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?

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