SOLUTION: A man walks 6 miles then rides a bike 42 miles. His total trip took 5 hours. If he rode the bike 8 MPH faster than he walked, how fast did he walk?

Question 307434: A man walks 6 miles then rides a bike 42 miles. His total trip took 5 hours. If he rode the bike 8 MPH faster than he walked, how fast did he walk?
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
A man walks 6 miles then rides a bike 42 miles. His total trip took 5 hours. If he rode the bike 8 MPH faster than he walked, how fast did he walk?
let the man walk at x mph
He rides at a speed of x+8 mph
time he walked was 6/x hours ( distance /rate )
time he rode = 42/ x+8
total time the trip took was 5 hours
time he walked + time he rode= 5
6/x + 42/ x+8 = 5
LCM = x(x+8)
6(x+8)+42x / x(x+8)*5
6x+48+ 42x= 5x^2 +40x
5x^2+40x -6x-42x-48=0
5x^2-8x-48=0
find the roots of the equation x1, x2
x1, x2 = -b +/- sqrt(b^2-4ac) / 2a
In this equation a= 5, b=-8 c= -48
x1= 8+sqrt(64+960) /10
x1=4
x2 =8-sqrt(64+960) /10
x2= -2.4
x1 = positive = 4 mph. The walking rate

he rode at x+4 mph = 4+8 =12
12 mph

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