document.write( "Question 1124612: Stacy rides her bike then the bus and finally the train to visit her uncle. She rides her bike at 14 mph. The bus travels at 42 mph while the train carries her at 105 mph. She rides the train for 1 hour 6 minutes longer than she rides the bus. She traveled 371 miles in 6 hours 36 minutes. How long was she riding her bike and reading on the bus?
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Algebra.Com's Answer #740938 by josgarithmetic(39617)\"\" \"About 
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document.write( "               SPEED      TIME         DISTANCE\r\n" );
document.write( "BIKE             14\r\n" );
document.write( "BUS              42           y\r\n" );
document.write( "TRAIN           105        y+1.1\r\n" );
document.write( "Total Travel               6.6            371\r\n" );
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\n" ); document.write( "\n" ); document.write( "The time on bike can be found from the total time and the other two time expressions. Bike time, \"5.5-2y\".\r
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\n" ); document.write( "\n" ); document.write( "Missing distance expressions can be filled-in:
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document.write( "               SPEED      TIME         DISTANCE\r\n" );
document.write( "BIKE             14        5.5-2y     14(5.5-2y)\r\n" );
document.write( "BUS              42           y       42y\r\n" );
document.write( "TRAIN           105        y+1.1      105(y+1.1)\r\n" );
document.write( "Total Travel               6.6            371\r\n" );
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\n" ); document.write( "\n" ); document.write( "Solve this equation in the one variable, y, the time while riding on the bus:\r
\n" ); document.write( "\n" ); document.write( "\"14%285.5-2y%29%2B42y%2B105%28y%2B1.1%29=371\" \n" ); document.write( "
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