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ps 17

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by dunkin77 » Sun Jun 17, 2007 8:52 am
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


I got x,y are 1/2 and 6 but could not find the answer...
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Re: ps 17

by jayhawk2001 » Sun Jun 17, 2007 1:48 pm
Rearranging, we get xy (xy - 1) = 6

Considering integer values for xy, we can have xy = 3 or xy = -2. Both
will give xy (xy - 1) = 6.

For xy = 3, we get y = 3/x (III is solved)

For xy = -2, we get y = -2/x (II is solved)

If we consider I, y = 1/2x or xy = 1/2. This does not satisfy xy(xy-1) =6.

So, II and III i.e. E ?
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by dunkin77 » Sun Jun 17, 2007 4:20 pm
Thanks, the answer is E. :D
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by f2001290 » Mon Jun 18, 2007 6:54 am
dunkin77

Use back-solving for such questions if you are un-able to get the answer.

Substitute y=(options)
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