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If xy ≠ 0 and x^2y^2 – xy = 6, which of the following could

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If xy ≠ 0 and x^2y^2 – xy = 6, which of the following could [#permalink] New post 09 Nov 2008, 20:47
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
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Re: x and y--28 [#permalink] New post 09 Nov 2008, 21:35
Jcpenny 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


A = xy, then A^2-A=6
A=3, A=-2

so y=3/x or y=-2/x

p.s. is the x^2y^2 the x^(2y)^2, the x^[2(y^2)], or the (x^2)(y^2)? hoho~, kidding~
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Re: x and y--28 [#permalink] New post 10 Nov 2008, 14:38
One more E.

X2y2-XY-6=0
(XY+2)(XY-3)=0
XY=-2 or XY=3
Y=-2/X or Y=3/X
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Re: x and y--28 [#permalink] New post 10 Nov 2008, 17:58
Another E.

You can reverse solve this by plugging in anwser choices if the straight approach does not occur.
Re: x and y--28   [#permalink] 10 Nov 2008, 17:58
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