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by beater » Sat Sep 27, 2008 2:22 pm
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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by manulath » Sat Sep 27, 2008 7:40 pm
Ans E.
what is OA?

Plugging in the choices yeilds II and III, I does not satisfy

neways if want to solve algebrically

as xy is not equal to 0, so both x and y are not equql to 0

divide equation by x^2....you get

y^2 - (y/x) - (6/x^2) = 0

y^2 - (3y/x) + (2y/x)- (6/x^2) = 0

y(y - 3/x) + 2/x(y - 3/x) = 0

(y - 3/x)(y + 2/x) = 0

hence
y=3/x...............III
y=-2/x..............II
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by stop@800 » Sun Sep 28, 2008 6:45 am
manulath has already given algebric solution.

In this case you can also solve by substituting these three values.
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