document.write( "Question 903162: Billy leaves home walking to school at 1.7 m/s. His sister Sally is running a bit late and leaves home 8.5 minutes late going 4.2 m/s on her bike. How long will it take for Sally to catch up to billy? How far away from home will they be when she catches him? \n" ); document.write( "
Algebra.Com's Answer #547832 by lwsshak3(11628)\"\" \"About 
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Billy leaves home walking to school at 1.7 m/s. His sister Sally is running a bit late and leaves home 8.5 minutes late going 4.2 m/s on her bike. How long will it take for Sally to catch up to billy? How far away from home will they be when she catches him?
\n" ); document.write( "***
\n" ); document.write( "1.7 m/s=60*1.7=102 m/min (Billy's speed)
\n" ); document.write( "4.2 m/s=60*4.2=252 m/min (Sally's speed)
\n" ); document.write( "..
\n" ); document.write( "let x=Sally's travel time
\n" ); document.write( "x+8.5=Billy's travel time
\n" ); document.write( "distance=speed*travel time(same for both Sally ans Billy)
\n" ); document.write( "..
\n" ); document.write( "252x=102(x+8.5)
\n" ); document.write( "252x=102x+867
\n" ); document.write( "150x=867
\n" ); document.write( "x=5.78 min
\n" ); document.write( "252x=1456.56 m
\n" ); document.write( "How long will it take for Sally to catch up to billy? 5.78 min
\n" ); document.write( "How far away from home will they be when she catches him? 1456.56 m \n" ); document.write( "
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