Algebra?? Please help me?? I don't understand??

Question

An airplane flew for 4 hours against a head wind of 40km/h. On the return flight the same wind was now a tail wind, and the f;ight took 3 hours.Find the speed of the airplane in still air?

Answer 1

Start with what you know.

Distance = rate(time) or d=rt

The plane takes the first trip in four hours with a headwind. This slows the plane down. The first trip can be represented by:

d=(r-40)4

The plane takes the return trip in three hours with a tailwind. This makes the plane go faster. This trip can be represented by:

d=(r+40)3

Since it's an outbound and return flight, these distances are equal:

(r-40)4=(r+40)3
4r-160=3r+120
r=120+160=280km/hr

Answer 2

Let speed in still air be S mph

Out Journey
Speed = S - 40 km/h
Time = 4 hours
Distance = 4(S -40) km

Return Journey
Speed = S + 40
Time = 3hours
Distance = 3(S + 400)

Thus 4(S - 40) = 3(S + 40)
4S - 160 = 3S + 120
S = 280 km /h