SOLUTION: Hello, The following problem was answered earlier, but I don't understand how you cam up with 56 and 49 in the equation portion of the problem to be divided by 525? Can you exp

Question 95356: Hello,
The following problem was answered earlier, but I don't understand how you cam up with 56 and 49 in the equation portion of the problem to be divided by 525? Can you explain.

I need help with this problem:
An airplane can fly a rate of 525 mi/h in still air. It can fly 2800 miles with the wind in the same time that it can fly 2450 miles against the wind. Find the rate if the wind.
Thank you!: Hello,
I need help with this problem:
An airplane can fly a rate of 525 mi/h in still air. It can fly 2800 miles with the wind in the same time that it can fly 2450 miles against the wind. Find the rate if the wind.
Thank you!
Answer by stanbon(9974) (Show Source):
You can put this solution on YOUR website!
An airplane can fly a rate of 525 mi/h in still air. It can fly 2800 miles with the wind in the same time that it can fly 2450 miles against the wind. Find the rate of the wind.
--------------
Let the rate of the wind be "w".
---------------
With the wind DATA:
Distance=2800 miles ; Rate = (525+w) mph ; Time = d/r = 2800/(525+w) hrs.
-----------------------------
Against the wind DATA:
Distance=2450 miles : Rate = (525-w) mph ; Time = d/r = 2450/(525-w) hrs.
------------------------------
EQUATION:
time = time
2800/(525+w) = 2450/(525-w)
56/(525+w) = 49/(525-w)
56*525-56w = 49*525+49w
105w = 7*525
w = 35 mph (the speed of the wind is 35 mph)
================

Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
the first line of the EQUATION section is divided by 50 to get the second line

the second line is cross multiplied to get the third line
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