.
Let w be the wind velocity (in miles per hour).
Then the plane' speed with the wind is (175 + w) miles per hour,
and the time to fly 1000 miles with the wind is hours.
Also, the plane' speed against the wind is (175 - w) miles per hour,
and the time to fly 1000 miles against the wind is hours.
Now you can write the "time" equation
- = 1 hour
saying that "it takes 1 hour longer to complete the return trip".
It is your basic equation. To solve it, multiply both sides by (175+w)*(175-w). You will get
1000(175+w) - 1000(175-w) = 175^2 - w^2, or
2*1000*w = 175^2 - w^2
w^2 + 2000w - 175^2 = 0
= = .
Only positive value of the roots is meaningful: w = = 15.197.
ANSWER. The wind' speed is 15.197 miles per hour.
CHECK. = 1 hour. ! Correct !
Solved, answered, checked and completed.
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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.