document.write( "Question 204385: A Jet can travel at a rate or 525 miles per hour in calm air. Traveling with the wind, the plane flew 1815 miles in the same time that it flew 1335 miles against the wind. Find the rate of the wind. \n" ); document.write( "
Algebra.Com's Answer #154324 by Earlsdon(6294)\"\" \"About 
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Using \"d+=+rt\" where d = distance traveled, r = rate/speed, and t = time of travel.
\n" ); document.write( "For the outbound trip: \"d%5B1%5D+=+1815\" miles, \"r%5B1%5D+=+%28525%2Bw%29\"mph.
\n" ); document.write( "For the return trip: \"d%5B2%5D+=+1335\" miles, \"r%5B2%5D+=+525-w%29\"mph.
\n" ); document.write( "So we can set up the two equations: (w = wind speed)
\n" ); document.write( "1) \"1815+=+%28525%2Bw%29%2At\"
\n" ); document.write( "2) \"1335+=+%28525-w%29%2At\" The time, t, is the same for both trips. Solve both equations fo t and set them equal to each other.
\n" ); document.write( "1) \"t+=+1815%2F%28525%2Bw%29\"
\n" ); document.write( "2) \"t+=+1335%2F%28525-w%29\" so...
\n" ); document.write( "\"1815%2F%28525%2Bw%29+=+1335%2F%28525-w%29\" Now we solve for w, the speed of the wind. Cross-multiply.
\n" ); document.write( "\"1815%28525-w%29+=+1335%28525%2Bw%29\"
\n" ); document.write( "\"952875-1815w+=+700875%2B1335w\" Subtract 700875 from both sides.
\n" ); document.write( "\"252000-1815w+=+1335w\" Add 1815w to both sides.
\n" ); document.write( "\"252000+=+3150w\" Finally, divide both sides by 3150.
\n" ); document.write( "\"highlight%28w+=+80%29\"mph.
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