document.write( "Question 9351: Please help me with this problem, I just can't seem to come up with the right formula. I will apreciate help in comming up with the correct formula. Thank you!\r
\n" ); document.write( "\n" ); document.write( "Two joggers left their homes, which are 21 miles apart, at the same time. They agree to meet at a point somewhere in between in 1.5 hours. If the rate of one of the joggers is 2 miles per hour faster than the other, find the rate of each jogger. \n" ); document.write( "
Algebra.Com's Answer #5078 by longjonsilver(2297)\"\" \"About 
You can put this solution on YOUR website!
OK, joggerA speed = x
\n" ); document.write( "so that joggerB speed = x+2.\r
\n" ); document.write( "\n" ); document.write( "Note also, if joggerA runs a distance d, then B must have run 21-d, since they have met somewhere.\r
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\n" ); document.write( "\n" ); document.write( "now speed = distance/time so\r
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\n" ); document.write( "\n" ); document.write( "for A: x = d/1.5 --> d = 1.5x --> eqn1\r
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\n" ); document.write( "\n" ); document.write( "for B: x+2 = (21-d)/1.5
\n" ); document.write( "--> 1.5x+3 = 21-d\r
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\n" ); document.write( "\n" ); document.write( "so, 1.5x + 3 = 21-(1.5x)
\n" ); document.write( "1.5x + 3 = 21-1.5x
\n" ); document.write( "3x = 18
\n" ); document.write( "so x = 6.\r
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\n" ); document.write( "\n" ); document.write( "So joggerA runs at 6mph and B runs at 8mph\r
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\n" ); document.write( "\n" ); document.write( "jon.
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