SOLUTION: A bicyclist took a ride to the country and it took 5 hours. To get back she went 5mph faster and it took 4 hours. What were the speeds, going and coming?

A bicyclist took a ride to the country and it took 5 hours.
To get back she went 5mph faster and it took 4 hours. What 
were the speeds, going and coming?

Make this chart

        Distance   Rate  Time    
Going  |        |       |    | 
Coming |        |       |    |

Let x = the speed going in mph

>>...To get back she went 5mph faster...<<

So the speed coming back = x+5 mph.  So fill in these 
speeds


        Distance   Rate  Time    
Going  |        |   x   |    | 
Coming |        |  x+5  |    | 

>>...A bicyclist took a ride to the country and it took 
5 hours...<<

>>...To get back...it took 4 hours...<<

So fill in 5 hours and 4 hours for the times:

        Distance   Rate  Time    
Going  |        |   x   |  5 | 
Coming |        |  x+5  |  4 | 

Now use Distance = Rate × Time to fill in the distances

        Distance   Rate  Time    
Going  |   5x   |   x   |  5 | 
Coming | 4(x+5) |  x+5  |  4 | 

Now since the distance going equals the distance back, we
form an equation by setting the two distances equal

       5x = 4(x+5)

Solve that and get x = 20 mph going, then since the speed
returning is 5mph more, the returning speed is 25 mph.

Edwin
AnlytcPhil@aol.com