Question

Решите систему уравнений подстановки

Answer 1

$\left\{\begin{array}{l}2x+y^2=6\\x+y=3\end{array}\right\ \ \left\{\begin{array}{l}2\, (3-y)+y^2=6\\x=3-y\end{array}\right\ \ \left\{\begin{array}{l}y^2-2y=0\\x=3-y\end{array}\right\ \ \left\{\begin{array}{l}y(y-2)=0\\x=3-y\end{array}\right\\\\\\\left\{\begin{array}{l}y_1=0\ ,\ y_2=2\\x_1=3\ ,\ x_2=1\end{array}\right\\\\\\Otvet:\ \ (3;0)\ ,\ (1;2)\ .$

Answer 2

(1,2) и (3,0)

Объяснение:

x = 3 - y, подставим:

2 * (3 - y) + y^2 = 6

6 - 2y + y^2 = 6

y^2 - 2y = 0

y(y-2)=0

y = 0 и y = 2

При y = 0:

x = 3 - 0

x = 3

При y = 2:

x = 3 - 2

x = 1