Example 6. When $a$ varies, find the range of solutions for the inequality $|x+a| + |x| < 2$.

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Solve by transforming the original inequality:
$$
\begin{array}{c}
|\mathrm{x}+\mathrm{a}|<2-|\mathrm{x}| . \\
\text { Let } \mathrm{y}_{1}=|\mathrm{x}+\mathrm{a}|, \\
\mathrm{y}_{2}=2-|\mathrm{x}| .
\end{array}
$$
(1) The graph of (1) is a set of parallel rays with endpoints at $(\cdots a, 0)$ and slopes $\mathrm{k}= \pm 1$ on the x-axis; (2) The graph of (2) is a set of parallel rays with endpoints at $(0,2)$ and slopes $\mathrm{k}= \pm 1$.
$$
\mathrm{x}_{\mathrm{B}}=-\frac{\mathrm{a}+2}{2} \text {. }
$$

From the function graph, we can see:
When $\mathrm{a} \leqslant-2, a \geqslant 2$, the original inequality has no solution;
When $-2<a<2$, the solution range of the original inequality is: $-\frac{a+2}{2}<x<\frac{2-a}{2}$.