$\begin{array}{l}\text { 1. Given } x+\lg x=10, y+10^{y}=10 \text {. Then } x+y \\ \text {. }\end{array}$
1. 10 .

Solution: Let $f(t)=t+10^{t}$, then $f(t)$ is a strictly increasing function, so there exists a unique real number $y$ such that $y+10^{y}=10$. Also, $\lg x=10$ $-x$, so $10^{10-x}=x$, which means $(10-x)+10^{10-x}=10$. Therefore, $10-x=y$, which implies $x+y=10$.