Question 34948
<pre><font size = 5><b>
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

<u>        Distance   Rate  Time</u>    
<u>Going  |        |       |    |</u> 
<u>Coming |        |       |    |</u>

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


<u>        Distance   Rate  Time</u>    
<u>Going  |        |   x   |    |</u> 
<u>Coming |        |  x+5  |    |</u> 

>>...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:

<u>        Distance   Rate  Time</u>    
<u>Going  |        |   x   |  5 |</u> 
<u>Coming |        |  x+5  |  4 |</u> 

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

<u>        Distance   Rate  Time</u>    
<u>Going  |   5x   |   x   |  5 |</u> 
<u>Coming | 4(x+5) |  x+5  |  4 |</u> 

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</pre>