Question

{2х^2-ух=-16 3х^2+ху=х+18​

Answer 1

$left { {{2x^2-yx=-16} atop {3x^2+xy=x+18}} right. +\2x^2+3x^2-yx+xy=-16+x+18;\5x^2-x-2=0;D=1+40\x=frac{1pm sqrt{41} }{10} \2x^2-yx+16=0\2*(frac{1-sqrt{41} }{10} )^2-frac{1-sqrt{41} }{10} y+16=0\frac{1+41-2sqrt{41} }{5} +160=(1-sqrt{41} )y;\frac{42-2sqrt{41} +800}{5(1-sqrt{41} )}=y;\frac{(842-2sqrt{41} )(1+sqrt{41} )}{5(1-41)} =y;\y=frac{760+840sqrt{41} }{-200} =-frac{19+21sqrt{41} }{5}$

И теперь найдём другую точку.

$2*(frac{1+sqrt{41} }{10} )^2-frac{1+sqrt{41} }{10} y+16=0\frac{1+41+2sqrt{41} }{5} +160=(1+sqrt{41} )y;\frac{42+800+2sqrt{41} }{5(1+sqrt{41} )} =y;\y=frac{(842+2sqrt{41} )(1-sqrt{41} )}{5(1-41)} =frac{760-840sqrt{41} }{-200} =frac{21sqrt{41} -19}{5}$

Ответ: $(frac{1-sqrt{41} }{10};-frac{19+21sqrt{41}}{5})+and+(frac{1+sqrt{41}}{10};frac{21sqrt{41}-19}{5})$