<PRE><B>Question 970277</B>
Colton went on a bike ride of 36 miles.
 He realized that if he had gone 3 mph faster, he would have arrived 6 hours sooner.
 How fast did he actually ride?
:
let s = actual speed
then
(s+3) = the faster time speed
:
Write a time equation, time = dist/speed
:
Actual time - faster time = 6 hrs
{{{36/s}}} - {{{36/((s+3))}}} = 6
multiply equation by s(s+3)
s(s+3)*{{{36/s}}} - s(s+3)*{{{36/((s+3))}}} = 6s(s+3)
cancel the denominators
36(s+3) - 36s = 6s^2 + 18s
36s + 108 - 36s = 6s^2 + 18s
Combine as a quadratic equation on the right
0 = 6s^2 + 18s - 108
simplify, divide by 6
s^2 + 3s - 18 = 0
Factors to 
(s + 6)(s - 3) = 0
the positive solution is all we want here
s = 3 mph his actual speed
:
:
Check this find the actual time at each speed
36/3 = 12 hrs
36/6 =  6 hrs; 6 hrs faster

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