Solving system of multivariable 2nd-degree polynomials x 2 + 3 x y − 9 =...

Roland Manning

Roland Manning

Answered

2022-06-25

Solving system of multivariable 2nd-degree polynomials
x 2 + 3 x y 9 = 0 ( 1 ) 2 y 2 4 x y + 5 = 0 ( 2 )
where ( x , y ) C 2 .
More generally, how would you solve any set of equations of the form:
a x 2 + b x y + c = 0 d y 2 + e x y + f = 0
where a , b , c , d , e , f Q and ( x , y ) C 2 .

Answer & Explanation

Abigail Palmer

Abigail Palmer

Expert

2022-06-26Added 30 answers

Multiply first equation by 5.
Multiply second equation by 9.
Add both Equations.
Divide this equation by y 2 .
Let t = x y .
You get a quadratic in t
The steps are[Generalization]:Reduce the two equations to one
a x 2 + b x y + c y 2 = 0
Then divide by y 2
a x 2 y 2 + b x y + c = 0
Replace x y = t
a t 2 + b t + c = 0

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