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The airplane effective speed, when flying into a head wind, is = 350 mph.
This speed is the difference u-v of the airplane speed in still air and the wind speed: u - v = 350 mph.
The airplane effective speed, when flying with the wind, is = 450 mph.
This speed is the sum u+v of the airplane speed in still air and the wind speed: u + v = 450 mph.
To determine the airplane speed in still air "u" and the wind speed "v", you need to solve this system of two equations:
u - v = 350, (1)
u + v = 450. (2)
Add the equations (1) and (2). You will get
2u = 350 + 450
2u = 800,
u = = 400 miles per hour. It is the airplane speed in still air.
Then from (1) you find v = u - 400 = 450 - 400 = 50 miles per hour. It is the wind speed.
Solved.
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It is a typical "tailwind and headwind" word problem.
See the lessons
- Wind and Current problems
- Wind and Current problems solvable by quadratic equations
- Selected problems from the archive on a plane flying with and against the wind
in this site, where you will find other similar solved problems with detailed explanations.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this textbook under the section "Word problems", the topic "Travel and Distance problems".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.