Two trains set out of two cities, A and B, simultaneously; One from A and one from B. Until the meeting one train has passed $108$ km more than the other. Later, one of the trains arrived at it's destination , in city B, $9$ hours after the meeting, while the other reached A after $16$ hours of the said meeting. Velocities of both trains were constant. What is the distance between the two cities?
I constructed a system of equations for this situation, with a final answer of $226.134$ km. Is this OK?
Train A leaves City A and travels $d_A$ km at $v_A$ km/hr to meet Train B, which left City B and traveled $d_b$ km at $v_B$ km/hr.
Since Train A arrives at its destination, it's the faster one. So,
$$d_A = v_At_{meet}; d_B = v_Bt_{meet}; d_A = d_B + 108.$$
Following that, Trains A and B travel the remaining leg of their journeys in 9 and 16 hours, respectively:
$$\frac{d_B}{v_A} = 9; \frac{d_A}{v_B} = 16.$$
Five equations; five unknowns: $v_A, v_B, d_A, d_B, t_{meet}.$
Substitute the first two equations into the last two to get that $v_A = (4/3)v_B,$ and hence $(v_A - v_B) = (1/3)v_B.$
Combine the first three to get $(v_A - v_B)t_{meet} = 108,$ and hence $(1/3)v_Bt_{meet} = (1/3)d_B = 108.$
So $d_B = 324$ km, meaning that $d_A = 432$ km, and the sum is $756$ km.