recall your d = rt, distance = rate * time.
let's say the rate of speed for Kala was "r" on the way over to the mountains.
Now, on the way over, it took 8 hours due to heavy traffic, whilst on the way back it took only 5 hours since there was no traffic.
The distance covered, say "d" miles, is the same on the way over as well as on the way back.
[tex]\bf \begin{array}{lccclll}
&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
\textit{to the mountains}&d&r&8\\
\textit{to back home}&d&r+21&5
\end{array}
\\\\\\
\begin{cases}
d=8r\implies \frac{d}{8}=\boxed{r}\\
d=5(r+21)\\
----------\\\\
d=5\left(\boxed{\frac{d}{8}}+21 \right)
\end{cases}
\\\\\\
d=5\left( \cfrac{d+168}{8} \right)\implies d=\cfrac{5d+840}{8}\implies 8d=5d+840
\\\\\\
3d=840\implies d=\cfrac{840}{3}\implies d=280[/tex]
