SOLUTION: Would someone please assist me with this problem? I cannot figure it out for anything. During the first part of a trip, a canoeist travels 84 miles at a certain speed. THe canoi

Question 604301: Would someone please assist me with this problem? I cannot figure it out for anything.
During the first part of a trip, a canoeist travels 84 miles at a certain speed. THe canoiest travels 22 miles on the second part of the trip at 5mph slower. he total trip is 2 hours. What is the speed on each part of the trip?
Speed on first part? Speed on second part?
Found 2 solutions by flame8855, mananth:
Answer by flame8855(424) (Show Source): You can put this solution on YOUR website!
for trip 1
distance = 84
speed : x
time t1
for tripe 2 :
distance = 22
speed = x-5
time = t2
t1+t2 =2
v =d/t
t = d/v
84/x + 22/(x-5) =2
84 (x-5) +22x = 2x(x-5)
x^2 - 58 x +210 = 0
x = 54.12 mph part 1
x-5 = 49.12 mph part 2

Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
First Part 84 miles
Second Part 22 miles

First Part x mph
Second Part x -5 mph
Total time 2 hours
First Part time 84 / x
Second Part time 22 / ( x -5 )

Time first part + time second part = 2 hours

84/x+22 /(x-5) =2
LCD = (x)*(x-5)
multiply the equation by the LCD
we get
84*(x -5)+22 x= 2
84x-420 +22x=2X^2-10
116 x+ -420 = 2 X^2
2 X^2 -116 x+ 420 = 0
2 X^2+ -116 x+ 420 = = 0
/ 2
2 X^2 -116 x+ 420 = 0

Find the roots of the equation by quadratic formula

a= 2 b= -116 c= 420

b^2-4ac= 13456 - 3360
b^2-4ac= 10096
= 100.48

x1=( 116 + 100.48 )/ 4
x1= 54.12
x2=( 116 -100.48 ) / 4
x2= 3.88
x2 is not possible
so speed = 54.12 mph I part
49.12 mph II part
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