7. If the real numbers $(x, y)$ satisfy $\left\{\begin{array}{l}x^{3}-6 x^{2}+12 x=13 \\ y^{3}+3 y-3 y^{2}=-4\end{array}\right.$, then $x+y=$
7. 3

Organize: $x^{3}-6 x^{2}+12 x=13$ to get $(x-2)^{3}=5$, so $(x-2)^{3}=(1-y)^{3}$
Interim conclusion: $y^{3}+3 y-3 y^{2}=-4 \quad$ to get $(y-1)^{3}=-5$, so $x-2=1-y$. Therefore, $x+y=3$