document.write( "Question 54163: At 1pm Tom leaves home on his bike heading to Sue's house at 9mph. An hour later, Sue leaves her house biking to Tom's at 6mph. Their houses are 39 miles apart. At what time will the meet? \n" ); document.write( "
Algebra.Com's Answer #36457 by stanbon(75887)\"\" \"About 
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At 1pm Tom leaves home on his bike heading to Sue's house at 9mph. An hour later, Sue leaves her house biking to Tom's at 6mph. Their houses are 39 miles apart. At what time will the meet?
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\n" ); document.write( "Tom DATA:
\n" ); document.write( "distance = x miles ; rate= 9 mph ; time= d/r=x/9 hrs.
\n" ); document.write( "Sue DATA:
\n" ); document.write( "distance = 39-x ; rate= 6 mph ; time = d/r = (39-x)/6
\n" ); document.write( "EQUATION:
\n" ); document.write( "Tom's time - Sue's time = 1 hr.
\n" ); document.write( "x/9 - (39-x)/6 = 1
\n" ); document.write( "6x -9(39-x) = 54
\n" ); document.write( "6x+9x-351 = 54
\n" ); document.write( "15x=405
\n" ); document.write( "x= 27 (Tom's distance is 27 miles)
\n" ); document.write( "At 9 mph this will take him 3 hours.
\n" ); document.write( "They will meet at 4 PM.
\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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