Question

{x^2+xy+y^2=19 {x+xy+y=1

Answer 1

$begin{cases}x^2+xy+y^2=19\x+xy+y=1end{cases}Rightarrowbegin{cases}left(frac{1-y}{1+y}right)^2+left(frac{1-y}{1+y}right)y+y^2=19\x=frac{1-y}{1+y}end{cases}\frac{1-2y+y^2+y+y^3+y^2+2y^3+y^4}{1+2y+y^2}=19\y^4+y^3+2y^2-y+1=19y^2+38y+19\y^4+y^3-17y^2-39y-18=0\(y^2-4y-3)(y^2+5y+6)=0\y_1=2-sqrt7\y_2=2+sqrt7\y_3=-2\y_4=-3$
$begin{cases}x_1=frac{-1+sqrt7}{3-sqrt7}\y_1=2-sqrt7end{cases}quadbegin{cases}x_2=-frac{1+sqrt7}{3+sqrt7}\y_2=2+sqrt7end{cases}quadbegin{cases}x_3=-3\y_3=-2end{cases}quadbegin{cases}x_4=-2\y_4=-3end{cases}$