SOLUTION: justin travels 160 miles away and then back home. his average speed on the way there is 9 mph faster than on his way home. he spends a total of 8 hours driving. find his two rate
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: justin travels 160 miles away and then back home. his average speed on the way there is 9 mph faster than on his way home. he spends a total of 8 hours driving. find his two rate
Log On
Question 981472: justin travels 160 miles away and then back home. his average speed on the way there is 9 mph faster than on his way home. he spends a total of 8 hours driving. find his two rates of speed. Found 2 solutions by josmiceli, MathTherapy:Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website! Let = his speed going back home in mi/hr = his speed going away from home in mi/hr
Let = his time going back home in hrs = his time in hrs going away from home
-----------------------------------------------
Equation for going away from home:
(1)
Equation for going back home:
(2)
-------------------
(1)
and
(2)
--------------------
Substitute (2) into (1)
(1)
Multiply both sides by
(1)
(1)
Use quadratic equation
You can finish- find , then
check my math -I'm not sure of calculations
I think the method is OK
You can put this solution on YOUR website!
justin travels 160 miles away and then back home. his average speed on the way there is 9 mph faster than on his way home. he spends a total of 8 hours driving. find his two rates of speed.
Let speed going be S
Then time taken to get to destination is:
Speed on return trip = S - 9
Time taken on return trip =
Since time taken for entire trip is 8 hours, we get:
160(S - 9) + 160S = 8S(S - 9) ------- Multiplying by LCD, S(S - 9)
(S - 45)(S - 4) = 0
S, or speed to destination = mph
Speed on return trip: 45 - 9, or mph
