$4 \cdot 35$ Solve the equation: $|x+1|+\frac{1}{2}|2 x-3|=5 x+2$.
[Solution] First, since the left side of the equation is not less than 0, the right side should also not be less than 0, i.e., $5 x+2$ $\geqslant 0, x \geqslant-\frac{2}{5}$.

Next, we discuss in two cases:
(1) When $x \geqslant \frac{3}{2}$, the original equation is:
$x+1+\frac{1}{2} \cdot(2 x-3)=5 x+2$,
$4 x-1=10 x+4, x=-\frac{5}{6}$, which is a contradiction, so there is no solution.
(2) When $-\frac{2}{5} \leqslant x<\frac{3}{2}$, the original equation is:
$$
\begin{array}{l}
x+1+\frac{1}{2}(3-2 x)=5 x+2, \\
\frac{5}{2}=5 x+2, \text { so } x=\frac{1}{10} .
\end{array}
$$

Therefore, the original equation has a unique solution $x=\frac{1}{10}$.