$4 \cdot 34$ Try to find the range of real values of $x$ that satisfy the following relation:
$$
|3 x-2|+|3 x+1|=3 \text{. }
$$
$[$ Solution] $\because|3 x-2|+|3 x+1|=|2-3 x|+|3 x+1|=3$,
and $\quad|2-3 x+3 x+1|=3$,
$\therefore \quad|2-3 x|+|3 x+1|=|(2-3 x)+(3 x+1)|$,
$\therefore 2-3 x$ and $3 x+1$ must have the same sign, or one of them is 0,
If $\left\{\begin{array}{l}2-3 x \geqslant 0 \\ 3 x+1 \geqslant 0,\end{array}\right.$, then $-\frac{1}{3} \leqslant x \leqslant \frac{2}{3}$.
If $\left\{\begin{array}{l}2-3 x \leqslant 0 \\ 3 x+1 \leqslant 0,\end{array}\right.$ no solution.
$\therefore$ The real value range of $x$ in the original equation is: $-\frac{1}{3} \leqslant x \leqslant \frac{2}{3}$.