Question 558130
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A passenger plane flew to moscow and back. The trip there took five hours and the trip back took nine hours.  It averaged 144 km/h faster on the trip there than on the return trip.  Find the passenger plane's average speed on the outbound trip.

O.k so I did have a go at it Trying to use a system of equations using the substitution method to eliminate one unknown but it didn't work.  I will try and explain what I was playing around with:

D1 = Distance there  D2 = Distance back ( but not sure if I needed this as it is same Distance )
Average Speed = S  

D1/5 + D2/9 = S

D1/5 = D2/9+144 ( I tried substituting this in )

D2/9+144 + D2/9 = S ( then I combined the like terms )

D2/18+144 = s ( but this is where I realized I had failed as I still had 2 variables.

Please help me someone I am so bad at word problems and I would love to be confident at them. 

P.S I wasn't sure about which was the outbound trip, I guessed it meant the trip back.


No!! The OUTBOUND trip is the one to Moscow, while the one back from Moscow is the INBOUND/RETURN trip.

Let OUTBOUND speed, be S
Then INBOUND/RETURN speed = S - 144
It took 5 hours to get there, so distance traveled to Moscow = 5S
It took 9 hours to get back, so distance on return trip = 9(S - 144)
Since distance there and back is the same, we get 5S = 9(S - 144)
                                                  5S = 9S - 9(144)
                                             4S - 9S = - 9(144)
                                                - 4S = - 9(144)
                Average speed on the OUTBOUND, or {{{highlight_green(matrix(1,9, S, "=", (- 9(144))/(- 4), "=", 9(36cross(- 144))/cross((- 4)), "=", 9(36), "=", highlight(matrix(1,2, 324, "km/h"))))}}}</pre>