document.write( "Question 834327: Tom and Liz used to live 23.5 miles apart. They would meet by Liz riding her bicycle 10 mph towards Tom while he rides his moped towards Liz at 18 mph. If Liz leaves 15 minutes before Tom, how long is she on her bicycle before she meets him?\r
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document.write( "Step by step please. \n" );
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Algebra.Com's Answer #502957 by ankor@dixie-net.com(22740) You can put this solution on YOUR website! Tom and Liz used to live 23.5 miles apart. \n" ); document.write( " They would meet by Liz riding her bicycle 10 mph towards Tom while he rides his moped towards Liz at 18 mph. \n" ); document.write( " If Liz leaves 15 minutes before Tom, how long is she on her bicycle before she meets him? \n" ); document.write( ": \n" ); document.write( "Change 15 min to .25 hrs (we are dealing mph) \n" ); document.write( ": \n" ); document.write( "Let t = Liz's traveling time \n" ); document.write( "then \n" ); document.write( "(t-.25) = Tom's traveling time (left 15 min later than Liz) \n" ); document.write( ": \n" ); document.write( "Write a Distance equation: Dist = speed * time \n" ); document.write( ": \n" ); document.write( "Liz's dist + Tom's dist = 23.5 mi \n" ); document.write( "10t + 18(t-.25) = 23.5 \n" ); document.write( "10t + 18t - 4.5 = 23.5 \n" ); document.write( "28t = 23.5 + 4.5 \n" ); document.write( "28t = 28 \n" ); document.write( "t = 28/28 \n" ); document.write( "t = 1 hr is Liz's traveling time \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check this by finding the dist each traveled \n" ); document.write( "10*1 = 10 mi \n" ); document.write( "18(.75) = 13.5 mi \n" ); document.write( "---------------- \n" ); document.write( "Total dist 23.5 \n" ); document.write( " |