Use the equation d = rt, once for each trip. I will use capitals for the variables and lowercase for subscripts.
D = R * T
Dr = Rr * Tr
where "r" is a subscript used to indicate that values for the return tripe.
We are given the time for each trip: 6 hours and then 8 hours for the return trip.
D = R * 6
Dr = Rr * 8
And the rate for both trips is based on x -- the speed in still air. It is x + 22 in one direction and x - 22 in the other.
D = (x + 22) * 6
Dr = (x - 22) * 8
Finally, we know that the distance is the same for both trips.
D = Dr
Therefore, through transitibity, we know that the right sides of both of equations are equal.
(x + 22) * 6 = (x - 22) * 8
or
6(x + 22) = 8(x - 22)
Solve that and you will get 154.
Use systems of lineear equations to solve:
An airplane flew for 6 hours with a 22-km/h tail wind. The return flight against the same wind took 8 hours. Find the speed of the plane in still air.
I know the answer is 154 km/hr (back of book resources :) )...I just don't understand how you would get to that answer.
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