Example 3, Solve the inequality: $|x-5|-|2x+3|$ < 1. (1978 Shanghai High School Mathematics Competition Question).

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Example 3, Solve the inequality: $|x-5|-|2x+3|$ < 1. (1978 Shanghai High School Mathematics Competition Question).
$$
\begin{array}{l}
\text { Solution: Let }|x-5|-|2 x+3|-1=0 \text { ( }) \\
\Rightarrow(x-5+2 x+3+1)(2 x-5-2 x-3 \\
+1)(x-5-2 x-3-1)(x-5+2 x+3-1) \\
= 0 \\
\Rightarrow \Rightarrow(3 x-1)(-x-7)(-x-9)(3 x-3) \\
\Rightarrow x=\frac{1}{3},-7,-9,1 .
\end{array}
$$

Upon verification, $x=-7, \frac{1}{3}$ are solutions to equation $(\cdot)$.
$$
\because f(x)=|x-5|-|2 x+3|-10, \\
f(1)=|1-5|-|2 \cdot 1+3|-10=-2\frac{1}{3}$.
$$