document.write( "Question 868456: A plane traveled 300 miles to houston and back. The trip there was with the wind. It took 3 hours. The trip back was into the wind. The trip back took 5 hours. Find the speed of the plane in stil air and the speed of the wind.
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Algebra.Com's Answer #523555 by mananth(16946)![]() ![]() You can put this solution on YOUR website! 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 \n" ); document.write( " \n" ); document.write( "Distance against 300 miles distance with 300 miles \n" ); document.write( "t=d/r against wind - \n" ); document.write( "300.00 / ( x - y )= 5.00 \n" ); document.write( "5.00 ( x - y ) = 3.00 \n" ); document.write( "5.00 x - 5.00 y = 300.00 ....................1 \n" ); document.write( " \n" ); document.write( "300.00 / ( x + y )= 3.00 \n" ); document.write( "3.00 ( x + y ) = 300.00 \n" ); document.write( "3.00 x + 3.00 y = 300.00 ...............2 \n" ); document.write( "Multiply (1) by 3.00 \n" ); document.write( "Multiply (2) by 5.00 \n" ); document.write( "we get \n" ); document.write( "15.00 x + -15.00 y = 900.00 \n" ); document.write( "15.00 x + 15.00 y = 1500.00 \n" ); document.write( "30.00 x = 2400.00 \n" ); document.write( "/ 30.00 \n" ); document.write( "x = 80.00 mph \n" ); document.write( " \n" ); document.write( "plug value of x in (1) y \n" ); document.write( "5.00 x -5.00 y = 300.00 \n" ); document.write( "400.00 -5.00 -400.00 = 300.00 \n" ); document.write( "-5.00 y = 300.00 \n" ); document.write( "-5.00 y = -100.00 mph \n" ); document.write( " y = 20.00 \n" ); document.write( "Plane speed 80.00 mph \n" ); document.write( "wind speed 20.00 mph \n" ); document.write( " \n" ); document.write( "m.ananth@hotmail.ca \n" ); document.write( " \n" ); document.write( " |