Question

Решите систему уравнений.
$left { {{x^2+2xy=16} atop {y^2-xy=3}} right.$

Answer 1

$left { {{x^2+2xy=16} atop {y^2-xy=3}} right.$

$x^2+2xy=16\ 2y^2-2xy=6\ \ x^2+2y^2=22\ x^2=22-2y^2\ x=sqrt{22-2y^2}\ \ 22-2y^2+2sqrt{22-2y^2}y=16\ -2y^2+2sqrt{22-2y^2}y=-6\ 2sqrt{22-2y^2}*y=2y^2-6\ 4y^2(22-2y^2)=4y^4-24y^2+36\ 88y^2-8y^4=4y^4-24y^2+36\ y=+-3\ y=+-frac{sqrt{3}}{3}\ x=+-2\ x=+-frac{8}{sqrt{3}}$