document.write( "Question 662097: A plane ttaveled 580 miles to Ankara and back. The trip there was with the wind. I t took 5 hours. The trip back was into the wind the trip back took 10 hours. Find the speed of the plabe in stillair and the speed of the wind?. \n" ); document.write( "

Algebra.Com's Answer #411925 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 10 hours
\n" ); document.write( "with wind x+y 5 hours
\n" ); document.write( "
\n" ); document.write( "Distance against wind 580 miles distance with wind 580 miles
\n" ); document.write( "t=d/r against wind
\n" ); document.write( "580 / ( x - y )= 10
\n" ); document.write( "10 ( x - y ) = 580
\n" ); document.write( "10 x - 10 y = 580 ....................1
\n" ); document.write( "
\n" ); document.write( "580 / ( x + y )= 5
\n" ); document.write( "5 ( x + y ) = 580
\n" ); document.write( "5 x + 5 y = 580 ...............2
\n" ); document.write( "Multiply (1) by 1
\n" ); document.write( "Multiply (2) by 2
\n" ); document.write( "we get
\n" ); document.write( "10 x + -10 y = 580
\n" ); document.write( "10 x + 10 y = 1160
\n" ); document.write( "20 x = 1740
\n" ); document.write( "/ 20
\n" ); document.write( "x = 87 mph
\n" ); document.write( "
\n" ); document.write( "plug value of x in (1)
\n" ); document.write( "10 x -10 y = 580
\n" ); document.write( "870 -10 y = 580
\n" ); document.write( "-10 y = 580 -870
\n" ); document.write( "-10 y = -290
\n" ); document.write( " y = 29 mph
\n" ); document.write( " plane speed 87 mph
\n" ); document.write( " wind speed= 29 mph
\n" ); document.write( "m.ananth@hotmail.ca
\n" ); document.write( " \n" ); document.write( "

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