13.217. The road between villages $A$ and $B$ first has an ascent, then a descent. A cyclist, moving downhill at a speed $a$ km/h faster than uphill, spends exactly $t$ hours on the journey from $A$ to $B$, and half of this time on the return journey from $B$ to $A$. Find the cyclist's speed uphill and downhill, if the distance between the villages is $b$ km.
Solution.

Let $x$ km/h be the cyclist's speed on the ascent, and $(x+a)$ km/h on the descent.

Let $y$ km be the length of the ascent, and $(b-y)$ km be the descent. Then:

$\left\{\begin{array}{l}\frac{y}{x}+\frac{b-y}{x+a}=t \\ \frac{b-y}{x}+\frac{y}{x+a}=\frac{t}{2}\end{array}\right.$

Solving the system, we find $x=\frac{4 b-3 a t+\sqrt{9 t^{2} a^{2}+16 b^{2}}}{6 t}$.

$$
\begin{aligned}
& x+a=\frac{4 b+3 a t+\sqrt{9 t^{2} a^{2}+16 b^{2}}}{6 t} . \\
& \text { Answer: } \frac{4 b \pm 3 a t+\sqrt{9 t^{2} a^{2}+16 b^{2}}}{6 t} \text{ km/h }; 4 b>3 a t .
\end{aligned}
$$