From city A, car a sets on it's road towards city B. 9 Hours later, car b, sets from city B towards city A. The two cars met along the way. At the point of the meeting, car a, passed 240 km more than car b had. car a arrived at city B 10 hours after the meeting. car b arrived at city A 9 hours after said meeting. Velocities of both cars were constant.
What is the distance that car b passed until the meeting? What are the velocities of both cars?
I got 3 equalities here, in 4 unknowns, which didn't allow me to reach a finite solution to the problem; \begin{cases} Va (t+9) = Vb(t) + 240 \\ Va (t+9) + Va (10) = S \\ Vb(9)+Vb(t) = S \end{cases} where Va, Vb, velocities of a and b, concordantly, and S the total distance between the cities. t the time b traveled until the meeting.
You have enough information for the unknowns.
Let $t_{m}$ be the meeting time, in hours. Let $d_A, d_B$ be the distances in kilometers that cars $A, B$ travel before they meet. Let $v_A, v_B$ be the speeds of the cars in km/hr. That's five unknowns.
Before the meeting, we know that $d_A = v_A t_m$ and $d_B = v_B (t_m - 9 \text{ hr}).$
We know also that $d_A - d_B = 240 \text{ km}.$
We know after the meeting that $d_B/v_A = 10 \text{ hr}$ and $d_A/v_B = 9 \text{ hr}.$
Five equations, five unknowns. Now solve for $d_B, v_A, v_B.$