SOLUTION: a train leaves the city heading west and travels at 50 miles per hour. Three hours later, a second train leaves from the same place and travels in the same diresction at 65 miles p

You can do it in your head. The first train goes 50 mph for 3 hours,
so it's 150 miles from the starting point when the second train starts.
The first train is 150 miles ahead of the second train, and the second 
train goes only 15 mph faster, so it'll take it 10 hours to catch
up the 150 miles it's behind.

But your teacher wants you to do it with algebra. OK

Let t = the time it will take the second train to catch up.

Make this chart:


                distance       rate       time
first train    
second train       


Let the answer be t, so fill in t for the second train's
time to catch up


                distance       rate       time
first train                             
second train                                t         

Next fill in the two give rates of speed:


                distance       rate       time
first train                     50         
second train                    65          t     


The first train travel 3 hours longer than the
second train, so add 3 to the time for the the 
first train, getting t+3 and fill that in for
the first train's time:P


                distance       rate       time
first train                     50         t+3
second train                    65          t 

No use distance = rate x time 

to fill in the two distances


                distance       rate       time
first train      50(t+3)        50         t+3
second train      65t           65          t 

Those distances must be equal when the second train 
catches up to overtake the first train: 

So the equation is 

          50(t+3) = 65t

Solve that and you'll get t = 10 which we got doing
it in our heads.

Edwin