠ 0 and x^2y^2 − xy = 6, which of the following could be y in te

I'm going to respond to your question with a question, woohoo921, because I'd rather teach you how to fish than give you a fish.
Say you know that xy(xy-1)=7.
In other words, the product of two numbers (xy and (xy - 1)) is 7.
Is it reasonable to infer that one of those numbers is 7? Why, or why not?

Side note: if we were told that the two numbers were positive integers, we could infer that one of them is 1 and the other is 7 (because 7 is a prime number). And, since we're on the subject, if you are told that the product of two different positive integers is the square of a prime number (p^2), you can infer that the two numbers are 1 and p^2.
How am I coming up with these "rules?" I'm just using quantitative reasoning.

If you like, come join me at one of my live AMA sessions on Zoom, where we can all practice this kind of reasoning together. Your first AMA is completely free of charge, using the free 7-day trial at quantreasoning.com

I see

I guess my question was if we must always try to use the Zero Factor Property to solve, or we can use your approach mentioned above. Thank you again.

I'm afraid I don't understand your question, woohoo921.

Hi altogether, I quickly had to revive this thread.

I was having the same problem as the person I am quoting until I realized the following:

xy² - xy - 6 = 0
rule: (a+b)*(a-b) = a²-ab+ba-b^2
but be CAREFUL: (xy-2)*(xy+3) !=(is not equal to) (xy-3)*(xy+2)

first is the right one: xy²+3xy-2xy-6 = 0
xy²+xy-6=0

and second is wrong: xy²+2xy-3xy-6
xy²-xy-6=0

Hope this helps.

Best,
Adrian

\(x^2y^2 − xy = 6\)

Or, \(a^2 - a - 6 = 0\) ; \(a = xy\)

Thus, \(a = 3\) , \(-2\)

Or, \(xy = 3\) ; \(y = \frac{3}{x}\) & \(xy = -2\) ; \(y = \frac{-2}{x}\)

Thus, Answer must be (E) II and III