document.write( "Question 887109: susan left her home at 7am driving her car at the rate of 45 mph. at 9am her sister marion drove after her along the same highway, travelling at the rate of 60 mph. in how many hours did marion passed susan? \n" ); document.write( "
Algebra.Com's Answer #536359 by Edwin McCravy(20059)\"\" \"About 
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document.write( "Two ways to work it:\r\n" );
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document.write( "First way:\r\n" );
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document.write( "At 9am Susan has been driving 2 hours since 7AM at the rate of 45 mph so she is \r\n" );
document.write( "90 miles from Marion when Marion starts.  Marion's catch-up rate is 60-45=15\r\n" );
document.write( "mph. Since TIME = DISTANCE/RATE, it will take Marion 90/15 = 6 hours to catch up\r\n" );
document.write( "and pass Susan.\r\n" );
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document.write( "Second way:\r\n" );
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document.write( "Make a chart.  Fill in the rates, and x for Marion's time.  Susan's time\r\n" );
document.write( "traveled when Marion catches up is 2 hours longer or x+2\r\n" );
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document.write( "          Distance = Rate · Time \r\n" );
document.write( "Susan                 45     x+2  \r\n" );
document.write( "Marion                60      x\r\n" );
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document.write( "Now fill in the distances by multiplying the rates times the times.\r\n" );
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document.write( "          Distance = Rate · Time \r\n" );
document.write( "Susan     45(x+2)     45     x+2  \r\n" );
document.write( "Marion       60x      60      x      \r\n" );
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document.write( "Their distances are the same, so set them equal:\r\n" );
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document.write( "     45(x+2) = 60x\r\n" );
document.write( "      45x+90 = 60x\r\n" );
document.write( "          90 = 15x\r\n" );
document.write( "           6 = x\r\n" );
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document.write( "Edwin
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