document.write( "Question 347413: Two cyclists, 108 miles apart, start riding toward each other at the same time. One cycles 3 times as fast as the other. If they meet 3 hours later, what is the speed (in mi/h) of the faster cyclist? \n" ); document.write( "

Algebra.Com's Answer #248399 by josmiceli(19441) $\"\"$ $\"About$

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If they are cycling towards each other, you can think of one as
\n" ); document.write( "being stationary, and the other as moving at the sum of their speeds.
\n" ); document.write( "Thinking of the faster one as the one in motion, the equation is
\n" ); document.write( " $\"d+=+r%2At\"$
\n" ); document.write( " $\"108+=+%28r+%2B+3r%29%2A3\"$
\n" ); document.write( " $\"108+=+12r\"$
\n" ); document.write( " $\"r+=+9\"$
\n" ); document.write( " $\"3r+=+27\"$
\n" ); document.write( "The faster cyclist's speed is 27 mi/hr
\n" ); document.write( "check:
\n" ); document.write( "The slower cyclist:
\n" ); document.write( " $\"d%5B1%5D+=+9%2A3\"$
\n" ); document.write( " $\"d%5B1%5D+=+27\"$ mi
\n" ); document.write( "The faster cyclist:
\n" ); document.write( " $\"d%5B2%5D+=+27%2A3\"$
\n" ); document.write( " $\"d%5B2%5D+=+81\"$ mi
\n" ); document.write( " $\"d%5B1%5D+%2B+d%5B2%5D+=+27+%2B+81\"$
\n" ); document.write( " $\"d%5B1%5D+%2B+d%5B2%5D+=+108\"$
\n" ); document.write( " \n" ); document.write( "

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