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how to solve [#permalink] New post 25 Feb 2011, 08:23
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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
[Reveal] Spoiler: OA

Last edited by naaga on 25 Feb 2011, 22:38, edited 1 time in total.
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Re: how to solve [#permalink] New post 25 Feb 2011, 09:11
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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.
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Re: how to solve [#permalink] New post 25 Feb 2011, 09:35
Fantastic approach. Thanks!

Bunuel wrote:
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.
Re: how to solve   [#permalink] 25 Feb 2011, 09:35
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