Question 4969
let the rate of airplane on the trip going = x+25.5km
Let the rate returning =x-25.5km 
the distance is the same both ways,
using rate*time=distance, we have two equations:
(x+25.5)*4.23=d
(x-25.5)*4.97=d
Set them equal to one another,
(x+25.5)*4.23= (x-25.5)*4.97

4.23x+107.865=4.97x-126.735
subtract 4.23x and add 126.735 to both sides:

234.6=.74x
divide by .74

317.027=x 
the airplane traveled 317.02kmh in still air.