document.write( "Question 831689: Two cyclists, 56 miles apart, start riding toward each other at the same time. One cycles 3 times as fast as the other. If they meet 2 hours later, what is the speed (in mi/h) of the faster cyclist?
\n" ); document.write( "
\n" ); document.write( "

Algebra.Com's Answer #501573 by hovuquocan1997(83)\"\" \"About 
You can put this solution on YOUR website!
Let the speed of the slow one be x
\n" ); document.write( "Let the speed of the fast one be y
\n" ); document.write( "Both travel and meet after 2 hours so we have the equation
\n" ); document.write( "2x + 2y = 56
\n" ); document.write( "Then you have the fast one is 3 times faster than the slow one, so you have
\n" ); document.write( "y = 3x =====>>>> 3x - y = 0
\n" ); document.write( "Then you solve the system of equation
\n" ); document.write( "\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "
Solved by pluggable solver: SOLVE linear system by SUBSTITUTION
Solve:
\n" ); document.write( " \"+system%28+%0D%0A++++2%5Cx+%2B+2%5Cy+=+56%2C%0D%0A++++3%5Cx+%2B+-1%5Cy+=+0+%29%0D%0A++\"We'll use substitution. After moving 2*y to the right, we get:
\n" ); document.write( " \"2%2Ax+=+56+-+2%2Ay\", or \"x+=+56%2F2+-+2%2Ay%2F2\". Substitute that
\n" ); document.write( " into another equation:
\n" ); document.write( " \"3%2A%2856%2F2+-+2%2Ay%2F2%29+%2B+-1%5Cy+=+0\" and simplify: So, we know that y=21. Since \"x+=+56%2F2+-+2%2Ay%2F2\", x=7.
\n" ); document.write( "
\n" ); document.write( " Answer: \"system%28+x=7%2C+y=21+%29\".
\n" ); document.write( "

\n" ); document.write( "\n" ); document.write( "y=21 so the speed of the faster one is 21mi/h
\n" ); document.write( "TA-DAH
\n" ); document.write( ":D
\n" ); document.write( "
\n" );