recall your d = rt, distance = rate * time.
going into the country, he went for 9 miles, coming back it was the same 9 miles.
let's say hiking his speed is "r" mph, and he took "t" hours hiking.
now, 3 hours and 20 minutes is just 3 hours and one third, or 3 and 1/3 hour.
if he took "t" hours hiking, driving in the car, it took the slack from 3 and 1/3 and "t", or 3 1/3 - t.
[tex]\bf \begin{array}{lccclll}
&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
walking&9&r&t\\
driving&9&27&3\frac{1}{3}-t
\end{array}
\\\\\\
\begin{cases}
9=rt\implies \frac{9}{r}=\boxed{t}\\\\ 9=27\left(3\frac{1}{3}-t \right)\\
----------\\
9=27\left(3\frac{1}{3}-\boxed{\frac{9}{r}} \right)
\end{cases}
\\\\\\
\cfrac{9}{27}=\cfrac{10}{3}-\cfrac{9}{r}\implies \cfrac{1}{3}=\cfrac{10r-27}{3r}\implies 3r=30r-81
\\\\\\
81=27r\implies \cfrac{81}{27}=r\implies 3=r[/tex]
