recall your d = rt, distance = rate * time.
A = first train.
B = second train leaving 2 hrs later.
if say by the time they meet, train B has traveled "t" hours, we know that train A has been traveling 2 hours more than that, because it left 2 hours earlier than train B, thus it has traveled "t + 2" hours.
keeping in mind that by the time they meet, they both have traveled "d" kilometers.
[tex]\bf \begin{array}{lcccl}
&\stackrel{km s}{distance}&\stackrel{km/h}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
\textit{Train A}&d&75&t\\
\textit{Train B}&d&45&t+2
\end{array}
\\\\\\
\begin{cases}
d=75t\implies \frac{d}{75}=\boxed{t}\\\\
d=45(t+2)\\
----------\\
d=45\left( \boxed{\frac{d}{75}}+2 \right)
\end{cases}[/tex]
[tex]\bf d=\cfrac{45d}{75}+90\implies d=\cfrac{3d}{5}+90\implies \stackrel{\textit{multiplying both sides by 5}}{5d=3d+450}
\\\\\\
2d=450\implies d=\cfrac{450}{2}\implies d=\stackrel{km s}{225}[/tex]
