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MGMAT - 4 those who love algebra ))

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Source: — Problem Solving |
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by rolrol19 » Thu Jan 29, 2009 11:09 am
My answer is D

x^2y^2=18 - 3xy

x^2y^2+3xy -18 = 0

Making it an equation (think of xy as one term)

(xy+6)(xy-3) = 0

which gives xy= -6 or xy= 3

Since the stem says that x and y are positive, we go with 3

replace xy by 3 in the second part of the question and you get : x^2y^2=18 - 3 (3)

x^2y^2=9

x^2=9/y^2

What do you think of the approach?
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by 4meonly » Thu Jan 29, 2009 12:10 pm
x^2y^2+3xy -18 = 0
(xy+6)(xy-3) = 0
I think your approach is correct.
I overlooked mentioned above equation but from
xy(xy+3)=18
concluded that xy=3 and than the same as yours
unfortunately, I dont know OA but will get it soon
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Re: MGMAT - 4 those who love algebra ))

by aroon7 » Thu Jan 29, 2009 9:19 pm
4meonly wrote:OA after B-)
Image
i plugged in:
x=3, y=1
only D suits the equation
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