Solving a distance, rate, time problem using a system of linear equations

Question

Flying against the jetstream, a jet travels 4550 miles in 7 hours. Flying with the jetstream, the same jet travels 3150 miles in 3 hours. What is the rate of the jet in still air and what is the rate of the jetstream? 

Answer 1

Let

V = speed of jet in still air

v = speed of jet stream

When flying against the jet stream, the jet's net speed is V-v. After 7 hours, it travels 4550 miles:

Dist = Rate * Time

4550 = (V-v)⋅ 7, which when simplified, is

V - v = 4550/7 = 650 mph *

When flying with the jet stream, the jet's net speed is V+v. After 3 hours, it travels 3150 miles:

Dist = Rate * Time

3150 = (V+v)⋅ 3, which when simplified, is

V + v = 3150/3 = 1050 mph **

Adding the two equations * and ** leads to:

2V = 1700, or simply V = 850 mph

Subtracting the two equations * and ** leads to:

2v = 400, or simply v = 200 mph

Answer 2

d=speed times time traveled 4550 = (x-j)7 x=speed in still air j=jet stream speed

3150 = (x+j)3

4550 = 7x - 7j multiply by 3

3150 =3x +3j multiply by 7 to get

13650=21x -21j

22050=21x +21j add them

35700=42x

x=35700/42 =850 mph = speed in still air

3150 = 3(850)+3j

3150 = 2550+3j

3j = 3150-2550

= 600

j=600/3=200 mph = speed of jet stream