3.1. A pedestrian left point $A$ for point $B$. At the same time, a cyclist left point $B$ for point $A$. After one hour, the pedestrian was three times farther from the cyclist than from point $A$. Another 30 minutes later, they met, after which both continued their journey. How many hours did it take the pedestrian to travel from $A$ to $B$?
Answer: 9

Solution. Let the distance from $A$ to $B$ be 1 km, the pedestrian moves at a speed of $x$ km/h, and the cyclist at $y$ km/h. Then in one hour, the pedestrian has walked $x$ km, the cyclist has traveled $y$ km, and the distance between them is $1-x-y$, which should be three times $x$. Therefore, $3 x=1-x-y, y=1-4 x$. In the next half hour, they will together cover $1-x-y$ km. Since their closing speed is $x+y$, we get the equation $\frac{1}{2}(x+y)=1-x-y$, from which $x+y=\frac{2}{3}$. Substituting $y=1-4 x$ into the last equation: $1-3 x=\frac{2}{3}, x=\frac{1}{9}$. Therefore, it will take the pedestrian 9 hours to travel the distance from $A$ to $B$.