2. $x, y \in \mathbf{R}$, and satisfy $\left\{\begin{array}{l}(x+1)^{\frac{3}{5}}+2023(x+1)=-2023, \\ (y+1)^{\frac{3}{5}}+2023(y+1)=2023\end{array}\right.$, then $x+y=$
Since the function $f(x)=x^{\frac{3}{5}}+2023 x$ is an odd function, and it is monotonically increasing on $(0,+\infty)$, then according to the problem, $f(x+1)+f(y+1)=0 \Rightarrow x+1+y+1=0 \Rightarrow x+y=-2$.