document.write( "Question 1164425: Two towns are 1050 miles apart. A group of hikers start from each town and
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Algebra.Com's Answer #788827 by Edwin McCravy(20054)\"\" \"About 
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one group travels 1 1/2 mile per hour faster than the other​ group
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document.write( "Let the SECOND quantity mentioned (i.e., the \"other group\"'s rate, i.e., the\r\n" );
document.write( "slower group's rate, be the one to represent by the letter for its unknown\r\n" );
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document.write( "So let r = the slower group's rate\r\n" );
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document.write( "Then use the sentence again to define the FIRST mentioned group's rate in\r\n" );
document.write( "terms of the second one mentioned in the sentence.\r\n" );
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one group travels 1 1/2 mile per hour faster than the other​ group
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document.write( "So the first (faster) group's rate is r + 1.5\r\n" );
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document.write( "Their approach rate is the sum of their rates, which is r + r + 1.5 or\r\n" );
document.write( "2r + 1.5\r\n" );
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document.write( "Then it's just \r\n" );
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document.write( "DISTANCE =    RATE   x TIME\r\n" );
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document.write( "  1050   = (2r + 1.5)∙(200)\r\n" );
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document.write( "Solve that for the rate of the slower group, then add 1.5 mi/h to get the\r\n" );
document.write( "rate of the faster group.\r\n" );
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document.write( "Edwin
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