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T= 5300÷x = 5300/x hrs
b) Since the aircraft flew 20km/h slower, its average speed is now (x-20)km/h
As such the time taken is now 5300/(x-20)
c) Since return trip is slower:
5300/(x-20) -1/5 = 5300/x
26500/(x-20) - 1 = 26500/x
1= 26500/(x-20) - 26500/x
x²-20x=26500x - 26500x + 530000