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# Maths: Y in terms of X ?

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Maths: Y in terms of X ? [#permalink]

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08 Feb 2008, 03:32
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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: Maths: Y in terms of X ? [#permalink]

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08 Feb 2008, 08:27
Expert's post
E

$$f=x^2y^2 - xy = 6$$

$$f=xy*(xy-1) = 6$$

I. 1/(2x): xy=1/2; f=-1/4 false.
II. – 2/x: xy=-2; f=6 true
III. 3/x: xy=3; f=6 true
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Re: Maths: Y in terms of X ? [#permalink]

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08 Feb 2008, 13:00
i just tried each of the options, and ended up with E. Is there a quicker way to do this ?
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Re: Maths: Y in terms of X ? [#permalink]

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08 Feb 2008, 13:12
suntaurian 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

could you clarify the problem. Is it (x^2)(y^2) or x^((2y)^2) or x^(2*(y^2))?
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Re: Maths: Y in terms of X ? [#permalink]

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08 Feb 2008, 15:08
I think it is (x^2)*(y^2) - x*y = 6 (I mean it better be. It is going to take me ages to solve x^(2y^2)...)

Just solve the equation assuming A = x*y, which yields A^2 - A = 6. Then solve for y in terms of x. Hope that helps!
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Re: Maths: Y in terms of X ? [#permalink]

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08 Feb 2008, 15:37
aimecon wrote:
I think it is (x^2)*(y^2) - x*y = 6 (I mean it better be. It is going to take me ages to solve x^(2y^2)...)

Just solve the equation assuming A = x*y, which yields A^2 - A = 6. Then solve for y in terms of x. Hope that helps!

Thanks, aimecon. Sometimes real difficulty is not solving the problem, but deciphering the problem.
Re: Maths: Y in terms of X ?   [#permalink] 08 Feb 2008, 15:37
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