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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

\(x^2*y^2-xy=6\) --> \((xy)^2-xy-6=0\) --> factor quadratics for xy: \((xy-3)(xy+2)=0\) (or just solve quadratics for xy) --> either \(xy-3=0\) and in this case \(y=\frac{3}{x}\) or \(xy+2=0\) and in this case \(y=-\frac{2}{x}\).

Re: If xy≠0 and x^2*y^2-xy=6, which of the following could be y [#permalink]

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Re: If xy≠0 and x^2*y^2-xy=6, which of the following could be y [#permalink]

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05 Feb 2016, 23:11

Baten80 wrote:

If xy not equal 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

A. I only B. II only C. I and II D. I and III E. II and III

Looking at the equation and below 3 options, we can figure out that xy and (xy-1) are two consecutive integers. There are only two possible combinations: 1) xy=3 and (xy-1)=2; y=3/x (Option III is correct); 2) xy=-2 and (xy-1)=-3; y=-2/x (Option II is correct).

If xy not equal 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

A. I only B. II only C. I and II D. I and III E. II and III

We are given the equation x^2y^2 – xy = 6, and although it may not be obvious, the equation is a quadratic-format equation. Thus, our first step is to set one side of the equation to zero. We then factor the other side as a quadratic. .

x^2y^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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