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# p1Q watch

1. could anyone be kind enough to show me this working out
solve the simultaneous equation
2x^2+3xy+y^2=0
3x+y=2
thanks
2. (Original post by vortex_199)
could anyone be kind enough to show me this working out
solve the simultaneous equation
2x^2+3xy+y^2=0
3x+y=2
thanks
3x+y=2
y=2-3x

substitute in the first equation:

2x^2+3x(2-3x)+(2-3x)^2=0
2x^2+6x-9x^2+(4-12x+9x^2)=0
2x^2+6x-9x^2+4-12x+9x^2=0
2x^2-6x+4=0
Divide by 2
x^2-3x+2=0

Factorise to get:

(x-1)(x-2)

Therefore:

x=1, y=-1
x=2, y=-4

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