<PRE><B>Question 512669</B>
On a 50 mile bike ride Irena averaged 4 mph faster for the first 36 miles than she did for the last 14 miles.
 The entire trip took 3 hours. Find her speed for the 36 miles
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let s = the speed for the 1st 36 mi
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Write a time equation; time = dist/speed
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36mi time + 14 mi time = 3 hrs
{{{36/s}}} + {{{14/((s-4))}}} = 3
Multiply by s(s-4)
s(s-4)*{{{36/s}}} + s(s-4)*{{{14/((s-4))}}} = s(s-4)*3
cancel the denominators
36(s-4) + 14s = 3s(s-4)
36s - 144 + 14s = 3x^2 - 12s
50s - 144 = 3x^2 - 12s
arrange as a quadratic equation on the right
0 = 3s^2 - 12s - 50s + 144 
3x^2 - 62s + 144 = 0
you can use the quadratic formula, but this will factor to:
(3s-8)(s-18) = 0
The reasonable solution
s = 18 mph for the 1st 36 mi
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:
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Check this by finding the time at each speed
36/18 = 2 hrs
14/14 = 1 hr
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