Система уравнений X+Y-8=0 X^2+Y^2=16+2XY

Решите задачу:

$\begin{cases}x+y-8=0\\x^2+y^2=16+2xy\end{cases}\Rightarrow\begin{cases}x=8-y\\x^2-2xy+y^2=16\end{cases}\Rightarrow\\\Rightarrow\begin{cases}x=8-y\\(x-y)^2=16\end{cases}\Rightarrow\begin{cases}x=8-y\\(8-y-y)^2=16\end{cases}\\(8-y-y)^2=16\\(8-2y)^2=16\\\begin{cases}8-2y=4\\8-2y=-4\end{cases}\Rightarrow\begin{cases}y_1=2\\y_2=6\end{cases}\\\begin{cases}x_1=6\\y_1=2\end{cases}\quad\quad\quad\begin{cases}x_2=2\\y_2=6\end{cases}\\OTBET:\;(6;\;2),\;(2;\;6)$