Question 69992
<pre><font size = 5 color = "darkblue"><b>
Steve traveled 600 miles at a certain speed.
Had he gone 20mph faster, the trip would have 
taken 1 hour less. Find the speed of his vehicle. 
Thank you in advance for the help!

I'll do it a different way from Stanbon:

Let r be the actual rate and t be the actual time.
Then the rate for the hypothetical trip would be r+20 and
the time for the hypothetical trip would be t-1.

So put the information into a DRT chart:

                   DISTANCE   RATE    TIME
Actual trip      |   600   |   r   |   t
Hypothetical trip|   600   |  r+20 |  t-1

Now use DISTANCE = (RATE)(TIME)

     600 = rt
     600 = (r+20)(t-1)

FOIL out the right side of the second equation:

     600 = rt - r + 20t - 20

Multiply the 1st equation by -1 and add it to
the second equation

     600 =  rt - r + 20t - 20
    -600 = -rt
   ---------------------------
       0 =      -r + 20t - 20
       r = 20t - 20

Substitute in the first equation

     600 = rt
     600 = (20t - 20)t
     600 = 20tē - 20t
       0 = 20tē - 20t - 600
Divide every term through by 20
       0 = tē - t - 30
       0 = (t - 6)(t + 5)

t - 6 = 0     t + 5 = 0
    t = 6 hrs     t = -5 hrs

We discard the negative answer.

To find r, substitute in
            
     600 = rt
     600 = r(6)
     600 = 6r
     100 = r

So he traveled at 100 mi/hr.
His vehicle must have been a
small plane.

Edwin</pre>