Question

{2х^2-ух=-16 3х^2+ху=х+18​

Answer 1

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

И теперь найдём другую точку.

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

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