document.write( "Question 403846: I have a problem that I can't seem to be able to set up. It is Pablo is 3 miles behind Marc. Pablo increases his speed to 5 miles per hour, while Marc, runs at 4mph. At this rate, how many miles will Pablo have to run to catch up to Marc? How long will it take? \n" ); document.write( "
Algebra.Com's Answer #285515 by ankor@dixie-net.com(22740)\"\" \"About 
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Pablo is 3 miles behind Marc.
\n" ); document.write( " Pablo increases his speed to 5 miles per hour, while Marc, runs at 4mph.
\n" ); document.write( " At this rate, how many miles will Pablo have to run to catch up to Marc?
\n" ); document.write( " How long will it take?
\n" ); document.write( ":
\n" ); document.write( "Let t = time for P to catch M
\n" ); document.write( "P has to run 3 more miles than M
\n" ); document.write( "Write a distance equation; dist = speed * time
\n" ); document.write( "5t = 4t + 3
\n" ); document.write( "5t - 4t = 3
\n" ); document.write( "t = 3 hrs for P to catch M
\n" ); document.write( ":
\n" ); document.write( "How many miles will P have to run to catch M
\n" ); document.write( "3 * 5 = 15 miles
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Check this, find the distance each runs
\n" ); document.write( "5*3 = 15 mi
\n" ); document.write( "4*3 = 12 mi
\n" ); document.write( "------------
\n" ); document.write( "diff: 3 mi \n" ); document.write( "
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