.
Let x = the jet rate at no wind (in mph), and
y = the rate of the wind.
Then the effective rate of the jet with the wind is x + y mph,
while the effective rate of the jet with the wind is x - y mph.
The condition says
= x - y (the effective speed against the wind)
= x + y (the effective speed with the wind)
After calculations, it takes the form
x - y = 762 (1)
x + y = 882 (2)
---------------------- Add the equations
2x = 762 + 882 = 1644 ====> x = = 822 mph. It is the rate of the jet at no wind.
Then from eq(2), y = 882 - 822 = 60 mph is the rate 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.
Then solve your problem by substituting your data.
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".
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.