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Ответ:

Решим систему методом подстановки.

[tex]\left\{\begin{array}{l}x=y^2-4y\\x+y=4\end{array}\right\ \ \left\{\begin{array}{l}x=y^2-4y\\(y^2-4y)+y=4\end{array}\right\ \ \left\{\begin{array}{l}x=y^2-4y\\y^2-3y-4=0\end{array}\right\\\\\\\left\{\begin{array}{l}x=y(y-4)\\(y+1)(y-4)=0\end{array}\right\ \ \left\{\begin{array}{l}x_1=-(-1-4)\ ,\ x_2=4\, (4-4)\\\ y_1=-1\ \ ,\ \ \ \ \ \ \ \ \, \ y_2=4\end{array}\right\ \[/tex]

[tex]\left\{\begin{array}{l}x_1=5\ ,\ \ \ x_2=0\\y_1=-1\ ,\ y_2=4\end{array}\right\ \ \ \ \Rightarrow \ \ \ \bf Otvet:\ \ (\, 5\, ;-1\, )\ ,\ (\, 0\, ;\, 4\, )\ .[/tex]