Verified answer

Ответ:

[tex]\left\{\begin{array}{l}x^2+y^2-10=0\\xy-3=0\end{array}\right\ \ \left\{\begin{array}{l}x^2+y^2=10\\xy=3\ \ \Big|\cdot 2\end{array}\right\ \ \oplus[/tex]  

Способ сложения: умножим 2 уравнение на 2 и прибавим к 1 уравнению.

[tex]\left\{\begin{array}{l}x^2+y^2+2xy=10+6\\xy=3\end{array}\right\ \ \left\{\begin{array}{l}(x+y)^2=16\\xy=3\end{array}\right\ \ \left\{\begin{array}{l}x+y=\pm 4\\xy=3\end{array}\right\ \ \ \Rightarrow[/tex]

Подучаем две системы уравнений а) и b) .

[tex]a)\ \ \left\{\begin{array}{l}x+y=4\\xy=3\end{array}\right\ \ \left\{\begin{array}{l}y=-x+4\\x(-x+4)=3\end{array}\right\ \ \left\{\begin{array}{l}y=-x+4\\-x^2+4x=3\end{array}\right\ \ \left\{\begin{array}{l}y=-x+4\\x^2-4x+3=0\end{array}\right\\\\\\\star \ \ x^2-4x+3=0\ \ \to \ \ x_1=1\ ,\ x_2=3\ \ (teorema\ Vieta)\ \ \star[/tex]

[tex]\left\{\begin{array}{l}y=-x+4\\x_1=1\ ,\ x_2=3\end{array}\right\ \ \left\{\begin{array}{l}y_1=3\ ,\ y_2=1\\x_1=1\ ,\ x_2=3\end{array}\right\ \ \Rightarrow \ \ \ (1;3)\ ,\ (3;1)[/tex]

[tex]b)\ \ \left\{\begin{array}{l}x+y=-4\\xy=3\end{array}\right\ \ \left\{\begin{array}{l}y=-x-4\\x(-x-4)=3\end{array}\right\ \ \left\{\begin{array}{l}y=-x-4\\-x^2-4x=3\end{array}\right\ \ \left\{\begin{array}{l}y=-x-4\\x^2+4x+3=0\end{array}\right\\\\\\\star \ \ x^2+4x+3=0\ \ \to \ \ x_1=-1\ ,\ x_2=-3\ \ (teorema\ Vieta)\ \ \star[/tex]

[tex]\left\{\begin{array}{l}y=-x-4\\x_1=-1\ ,\ x_2=-3\end{array}\right\ \ \left\{\begin{array}{l}y_1=-3\ ,\ y_2=-1\\x_1=-1\ ,\ x_2=-3\end{array}\right\ \ \Rightarrow \ \ \ (-1;-3)\ ,\ (-3;-1)\\\\\\\bf Otvet:\ (1;3)\ ,\ (3;1)\ ,\ (-1;-3)\ ,\ (-3;-1)\ .[/tex]