Answer

Hi Laura,

This equation can be factorised as follows

x^2+3xy-18y^2 = (x+6y)(x-3y)

I factorised this in a similar way to how you'd factorise a quadratic. You can just treat the y as a part of the coefficients ie

x^2+3xy-18y^2 = x^2 + (3y)x - (18y^2)

a = 1
b = 3y
c = -18y^2

I then looked for values m and n that satify

m+n = 3y

and

mn = -18y^2

these were m =6y and n=-3y

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If you were to plot the solutions of say

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

that would mean you have to plot both

x+6y=0

and

x-3y=0

ie

y = -x/6 (straight line through the origin with gradient -1/6)

amd

y = x/3 (straight line through the origin with gradient 1/3)

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The crux with these equations is often to try and consider it with just one variable (say x) and treat the other as just a constant.

If you have any further questions about these feel free to ask. All the best,

Ben

Expert:	Ben Brown
Pos. Feedback:	100.0 %
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Answered:	1/6/2009

Mathematician

First Class Honours in Mathematics from Oxford University