Bunuel wrote:
If xy ≠ 0 and x^2y^2 − 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
Kudos for a correct solution.
We are given the equation, (x^2)(y^2) – xy = 6, and, although it may not be obvious, the equation is a quadratic-format equation. Thus, our first step is to set the equation equal to zero. We then factor the left side and, finally, solve for xy.
(x^2)(y^2) – xy - 6 = 0
(xy – 3)(xy + 2) = 0
xy – 3 = 0 or xy + 2 = 0
xy = 3 or xy = -2
Since we need y in terms of x, we can isolate y in both of our equations.
y = 3/x or y = -2/x
Thus, the expressions in II and III are correct.
Answer: E
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