recall your d = rt, distance = rate * time
we know the total trip was 352miles, so, let's say he was going for the first 2hrs at "r" speed, he went "d" miles, then he slowed down to "r - 20" and went 6 more hours, since the total trip is 352 miles, the 6hrs take the slack from 352 - d, and that's how many miles he covered on those 6hours
[tex]\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
\textit{before slowing}&d&r&2\\
\textit{after slowing}&352-d&r-20&6
\end{array}
\\\\\\
\begin{cases}
\boxed{d}=2r\\
352-d=(r-20)6\\
----------\\
352-\boxed{2r}=(r-20)6
\end{cases}[/tex]
solve for "r".
