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The effective speed of the plane flying with the wind is = 530 miles per hour.
From the other side, it is the sum of the plane speed at no wind PLUS the speed of the wind.
It gives you your first equation
u + v = 530. (1)
The effective speed of the plane flying against the wind is = 450 miles per hour.
From the other side, it is the difference of the plane speed at no wind and the speed of the wind.
It gives you your second equation
u - v = 450. (2)
So you have the system of two equations in two unknowns (1) and (2).
To solve the system, add the two equations. You will get
2u = 530 + 450 = 980
u = 980/2 = 490.
Then from eq(2), v = 530 - 490 = 40.
Answer. The speed of the plane at no wind is 490 miles per hour.
The speed of wind is 40 miles per hour.
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.