document.write( "Question 985097: Against the wind a commercial airline in South America flew 840 miles in 4 hours. With a tailwind the return trip took 3.5 hours. What was the speed of the plane in the air? What w \n" ); document.write( "
Algebra.Com's Answer #605945 by ikleyn(52794)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "Let  u  be the plane speed in the still air  (in miles per hour)  and let  v  be the wind speed  (also in  \"mi%2Fh\").\r
\n" ); document.write( "\n" ); document.write( "Then the speed of the plane against the wind is  u - v  (relative to the earth),  while the speed with the tailwind is  u + v.\r
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\n" ); document.write( "\n" ); document.write( "Thus you have the system of two equation for two unknowns\r
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\n" ); document.write( "\n" ); document.write( "\"system%28840+=+4%2A%28u+-+v%29%2C%0D%0A840+=+3.5%2A%28u+%2B+v%29%29\",\r
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\n" ); document.write( "\n" ); document.write( "or,  dividing the first equation by  4  and the second equation by  3.5\r
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\n" ); document.write( "\n" ); document.write( "\"system%28u+-+v+=+210%2C%0D%0Au+%2B+v+=+240%29\".\r
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\n" ); document.write( "\n" ); document.write( "Add the equations in the last system.  You will get\r
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\n" ); document.write( "\n" ); document.write( "2u = 450.\r
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\n" ); document.write( "\n" ); document.write( "Hence,  u = \"450%2F2\" = 225 \"mi%2Fh\".  It is the speed of the plane in still air.\r
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\n" ); document.write( "\n" ); document.write( "Next,  substitute this value of  u  into the second equation,  and you will get  v = 15 \"mi%2Fh\"  for the wind speed. \r
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\n" ); document.write( "\n" ); document.write( "Answer.  The plane speed is  225 \"mi%2Fh\"  in the still air.  The wind speed is  15 \"mi%2Fh\".\r
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\n" ); document.write( "\n" ); document.write( "For more wind and current problems see my lessons in this site\r
\n" ); document.write( "\n" ); document.write( "    - Wind and Current problems, \r
\n" ); document.write( "\n" ); document.write( "    - Wind and Current problems solvable by quadratic equations.\r
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