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algebra...

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Re: algebra...

by sureshbala » Mon Feb 09, 2009 3:03 pm
willbeatthegmat wrote:If x & y are +ve & x^2y^2= 18-3xy, then x^2 =??

(18-3y)/y^3
18/y^2
18/(y^2+3y)
9/y^2
36/y^2


OA D
I think the question has to be "If x and y are positive integers and x^2.y^2 = 18-3xy, then x^2 = ?

Given x^2 . y^2 = 18-3xy
i.e x^2.y^2 + 3xy = 18.
i.e xy(xy+3) = 18.

So 18 must be written as product of two positive numbers which differ by 3. i.e 18 = 3x6

So xy(xy+3)= 3x6

Thus xy =3 and hence x^2.y^2 = 9. So x^2 = 9/y^2
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by utibay » Wed Feb 11, 2009 2:05 am
The question has to be that X & Y are positive integers.

X^2*Y^2=18-3XY
X^2*Y^2+3XY-18=0

Think of XY as the variable B

B^2+3B-18=0
(B+6)(B-3)=0

Which is really,

(XY+6)(XY-3)=0
XY = 3 or -6

Since X & Y are both positive, XY cannot be negative, since that would mean that either X or Y would be negative.

So, XY=3
X=3/Y
X^2=9/Y^2
"Being ignorant is not so much a shame, as being unwilling to learn." -Benjamin Franklin
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by willbeatthegmat » Wed Feb 11, 2009 5:50 am
well, integer is not mentioned in the question...ther might b a misprint...neways thank u guys...
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