Решаем способом подстановки:
[m]\left\{\begin {matrix}(2y+6+2)(y+1)=36\\x=2y+6\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}2y^2+10y-28=0\\x=2y+6\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}y^2+5y-14=0\\x=2y+6\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}y=-7\\x=2\cdot (-7)+6\end {matrix}\right.[/m] или [m]\left\{\begin {matrix}y=2\\x=2\cdot 2+6\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}y=-7\\x=-8\end {matrix}\right.[/m] или [m]\left\{\begin {matrix}y=2\\x=10\end {matrix}\right.[/m]