document.write( "Question 931150: An airplane can fly with the wind a distance of 900 miles in 3 hours. However, the return trip against the wind takes 5 hours. Find the speed of the plane in still air and the speed of the wind. \r
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Algebra.Com's Answer #565453 by mananth(16946)\"\" \"About 
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plane speed =x mph
\n" ); document.write( "wind speed =y mph
\n" ); document.write( "against wind 5 hours
\n" ); document.write( "with wind 3 hours
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\n" ); document.write( "Distance with wind 900 miles distance against wind 900
\n" ); document.write( "t=d/r against wind (x-y)
\n" ); document.write( "900.00 / ( x - y )= 5.00
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\n" ); document.write( "5.00 x - -5.00 y = 900.00 ....................1
\n" ); document.write( "with wind (x+y)
\n" ); document.write( "900.00 / ( x + y )= 3.00
\n" ); document.write( "3.00 ( x + y ) = 900.00
\n" ); document.write( "3.00 x + 3.00 y = 900.00 ...............2
\n" ); document.write( "Multiply (1) by 3.00
\n" ); document.write( "Multiply (2) by 5.00
\n" ); document.write( "we get 2.00
\n" ); document.write( "15.00 x + -15.00 y = 2700.00
\n" ); document.write( "15.00 x + 15.00 y = 4500.00
\n" ); document.write( "30.00 x = 7200.00
\n" ); document.write( "/ 30.00
\n" ); document.write( "x = 240.00 mph
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\n" ); document.write( "plug value of x in (1) y
\n" ); document.write( "5.00 x -5.00 y = 900.00
\n" ); document.write( "1200.00 -5.00 -1200.00 = 900.00
\n" ); document.write( "-5.00 y = 900.00
\n" ); document.write( "-5.00 y = -300.00 mph
\n" ); document.write( " y = 60.00
\n" ); document.write( "plane 240.00 mph
\n" ); document.write( "wind 60.00 mph
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