document.write( "Question 1132807: A salesman drives from Ajax to Barrington, a distance of 147 mi, at a steady speed. He then increases his speed by 15 mi/h to drive the 190 mi from Barrington to Collins. If the second leg of his trip took 5 min more time than the first leg, how fast was he driving between Ajax and Barrington? \n" ); document.write( "
Algebra.Com's Answer #749945 by josgarithmetic(39618)\"\" \"About 
You can put this solution on YOUR website!
r, speed of first leg of trip\r
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\n" ); document.write( "\n" ); document.write( "\"190%2F%28r%2B15%29-147%2Fr=1%2F12\"\r
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\n" ); document.write( "\n" ); document.write( "\"highlight%28r=60%29\"\r
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document.write( "                          SPEED       TIME        DISTANCE\r\n" );
document.write( "Ajax to Barrington        r           147/r        147\r\n" );
document.write( "Barrington to Collins    r+15         190/(r+15)   190\r\n" );
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document.write( "Difference                          5/60=1/12\r\n" );
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\r
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\n" ); document.write( "\"r%5E2%2B%2815-12%2A190%2B12%2A147%29r%2B15%2A12%2A147=0\"
\n" ); document.write( "\"r%5E2-501r%2B26460=0\"\r
\n" ); document.write( "\n" ); document.write( "\"discriminant\"\"501%5E2-4%2A26460=145161=381%5E2\"\r
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\n" ); document.write( "\n" ); document.write( "\"r=%28501%2B-+381%29%2F2\"
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\n" ); document.write( "\"Lesser\"\"value\"\"more\"\"reasonable\"\"car\"\"speed\"
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