so hmmm notice, let's say two, trucks, truck A and truck B,
so truck B is going faster by 1mph than truck A
notice, the distance between them is 291 after 3hrs,
that means, both truck drivers, have been driving for 3hrs,
and covered both combined 291,
let us say, A covered distance "d", so, B covered the rest of it,
or 291 - d,
though their distance and rate differs some, their driving time is
the same 3hrs though
thus [tex]\bf \begin{array}{ccccllll}
&distance&rate&time(hrs)\\
&\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\\
A&d&r&3\\
B&291-d&r+1&3
\end{array}
\\\\
\begin{cases}
d=(r)(3)\to d=\boxed{3r}
\\\\
291-d=(r+1)(3)\\
--------------\\
291-\boxed{3r}=(r+1)(3)
\end{cases}[/tex]
solve for "r", that's how fast truck A was going
what about truck B? well, B was going r + 1
