document.write( "Question 799008: Flying against the wind, an airplane travels 5100 km in 6 hours. Flying with the wind, the same plane travels 5750 km in 5 hours. What is the rate of the plane in still air and what is the rate of the wind? \n" ); document.write( "
Algebra.Com's Answer #482451 by Finavon(81)![]() ![]() ![]() You can put this solution on YOUR website! Wind speed w; plane airspeed p \n" ); document.write( "Dist=speed * time \n" ); document.write( "with wind: (p+w)*5 = 5750 (1) \n" ); document.write( "against: (p-w)*6 = 5100 (2) \n" ); document.write( "(1) 5p+5w=5750 \n" ); document.write( "(2)+650: 650+6p-6w = 5100+650 = 5750 = (1) \n" ); document.write( "so 5p+5w=650+6p-6w \n" ); document.write( "11w=650+p or p=(11w-650) (3) \n" ); document.write( "subst for p in (1): 5*(11w-650)+5w = 5750 \n" ); document.write( "55w+5w-3250 = 5750 or 60w = 9000 \n" ); document.write( "w=150 \n" ); document.write( "subst in (3): p=11*150 - 650 = 1650-650 = 1000 \n" ); document.write( "Plane 1000 km/hr; Wind 150 km/hr\r \n" ); document.write( "\n" ); document.write( "Check: \n" ); document.write( "with: 1150*5= 5750 \n" ); document.write( "against: 850*6= 5100 \n" ); document.write( " |