document.write( "Question 107682: Ricky drove from Town A to Town B in 3 hours. His return trip from Town B to Town A took 5 hours because he drove 15 miles per hour slower on the return trip.
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Algebra.Com's Answer #78441 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
You must begin with the firmly held conviction that nobody picked up Town A and moved it during the 8+ hours that Ricky was gone.\r
\n" ); document.write( "\n" ); document.write( "We know that \"d=rt\"\r
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\n" ); document.write( "\n" ); document.write( "For the initial trip, we only know that t = 3, so:\r
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\n" ); document.write( "\n" ); document.write( "\"d=3r\"\r
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\n" ); document.write( "\n" ); document.write( "For the return trip, we know that t = 5 and r is 15 mph slower, so:\r
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\n" ); document.write( "\n" ); document.write( "\"d=5%28r-15%29\"\r
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\n" ); document.write( "\n" ); document.write( "Since the distance going is the same as the distance returning, we can say:\r
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\n" ); document.write( "\n" ); document.write( "\"3r=5%28r-15%29\"\r
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\n" ); document.write( "\n" ); document.write( "Simplify and solve:\r
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\n" ); document.write( "\n" ); document.write( "\"3r=5r-75\"
\n" ); document.write( "\"-2r=-75\"
\n" ); document.write( "\"r=37.5\" mph\r
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\n" ); document.write( "\n" ); document.write( "Check:\r
\n" ); document.write( "\n" ); document.write( "Does \"3%2A37.5=5%2837.5-15%29\"? I'll leave the arithmetic for you.
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