Let the train’s speed be x km/h and the time elapsed be y hours.
The entire distance is xy kilometers.
Scenario 1:
The speed improves by 10 kilometers per hour, while the time required decreases by 2 hours. The traveled distance remains xy kilometers.
(x+10)(y-2)=xy
Therefore, xy-2x+10y-20=xy
-2x+10y-20=0
-2x+10y=20…………………(1)
Scenario 2:
The time required rises by 3 hours when the speed lowers by 10 km/h.
However, the distance is still xy kilometers.
(x-10)(y+3)=xy
Therefore, xy+3x-10y-20=xy
3x-10y-30=0
3x-10y=30…………………(2)
x=50km/h
By this (2)
150-10y=30
y=12
So the distance is 50×12=600 km.