Найти решение системы уравнений:
2х(2)-ху=24
4х-у=16

[tex] \left \{ {{2x^2-xy=24} \atop {4x-y=16}} \right. \; \left \{ {{2x^2-x(4x-16)=24} \atop {y=4x-16}} \right. \; \left \{ {{2x^2-4x^2+16x-24=0} \atop {y=4x-16}} \right. \; \left \{ {{-2x^2+16x-24=0} \atop {y=4x-16}} \right. \\\\-2x^2+16x-24=0\, |:(-2)\\\\x^2-8x+12=0\\\\x_1=2,\; \; x_2=6\; \; (teor.\; Vieta)\\\\y_1=4\cdot 2-16=-8,\; \; y_2=4\cdot 6-16=8\\\\Otvet:\; \; (2,-8)\; ,\; (6,8).[/tex]

Решим систему методом подстановки.
[tex] \left \{ {{(2x^2-xy=24} \atop {4x-y=16}} \right. [/tex]
[tex] \left \{ {{2x^2-xy=24} \atop {-y=4x-16}} \right. [/tex]
[tex] \left \{ {{2x^2-x(4x-16)=24} \atop {y=4x-16}} \right. [/tex]
[tex] \left \{ {{-2x^2+16x-24=0} \atop {y=4x-16}} \right. [/tex]

-2х²+16х-24=0
-2(х²-8х+12)=0
х²-8х+12=0
D=b²-4ac=(-8)²-4*(1*12)=16
[tex]x1,2= \frac{-b+- \sqrt{D} }{2a} = \frac{8+- \sqrt{16} }{2} [/tex]
x₁=2;х₂=6
у=4х-16
y₁=4*2-16=8-16=-8 ;y₂=4*6-16=24-16=8
Ответ:(2;-8);(6;8)