Answer
the speed of the faster train, $x_1$, is $
\dfrac{191}{3} \text{ }mph
$ while the speed of the slower train, $x_2,$ is $
45 \text{ }mph$
Work Step by Step
Let $x_1$ be the speed of the faster train and $x_2$ the speed of the slower train.
Using $D=rt,$ then for the faster train,
\begin{array}{l}\require{cancel}
382=x_1(6)
\\
\dfrac{382}{6}=x_1
\\
x_1=\dfrac{\cancel2\cdot191}{\cancel2\cdot3}
\\
x_1=\dfrac{191}{3}
.\end{array}
Using $D=rt,$ then for the slower train,
\begin{array}{l}\require{cancel}
382-112=x_2(6)
\\
270=x_2(6)
\\
\dfrac{270}{6}=x_2
\\
x_2=45
.\end{array}
Hence, the speed of the faster train, $x_1$, is $
\dfrac{191}{3} \text{ }mph
$ while the speed of the slower train, $x_2,$ is $
45 \text{ }mph.$
