document.write( "Question 361015: Two friends live 60 miles apart. The decide to ride their bikes to meet each other. Friend 1 rides at 21 mph. Friend 2 rides at 15 mph. They leave at the same time. How long will it take them to meet and how far did each one ride? \n" ); document.write( "

Algebra.Com's Answer #257559 by josmiceli(19441) $\"\"$ $\"About$

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The key is that if they will always meet at
\n" ); document.write( "the same time on an observers watch.
\n" ); document.write( "since they leave at the same time, each
\n" ); document.write( "friend's equation of motion has the same $\"t\"$ , elapsed time.
\n" ); document.write( "given:
\n" ); document.write( "friend 1:
\n" ); document.write( " $\"d+=+21t\"$
\n" ); document.write( "Friend 2:
\n" ); document.write( " $\"60+-+d+=+15t\"$
\n" ); document.write( "By substitution:
\n" ); document.write( " $\"60+-+21t+=+15t\"$
\n" ); document.write( " $\"36t+=+60\"$
\n" ); document.write( " $\"3t+=+5\"$
\n" ); document.write( " $\"t+=+5%2F3\"$ hr
\n" ); document.write( "It takes 1 hr and 40 min for them to meet
\n" ); document.write( "friend 1:
\n" ); document.write( " $\"d+=+21%2A%285%2F3%29\"$
\n" ); document.write( " $\"d+=+35\"$ mi
\n" ); document.write( "friend 2:
\n" ); document.write( " $\"60+-+d+=+15%2A%285%2F3%29\"$
\n" ); document.write( " $\"60+-+d+=+25\"$ mi
\n" ); document.write( "friend 1 goes 35 mi and friend 2 goes 25 mi \n" ); document.write( "

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