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...
ps 17
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- jayhawk2001
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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 ?
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 ?