SOLUTION: Against the wind a commercial airline in South America flew 840 miles in 4 hours. With a tailwind the return trip took 3.5 hours. What was the speed of the plane in the air? What w

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Question 985097: Against the wind a commercial airline in South America flew 840 miles in 4 hours. With a tailwind the return trip took 3.5 hours. What was the speed of the plane in the air? What w
Answer by ikleyn(52790) About Me  (Show Source):
You can put this solution on YOUR website!

Let  u  be the plane speed in the still air  (in miles per hour)  and let  v  be the wind speed  (also in  mi%2Fh).
Then the speed of the plane against the wind is  u - v  (relative to the earth),  while the speed with the tailwind is  u + v.

Thus you have the system of two equation for two unknowns

system%28840+=+4%2A%28u+-+v%29%2C%0D%0A840+=+3.5%2A%28u+%2B+v%29%29,

or,  dividing the first equation by  4  and the second equation by  3.5

system%28u+-+v+=+210%2C%0D%0Au+%2B+v+=+240%29.

Add the equations in the last system.  You will get

2u = 450.

Hence,  u = 450%2F2 = 225 mi%2Fh.  It is the speed of the plane in still air.

Next,  substitute this value of  u  into the second equation,  and you will get  v = 15 mi%2Fh  for the wind speed.

Answer.  The plane speed is  225 mi%2Fh  in the still air.  The wind speed is  15 mi%2Fh.

For more wind and current problems see my lessons in this site
    - Wind and Current problems,
    - Wind and Current problems solvable by quadratic equations.