, 2020

Since $x^2+3xy+y^2 = 909$ and $3x^2+xy+3y^2=1287$, then $$\begin{aligned} (x^2+3xy+y^2)+(3x^2+xy+3y^2) & = 909 + 1287 \\ 4x^2 + 4xy + 4y^2 & = 2196 \\ x^2 + xy + y^2 & = 549\end{aligned}$$ Since $x^2+3xy+y^2 = 909$ and $x^2 + xy + y^2 = 549$, then $$\begin{aligned} (x^2+3xy+y^2) - (x^2 + xy + y^2) & = 909 - 549 \\ 2xy & = 360 \\ xy & = 180\end{aligned}$$ Since $x^2 + 3xy + y^2 = 909$ and $xy = 180$, then $$\begin{aligned} (x^2 + 3xy + y^2) - xy & = 909 - 180 \\ x^2 + 2xy + y^2 & = 729 \\ (x+y)^2 & = 27^2\end{aligned}$$ Therefore, $x+y = 27$ or $x+y = -27$. This also shows that $x+y$ cannot equal any of 39, 29, 92, and 41.
(We can in fact solve the system of equations $x+y = 27$ and $xy = 180$ for $x$ and $y$ to show that there do exist real numbers $x$ and $y$ that are solutions to the original system of equations.)
Therefore, a possible value for $x+y$ is (A) 27.