Suppose that $x$ and $y$ are real numbers that satisfy the two equations:

$$
\begin{gathered}
x^{2}+3 x y+y^{2}=909 \\
3 x^{2}+x y+3 y^{2}=1287
\end{gathered}
$$

What is a possible value for $x+y$ ?
(A) 27
(B) 39
(C) 29
(D) 92
(E) 41
Since $x^{2}+3 x y+y^{2}=909$ and $3 x^{2}+x y+3 y^{2}=1287$, then

$$
\begin{aligned}
\left(x^{2}+3 x y+y^{2}\right)+\left(3 x^{2}+x y+3 y^{2}\right) & =909+1287 \\
4 x^{2}+4 x y+4 y^{2} & =2196 \\
x^{2}+x y+y^{2} & =549
\end{aligned}
$$

Since $x^{2}+3 x y+y^{2}=909$ and $x^{2}+x y+y^{2}=549$, then

$$
\begin{aligned}
\left(x^{2}+3 x y+y^{2}\right)-\left(x^{2}+x y+y^{2}\right) & =909-549 \\
2 x y & =360 \\
x y & =180
\end{aligned}
$$

Since $x^{2}+3 x y+y^{2}=909$ and $x y=180$, then

$$
\begin{aligned}
\left(x^{2}+3 x y+y^{2}\right)-x y & =909-180 \\
x^{2}+2 x y+y^{2} & =729 \\
(x+y)^{2} & =27^{2}
\end{aligned}
$$

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 $x y=180$ 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 .

ANswer: (A)

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