ps 17

dunkin77 · Posted: Sun Jun 17, 2007 9:52 am Post subject: ps 17

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...

jayhawk2001 · Posted: Sun Jun 17, 2007 2:48 pm Post subject: Re: ps 17

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 ?

dunkin77 · Posted: Sun Jun 17, 2007 5:20 pm Post subject:

Thanks, the answer is E. Very Happy

f2001290 · Posted: Mon Jun 18, 2007 7:54 am Post subject:

dunkin77

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

Substitute y=(options)