One train leaves City A heading for City B which is 390 mile

One train leaves City A heading for City B which is 390 miles away. At the same time a secind train leaves City B heading for City A, going 15 mph faster than the first train. If they meet in 3 hours and 20 minutes, how fast are the trains traveling?

Solution

D=RT
390=(X+X+15)10/3
390=(2X+15)10/3
390=(20X+150)/3
390*3=20X+150
20X=1,170-150
20X=1,020
X=1,020/20
X=51 MPH FOR THE SLOWER TRAIN.
51+15=66 MPH IS THE SPEED OF THE FASTER TRAIN.
PROOF:
390=(51+66)10/3
390=117*10/3
390=1,170/3
390=390