Question

Найти решение системы уравнений:{2х2-ху=24 {4х-у=16

Answer 1

$left { {{2x^{2}- xy=24} atop {4x-y=16}} right.$
$left { {{y=4x-16} atop {2x^{2}-x(4x-16) =24}} right.$
2x²-4x²+16x-24=0
-2x²+16x-24=0 |:(-2)
x²-8x+12=0
D= 64 - 48 = 16
x₁= 8+4/2=6
x₂=8-4/2 = 2

y₁= 4*6-16 = 8
y₂= 4*2-16= -8

Ответ: (6;8), (2;-8).

Answer 2

$left { {{2x^2-xy=24} atop {4x-y=16}} right. \ y=4x-16 \ 2x^2-x(4x-16)=24 \ 2x^2-4x^2+16x=24 Rightarrow -2x^2+16x-24=0 Rightarrow x^2-8x+12=0 \ Delta=64-48=16 textgreater 0 \ x_1= frac{8-4}{2}=2 \ x_2= frac{8+4}{2}=6 \ y_1=4x_1-16=8-16=-8 \ y_2=4x_2-16=24-16=8 \ (x_1;y_1)=(2;-8);(x_2;y_2)=(6;8)$