document.write( "Question 330782: At 8:00am Dave leaves his home on his moped, traveling east at 18 mph. At the same time and 80 mile away, Cathy leaves her home on her bicycle, traveling west toward Dave at 14 mph. At what time will they meet? \n" ); document.write( "
Algebra.Com's Answer #237134 by texttutoring(324)\"\" \"About 
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You know that D=VT, where D=distance, V=velocity (speed) and T = time.\r
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\n" ); document.write( "\n" ); document.write( "Rearranging for time, we have D/V = T\r
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\n" ); document.write( "\n" ); document.write( "We know that their times are the same, so Cathy's time = Dave's time\r
\n" ); document.write( "\n" ); document.write( "Tc = Td
\n" ); document.write( "Dc/Vc = Dd/Vd\r
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\n" ); document.write( "\n" ); document.write( "Where Dc=Cathy's distance when they meet, Vc = Cathy's velocity, and so on...\r
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\n" ); document.write( "\n" ); document.write( "We know that Dc+Dd = 80, (or Dc =80-Dd) because they start 80 miles apart and are heading towards each other.\r
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\n" ); document.write( "\n" ); document.write( "Put in what you know:\r
\n" ); document.write( "\n" ); document.write( "Dc/14 = Dd/18
\n" ); document.write( "80-Dd/14 = Dd/18\r
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\n" ); document.write( "\n" ); document.write( "Solve for Dd by cross multiplying:\r
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\n" ); document.write( "\n" ); document.write( "18(80-Dd) = 14Dd
\n" ); document.write( "32Dd = 1440
\n" ); document.write( "Dd = 45\r
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\n" ); document.write( "\n" ); document.write( "So Dave travels 45 miles. Now find his time:\r
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\n" ); document.write( "\n" ); document.write( "T = Dd/Vd
\n" ); document.write( "T = 45/18
\n" ); document.write( "T = 2.5\r
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\n" ); document.write( "\n" ); document.write( "So they will meet after 2.5 hours. You can solve for T in the equation for Cathy, and it will be the same.
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