SOLUTION: Julio rides his bike for 6 mi and gets a flat tire. Then he has to walk with the bike for another mile. His speed walking is 6 mph less than his speed riding the bike. If the total

Question 603522: Julio rides his bike for 6 mi and gets a flat tire. Then he has to walk with the bike for another mile. His speed walking is 6 mph less than his speed riding the bike. If the total time is 1hr, find his speed riding the bike and his speed walking?
Found 2 solutions by josmiceli, stanbon:
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let = his speed on the bike
Let = his time riding the bike
--------
Equation for riding the bike:
(1)
Equation walking:
(2)
----------------------
(2)
from (1):
(1)
Substitute (1) into (2)
(2)
(2)
Multiply both sides by
(2)
(2)
Use the quadratic formula

mi/hr
and

( this doesn't work, ends up negative )
and
mi/hr
mi/hr
His speed on the bike is 9 mi/hr
His speed walking is 3 mi/hr
check:
(1)
(1) hrs
(1) hrs
and
(2)
(2)
(2)
(2)
(2)
(2) hrs
OK
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Julio rides his bike for 6 mi and gets a flat tire. Then he has to walk with the bike for another mile. His speed walking is 6 mph less than his speed riding the bike. If the total time is 1hr, find his speed riding the bike and his speed walking?
------
Bike DATA:
distance = 6 miles ; rate = x mph ; time = 6/x hrs
-----
Walking DATA:
distance = 1 mile ; rate = (x-6) mph ; time = 1/(x-6) hrs
-----
Equation:
time + time = 1 hr
6/x + 1/(x-6) = 1
Multiply thru by x(x-6)
6(x-6) + x = x(x-6)
6x-36+x = x^2-6x
x^2-13x+36 = 0
---
Factor:
(x-4)(x-9) = 0
Usable solution:
x = 9 mph (bike speed)
x-6 = 3 mph (walk speed)
==========================
Cheers,
Stan H.
==============
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