document.write( "Question 339383: At 6:00 a.m. a man left a town A and travelled toward a town B 19 miles away. At 6:30 another man started from B and travelled toward A. The two men met on the road at 8:30 a.m. If the rate of the first traveller is 1/2 mile per hour less than that of the second,find the rate of each.\r
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Algebra.Com's Answer #243215 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
Write an equation for both men:
\n" ); document.write( "\"d%5BA%5D+=+r%5BA%5D%2At%5BA%5D\"
\n" ); document.write( "and
\n" ); document.write( "\"d%5BB%5D+=+r%5BB%5D%2At%5BB%5D\"
\n" ); document.write( "given:
\n" ); document.write( "\"d%5BA%5D+%2B+d%5BB%5D+=+19\"
\n" ); document.write( "\"t%5BA%5D+=+2.5\"
\n" ); document.write( "\"t%5BB%5D+=+2\"
\n" ); document.write( "\"r%5BA%5D+=+r%5BB%5D+-+.5\"
\n" ); document.write( "---------------
\n" ); document.write( "I can rewrite the equations of motion
\n" ); document.write( "\"d%5BA%5D+=+r%5BA%5D%2A2.5\"
\n" ); document.write( "and
\n" ); document.write( "\"19+-+d%5BA%5D+=+%28r%5BA%5D+%2B+.5%29%2A2\"
\n" ); document.write( "\"-d%5BA%5D+=+-19+%2B+2r%5BA%5D+%2B+1\"
\n" ); document.write( "\"d%5BA%5D+=+18+-+2r%5BA%5D\"
\n" ); document.write( "and, since both are equal to \"d%5BA%5D\"
\n" ); document.write( "\"2.5r%5BA%5D+=+18+-+2r%5BA%5D\"
\n" ); document.write( "\"4.5r%5BA%5D+=+18\"
\n" ); document.write( "\"r%5BA%5D+=+4\"
\n" ); document.write( "\"r%5BB%5D+=+r%5BA%5D+%2B+.5\"
\n" ); document.write( "\"r%5BB%5D+=+4.5\"
\n" ); document.write( "The 1st man's rate is 4 mi/hr
\n" ); document.write( "The 2nd man's rate is 4.5 mi/hr
\n" ); document.write( "check:
\n" ); document.write( "\"d%5BA%5D+=+r%5BA%5D%2A2.5\"
\n" ); document.write( "\"d%5BA%5D+=+4%2A2.5\"
\n" ); document.write( "\"d%5BA%5D+=+10\" mi
\n" ); document.write( "and
\n" ); document.write( "\"d%5BB%5D+=+r%5BB%5D%2A2\"
\n" ); document.write( "\"d%5BB%5D+=+4.5%2A2\"
\n" ); document.write( "\"d%5BB%5D+=+9\"
\n" ); document.write( "\"d%5BA%5D+%2B+d%5BB%5D+=+19\"
\n" ); document.write( "OK\r
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