Solving distance and time using system of linear eqations

Question

An airplane takes 6 hours to travel a distance of 4140 k7lometers against the wind. The return trip takes 5 hours with the wind whst is the rate of the plane in still air and what is the rate of the wind?

Answer 1

Let r = rate in still air and let w = speed of the wind

Rate with the wind = r + w and rate against the wind = r - w

5(r + w) = 4140

6(r - w) = 4140

r + w = 828

r - w = 690

Add the equations and then solve for r and w.

Answer 2

d=rt

distance == rate of speed times time

d/r= t

d=4140 km, r=p-w when t=6 and r=p+2 when t=5

p = plane speed in still air, w = wind speed

p-w = plane speed against the wind

p+w = plane speed with the wind

4140/(p-w) = 6

4140/(p+w) = 5

p+w = 4140/5= 828

p-w = 4140/6 = 690

2p = 828+690= 1518

p = 1518/2 = 759 km/hr= plane speed in still air

w = 828-759= 69 km/hr = wind speed

6(759-69) =6(690) = 4140

5(759+69) =5(828)= 4140