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

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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




Answer by MathTherapy(10557)   (Show Source): 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 

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