document.write( "Question 478622: 1. A plane travels from Orlando to Denver and back again. On the five-hour trip from Orlando to Denver, the plane has a tailwind of 40 miles per hour. On the return trip from Denver to Orlando, the plane faces a headwind of 40 miles per hour. This trip takes six hours. What is the speed of the airplane in still air? \r
\n" ); document.write( "\n" ); document.write( "Show full equation please, I'd like to fully understand it.
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Algebra.Com's Answer #327938 by ankor@dixie-net.com(22740)\"\" \"About 
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A plane travels from Orlando to Denver and back again.
\n" ); document.write( "On the five-hour trip from Orlando to Denver, the plane has a tailwind of 40 miles per hour.
\n" ); document.write( " On the return trip from Denver to Orlando, the plane faces a headwind of 40 miles per hour.
\n" ); document.write( " This trip takes six hours. What is the speed of the airplane in still air?
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\n" ); document.write( "Let s = plane speed in still air
\n" ); document.write( "then
\n" ); document.write( "(s+40) = effective speed with the wind
\n" ); document.write( "and
\n" ); document.write( "(s-40) = effective speed against
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\n" ); document.write( "The two trips are the same distance, write a distance equation Dist = speed * time
\n" ); document.write( "6(s-40) = 5(s+40)
\n" ); document.write( "6s - 240 = 5s + 200
\n" ); document.write( "6s - 5s = 200 + 240
\n" ); document.write( "s = 440 mph in still air
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\n" ); document.write( "Check this by finding the actual distances, they should be the same
\n" ); document.write( "6*400 = 2400 mi
\n" ); document.write( "5*480 = 2400 mi
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