document.write( "Question 418530: a fast plane making a 2400 mile trip was delayed 14 minutes before take off because of bad weather. To make up for the delay, the pilot increased his airspeed and took a favorable tailwind. Thus he was able to average 6 miles per hour more than usual. Find the usual average ground speed and flying time for the trip.\r
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document.write( "I know the first equation is (x+6)*(y+(14/60))=2400 but i can't figure out the second equation to do a system of equations.
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document.write( "(x=average airspeed y=flying time) \n" );
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Algebra.Com's Answer #292896 by mananth(16946)![]() ![]() You can put this solution on YOUR website! average speed =x \n" ); document.write( "time = y\r \n" ); document.write( "\n" ); document.write( ".. \n" ); document.write( "xy=2400 \n" ); document.write( "x=2400/y \n" ); document.write( "(x+6)(y-14/60)= 2400 \n" ); document.write( "(x+6)(60y-14)/60=2400 \n" ); document.write( "(x+6)(60y-14)=60*2400 \n" ); document.write( "plug x \n" ); document.write( "(2400/y +6)(60y-14)=60*2400 \n" ); document.write( "(2400+6y)(60y-14)/y=144000 \n" ); document.write( "(2400+6y)(60y-14)=144000y \n" ); document.write( "6(400+y)*2(30y-7)=144000y \n" ); document.write( "(400+y)(30y-7)=144000y/12 \n" ); document.write( "12000y-2800+30y^2-7y=12000y \n" ); document.write( "30y^2-7y-2800=0\r \n" ); document.write( "\n" ); document.write( "solve using quadratic formula \n" ); document.write( "b^2-4ac=336049 \n" ); document.write( "... \n" ); document.write( "y=9.78 hours \n" ); document.write( "x=2400/9.78 = 245.4 mph \n" ); document.write( " |