document.write( "Question 215307: A family drove 1080 miles to their vacation lodge. Bercause of increased traffic denisty, their average speed on the return trip was decreased by 6 miles per hour and the trip took 2 1/2 hours longer. Determine their average speed on the way to the lodge. \n" ); document.write( "

Algebra.Com's Answer #162685 by ptaylor(2198) $\"\"$ $\"About$

You can put this solution on YOUR website!
Distance(d) equals Rate(r) times Time(t) or d=rt;r=d/t and t=d/r
\n" ); document.write( "Let r=average speed to their vacation lodge
\n" ); document.write( "time required to get to their vacation lodge=1080/r
\n" ); document.write( "r-6=average speed on the return trip
\n" ); document.write( "time required for their return trip=1080/(r-6) and we are told that this time is 2.5 hours longer then the time to get to their vacation lodge, so:\r
\n" ); document.write( "\n" ); document.write( "(1080/r) + 2.5=1080/(r-6) multiply each term by r(r-6)
\n" ); document.write( "1080(r-6)+2.5r(r-6)=1080r get rid of parens
\n" ); document.write( "1080r-6480+2.5r^2-15r=1080r subtract 1080r from each side
\n" ); document.write( "1080r-1080r-6480+2.5r^2-15r=1080r-1080r collect like terms
\n" ); document.write( "2.5r^2-15r-6480=0 divide each term by 2.5
\n" ); document.write( "r^2-6r-2592=0 Quadratic in standard form and it can be factored
\n" ); document.write( "(r-54)(r+48)=0
\n" ); document.write( "r=54 mph----avarage speed to their vacation lodge
\n" ); document.write( "and
\n" ); document.write( "r=-48 mph---DISREGARD! In this case speeds are positive\r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "time to their vacation lodge=1080/54=20 hours\r
\n" ); document.write( "\n" ); document.write( "Time for return trip=1080/48=22.5 hours
\n" ); document.write( "22.5-20=2.5
\n" ); document.write( "2.5=2.5\r
\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor \n" ); document.write( "

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