<PRE><B>Question 612723</B>
During the first part of a trip, a canoeist travels 49 miles at a certain speed.
 The canoeist travels 4 miles on the second part of the trip at a speed of 5 mph slower.
 The total time for the trip is 2 hours.
 What was the speed on each part of the trip?
:
Let s = speed on the 1st 49 mi
then
(s-5) = speed on the 2nd part (4 mi)
:
Write a time equation: time = dist/speed
:
{{{49/s}}} + {{{4/((s-5))}}} = 2
Multiply by s(s-5)
s(s-5)*{{{49/s}}} + s(s-5)*{{{4/((s-5))}}} = 2s(s-5)
Cancel the denominators
49(s-5) + 4s = 2s^2 - 10s
49s - 245 + 4s = 2s^2 - 10s
53s - 245 = 2s^2 - 10s
Combine as a quadratic equation on the right
0 = 2s^2 - 10s - 53s + 245
2s^2 - 63s + 245 = 0
We have to use the quadratic formula to find s
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this problem; x=s; a=2; b=-63; c= 245
{{{s = (-(-63) +- sqrt(-63^2-4*2*245 ))/(2*2) }}}
Two solutions, but only one will make sense
s ~ 26.955 mi for the 1st 49 mi 
then
26.955 - 5 = 21.955 mph for the last 4 mi
:
:
Confirm this; find the time with these solutions
{{{49/26.955}}} + {{{4/20.955}}} = 2 </PRE>