document.write( "Question 802658: against the wind a commercial airline in south america flew 420 miles in 3 hours. with a tailwind the return trip took 2.5 hours. what was the speed of the plane in still air? what was the speed of the wind? \n" ); document.write( "
Algebra.Com's Answer #483965 by mananth(16946)\"\" \"About 
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Plane speed =x mph
\n" ); document.write( "wind speed =y mph
\n" ); document.write( "against wind 3 hours
\n" ); document.write( "with wind 2.5 hours
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\n" ); document.write( "Distance against 420 miles distance with 420 miles
\n" ); document.write( "t=d/r against wind -
\n" ); document.write( "420.00 / ( x - y )= 3.00
\n" ); document.write( "3.00 ( x - y ) = 2.50
\n" ); document.write( "3.00 x - 3.00 y = 420.00 ....................1
\n" ); document.write( "
\n" ); document.write( "420.00 / ( x + y )= 2.50
\n" ); document.write( "2.50 ( x + y ) = 420.00
\n" ); document.write( "2.50 x + 2.50 y = 420.00 ...............2
\n" ); document.write( "Multiply (1) by 2.50
\n" ); document.write( "Multiply (2) by 3.00
\n" ); document.write( "we get
\n" ); document.write( "7.50 x + -7.50 y = 1050.00
\n" ); document.write( "7.50 x + 7.50 y = 1260.00
\n" ); document.write( "15.00 x = 2310.00
\n" ); document.write( "/ 15.00
\n" ); document.write( "x = 154.00 mph
\n" ); document.write( "
\n" ); document.write( "plug value of x in (1) y
\n" ); document.write( "3.00 x -3.00 y = 420.00
\n" ); document.write( "462.00 -3.00 -462.00 = 420.00
\n" ); document.write( "-3.00 y = 420.00
\n" ); document.write( "-3.00 y = -42.00 mph
\n" ); document.write( " y = 14.00
\n" ); document.write( "Plane speed 154.00 mph
\n" ); document.write( "wind speed 14.00 mph
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\n" ); document.write( "m.ananth@hotmail.ca
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