Two cities are 45 miles apart. two trains, with speeds of 70 mph and 60 mph, leave the two cities at the same time so that one is catching up to the other. how long after the trains leave will they be 10 miles apart for the first time? how long after the trains leave will they be 10 miles apart for the second time?

T1  - the time the trains are moving to each other.
For time t1, first train will go the distance 70*t.
The second train, for time t will go the distance 60*t.
Both for time t1  - 70t+60(t1)=(70+60)*(t1).
45 - (70+60)*t is the distance between trains that left after t hours.
For the first time they will be 10 miles apart when they are moving to each other
45 - (70+60)*t = 10
45-10=(70+60)*t
35=130*t
t=35/130 h =7/26 h≈ 0.27 h

The train meet when distance=0
45 - (70+60)*t=0
45=130t
t=45/130 h = 9/26 h ≈0.35 h when trains met.

Trains began to move apart from each other after 9/26 h moving,
After t2 time trains will be (70+60)*(t2)=130*(t2) mi apart from each other.

130*(t2)=10
t2=1/13h≈0.08h after they met, they are apart from each other 10 mi.

so after beginning of the movement it will be 
time when they met + time when they are moved  10 mi apart from each other
9/26 h+1/13h = 11/26 h≈ 0.42 h

1. isolate the variable, in this case, d. subtract 35.5 on both sides of the equation.2. 52 subtracted by 35.5 is 16.53. therefore your answer would be d= 16.5this is the basics of pre algebra, isolating variable through adding, subtracting, multiplying, and dividing, if you have any other questions to ask just comment!