$A$ and $B$ - although one lives $3 \mathrm{~km}$ farther from the city than the other - arrived in the city at the same time, with $A$ in a car and $B$ in a truck. The coachman and the driver picked them up along the way. Both $A$ and $B$ started walking from their homes at the same time, and both had covered half of their respective journeys when they were picked up by the aforementioned vehicles. $A$ walked $1.5$ times faster than $B$, but the truck that picked up $B$ traveled $1.5$ times faster than the car in which $A$ was riding. Furthermore, the car traveled at twice the speed at which $A$ walked. How far from the city do $A$ and $B$ live?
First, let's decide which of $A$ and $B$ lived farther from the city. If they lived at the same distance, by the time $B$ reached halfway, $A$ would have walked three-quarters of the way, and by car, $A$ would have just arrived at the city. If $B$ lived farther, $A$ would have been in the city even earlier. Therefore, $A$'s journey must be longer.

Let $B$'s journey be $x$, then $A$'s journey is $x+3$. If $B$'s walking speed is $v$, then $A$'s walking speed is $1.5v$, and by car, it is $3v$. Thus, $B$'s driving speed is $4.5v$. $B$'s travel time walking is $\frac{x}{2v}$, and by car, it is $\frac{x}{9v}$. The travel time for $A$ walking is $\frac{x+3}{3v}$, and by car, it is $\frac{x+3}{6v}$. According to the problem, $A$ and $B$ started and arrived at the same time, so:

$$
\frac{x}{2v} + \frac{x}{9v} = \frac{x+3}{3v} + \frac{x+3}{6v}
$$

Multiplying the equation by $18v$ throughout, we get:

$$
11x = 9x + 27
$$

From this,

$$
x = 13.5
$$

Therefore, $B$ lives $13.5 \text{ km}$ from the city, and $A$ lives $16.5 \text{ km}$ from the city.