document.write( "Question 579243: With the wind, a plane flew 1780 miles in 6 hours. On the return trip, the pilot was forced to land after 2 hours, having traveled only 440 miles. Find the rate of the plane in still air and the rate of the wind. \n" ); document.write( "
Algebra.Com's Answer #370928 by mananth(16946)\"\" \"About 
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Plane speed =x mph
\n" ); document.write( "wind speed =y mph
\n" ); document.write( "against wind x-y 2 hours
\n" ); document.write( "with wind x+y 6 hours
\n" ); document.write( "
\n" ); document.write( "Distance = 1780 miles distance= 440
\n" ); document.write( "t=d/r against wind
\n" ); document.write( "440.00 / ( x - y )= 2.00
\n" ); document.write( "
\n" ); document.write( "2.00 x - -2.00 y = 440.00 ....................1
\n" ); document.write( "
\n" ); document.write( "1780.00 / ( x + y )= 6.00
\n" ); document.write( "6.00 ( x + y ) = 1780.00
\n" ); document.write( "6.00 x + 6.00 y = 1780.00 ...............2
\n" ); document.write( "Multiply (1) by 3.00
\n" ); document.write( "Multiply (2) by 1.00
\n" ); document.write( "we get
\n" ); document.write( "6.00 x + -6.00 y = 1320.00
\n" ); document.write( "6.00 x + 6.00 y = 1780.00
\n" ); document.write( "12.00 x = 3100.00
\n" ); document.write( "/ 12.00
\n" ); document.write( "x = 258.33 mph
\n" ); document.write( "
\n" ); document.write( "plug value of x in (1)
\n" ); document.write( "2 x -2 y = 440
\n" ); document.write( "516.67 -2 y = 440
\n" ); document.write( "-2 y = 440 -516.67
\n" ); document.write( "-2 y = -76.67
\n" ); document.write( " y = 38.33 mph \r
\n" ); document.write( "\n" ); document.write( "Plane speed = 258.33 mph
\n" ); document.write( "wind speed = 38.33 mph\r
\n" ); document.write( "\n" ); document.write( "m.ananth@hotmail.ca
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