so hmm recall your d = rt, or distance = rate * time
so, Harry drove 5hrs, at say a speed rate of "r", then he slowed down by 20mph, so, if he was driving at "r" mph firstly, then after he slowed down, he's driving at " r - 20"
now, we know, the total trip on the freeway and country road was 518 miles, so, if on the freeway he covered say hmmm "d" distance, on the country road he covered the slack, or 518 - d
thus [tex]\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
\textit{freeway}&d&r&5\\
\textit{country road}&518-d&r-20&6
\end{array}\\\\
-----------------------------\\\\
\begin{cases}
\boxed{d}=5r\\
518-d=6(r-20)\\
----------\\
518-\boxed{5r}=6(r-20)
\end{cases}[/tex]
solve for "r"