Example 2: Person A and Person B start from points $A$ and $B$ respectively at the same time and walk towards each other. They meet at point $C$, which is 10 kilometers away from $A$. After meeting, both continue at the same speed, reach $B$ and $A$ respectively, and immediately return. They meet again at point $D$, which is 3 kilometers away from $B$. Find the distance between $A$ and $B$.

Analysis: As shown in Figure 3, the time it takes for A to travel from $A$ to $C$ is equal to the time it takes for B to travel from $B$ to $C$; the time it takes for A to travel from $A$ to $B$ and then to $D$ is equal to the time it takes for B to travel from $B$ to $A$ and then to $D$.
Solution: Let the distance between $A$ and $B$ be $x$ kilometers, the speed of person A be $v_{1}$ kilometers/hour, and the speed of person B be $v_{2}$ kilometers/hour, then
$$
\left\{\begin{array}{l}
\frac{10}{v_{1}}=\frac{x-10}{v_{2}}, \\
\frac{x+3}{v_{1}}=\frac{2 x-3}{v_{2}} .
\end{array}\right.
$$

Eliminating the parameter $v_{1} \backslash v_{2}$, we get
$$
\frac{x-10}{10}=\frac{2 x-3}{x+3} \text {. }
$$

Solving for $x$ yields $x=27$ (kilometers).
Answer: Omitted.
(3) Speed equivalence relationship