.
Let u be the airspeed and v be the speed of the wind.
Then the ground-speed flying east (with the wind) is (u+v),
while ground-speed flying west (against the wind) is (u-v).
The condition gives you these two equations:
= u - v (1) (flying west against the wind) and
= u + v (2) (flying east against the wind).
It is the same as
u - v = 200, (3)
u + v = 300. (4)
Add equations (3) and (4) (both sides). You will get
2u = 200 + 300 = 500. Hence, u = = 250 miles per hour is the airspeed.
Then from (4) u = 300 - 250 = 50 miles per hour is the speed of the wind.
Solved.
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.
In these lessons you will find the detailed solutions of many similar problems.
Consider them as samples. Read them attentively.
In this way you will learn how to solve similar problems once and for all.
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".