document.write( "Question 617499: Two joggers set out at the same time from their homes 30 miles apart. They agree to meet at a point somewhere in between in two and a half hours. If the rate of one is 2 mph faster than the rate of the other, find the rate of each \n" ); document.write( "
Algebra.Com's Answer #388361 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
Let \"+s+\" = the speed of the slower jogger in mi/hr
\n" ); document.write( "\"+s+%2B+2+\" = the speed of the faster jogger in mi/hr
\n" ); document.write( "Let \"+d+\" = the distance the slower jogger has to
\n" ); document.write( "run until they meet
\n" ); document.write( "\"+30+-+d+\" = the distance the faster jogger runs
\n" ); document.write( "-----------------
\n" ); document.write( "Equation for slower jogger:
\n" ); document.write( "(1) \"+d+=+s%2A2.5+\"
\n" ); document.write( "Equation for faster jogger:
\n" ); document.write( "(2) \"+30+-+d+=+%28+s+%2B+2+%29%2A2.5+\"
\n" ); document.write( "-------------------------------
\n" ); document.write( "Substitute (1) into (2)
\n" ); document.write( "(2) \"+30+-+2.5s+=+%28+s+%2B+2+%29%2A2.5+\"
\n" ); document.write( "(2) \"+30+-+2.5s+=+2.5s+%2B+5+\"
\n" ); document.write( "(2) \"+5s+=+30+-+5+\"
\n" ); document.write( "(2) \"+5s+=+25+\"
\n" ); document.write( "(2) \"+s+=+5+\"
\n" ); document.write( "and
\n" ); document.write( "\"+s+%2B+2+=+7+\"
\n" ); document.write( "5 mi/hr is the speed of the slower jogger
\n" ); document.write( "7 mi/hr is the speed of the faster jogger
\n" ); document.write( "check:
\n" ); document.write( "(2) \"+30+-+d+=+%28+s+%2B+2+%29%2A2.5+\"
\n" ); document.write( "(2) \"+30+-+d+=+7%2A2.5+\"
\n" ); document.write( "(2) \"+d+=+30+-+17.5+\"
\n" ); document.write( "(2) \"+d+=+12.5+\"
\n" ); document.write( "and
\n" ); document.write( "(1) \"+d+=+s%2A2.5+\"
\n" ); document.write( "(1) \"+d+=+5%2A2.5+\"
\n" ); document.write( "(1) \"+d+=+12.5+\"
\n" ); document.write( "OK
\n" ); document.write( "
\n" );