[tex]\bf \begin{array}{ccccllll}
&distance&rate(km/hr)&time(hrs)\\
&\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\\
A&d&60&t\\
B&600-d&v&t+3
\end{array}[/tex]
[tex]\bf \textit{meaning}\implies
\begin{cases}
d=(60)(t)
\\ \quad \\
600-d=(v)(t+3)\\
------------\\
d=\boxed{60t}\qquad thus
\\ \quad \\
600-\boxed{60t}=v(t+3)\leftarrow \textit{solve for "t"}
\end{cases}[/tex]
keep in mind, that "t" is the time when the train at A station, left towards B station
they met, at some time "t", and by the time that happened, train from A
which started 3 hours earlier, had already covered "d" distance,
whatever that is
and the train coming from B, covered, 600-d, or the difference