document.write( "Question 88059: I need help with another problem please help.
\n" ); document.write( "Martin leaves home at 9 a.m. bicycling at a rate of 24mi/hr.Two hours later,John leaves,driving at the rate of 48mi/hr.At what time will John cath up with Martin? \n" ); document.write( "
Algebra.Com's Answer #64055 by ankor@dixie-net.com(22740)\"\" \"About 
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Martin leaves home at 9 a.m. bicycling at a rate of 24mi/hr. Two hours later,John leaves,driving at the rate of 48mi/hr.At what time will John catch up with Martin?
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\n" ); document.write( "Let x = time (in hrs) that M is on the road
\n" ); document.write( "Then
\n" ); document.write( "(x-2) = time J is on the road, when he catches up
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\n" ); document.write( "We know when this happens they will have traveled the same distance.
\n" ); document.write( "We can make a simple equationn from this fact: Dist = speed * time
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\n" ); document.write( "J's dist = M's dist
\n" ); document.write( "48(x-2) = 24x
\n" ); document.write( "48x - 96 = 24x
\n" ); document.write( "48x - 24x = +96
\n" ); document.write( "24x = 96
\n" ); document.write( "x = 96/24
\n" ); document.write( "x = 4 hrs; from 9 am would be 1 pm when J catches up with M
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\n" ); document.write( "Check solution using distance, J's time is 2hrs:
\n" ); document.write( "48(2) = 24(4)
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\n" ); document.write( "Not that hard, right? \n" ); document.write( "
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