document.write( "Question 151753: Mary Beth and Michael leave school traveling in opposite directions. Michael is walking and Mary Beth is biking, averaging 6km/h more than Michael. If they are 18km apart after 1.5 hours, what is Michael's rate? \n" ); document.write( "
Algebra.Com's Answer #111563 by ankor@dixie-net.com(22740)\"\" \"About 
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Mary Beth and Michael leave school traveling in opposite directions. Michael is walking and Mary Beth is biking, averaging 6 km/h more than Michael. If they are 18 km apart after 1.5 hours, what is Michael's rate?
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\n" ); document.write( "Let s = Michael's rate walking
\n" ); document.write( "Then
\n" ); document.write( "(s+6) = Mary's rate riding
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\n" ); document.write( "Write a distance equation: Dist = time * speed
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\n" ); document.write( "M's dist + M.B's dist = 18
\n" ); document.write( "1.5s + 1.5(s+6) = 18
\n" ); document.write( "1.5s + 1.5s + 9 = 18
\n" ); document.write( "3s = 18 - 9
\n" ); document.write( "s = \"9%2F3\"
\n" ); document.write( "s = 3 km/hr is Micheal's walking rate
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\n" ); document.write( "Check solution (9 km/hr is M.B.'s rate)
\n" ); document.write( "1.5(3) + 1.5(9) =
\n" ); document.write( "4.5 + 13.5 = 18 \n" ); document.write( "
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