8. Given real numbers $x, y$ satisfy
$$
\left\{\begin{array}{l}
(x-11)^{5}+15(x-11)=5, \\
(y-4)^{5}+15(y-4)=-5 .
\end{array}\right.
$$

Then $x+y$ equals $\qquad$ .
8. 15 .

Since the function $f(t)=t^{5}+15 t$ is strictly monotonically increasing in the real number domain, and $f(x-11)=f(4-y)$, then $x-11=4-y$. Therefore, $x+y=15$.