Example 12. Solve the inequality

$$
|x-1|+|2-x|>3+x
$$
Solution. The points $x=1$ and $x=2$ divide the number line (the domain of inequality (10)) into three intervals: $x<1$, $1 \leqslant x \leqslant 2$, and $x>2$. We will solve the given inequality on each of these intervals.

If $x<1$, then $1-x>0$ and $2-x>0$. Inequality (10) becomes $1-x+2-x>3+x$, i.e., $x < 0$.
$$
\begin{array}{c}
\left\{\begin{array}{l}
x<1 \\
1-x+2-x>3+x
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x<1 \\
x<0
\end{array} \Leftrightarrow x<0\right.\right.
\end{array}
$$

If $1 \leqslant x \leqslant 2$, then $x-1 \geqslant 0$ and $2-x \geqslant 0$. Inequality (10) becomes $x-1+2-x>3+x$, i.e., $1>3+x$, which is impossible.
$$
\begin{array}{c}
\left\{\begin{array}{l}
1 \leqslant x \leqslant 2 \\
1-x+2-x>3+x
\end{array} \Leftrightarrow \left\{\begin{array}{l}
1 \leqslant x \leqslant 2 \\
1>3+x
\end{array} \Leftrightarrow \text { no solution }\right.\right.
\end{array}
$$

If $x>2$, then $x-1>0$ and $2-x<0$. Inequality (10) becomes $x-1+x-2>3+x$, i.e., $x>6$.
$$
\begin{array}{c}
\left\{\begin{array}{l}
x>2 \\
x-1+x-2>3+x
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x>2 \\
x>6
\end{array} \Leftrightarrow x>6\right.\right.
$$

Combining the solutions found on all parts of the domain of inequality (10), we get its solution set $(-\infty ; 0) \mathrm{U}$ $U(6 ;+\infty)$.