document.write( "Question 143652: if two friends leave from the same place traveling to a destination of 720 miles away, and one leaves 2 hours before the other traveling at a rate of 60 miles an hour and the other leaves traveling at a rate of 80 miles an hour, at what time would they pass eachother on the road? \n" ); document.write( "
Algebra.Com's Answer #104544 by josmiceli(19441)\"\" \"About 
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Imagine you have a stopwatch. Start the stopwatch the instant
\n" ); document.write( "the 2nd friend leaves. Then ask: \"where is the 1st one at that time?\"
\n" ); document.write( "He has gone \"60%2A2+=+120\"mi
\n" ); document.write( "When will they pass eachother?
\n" ); document.write( "Call the distance the 1st one has to go \"d\"
\n" ); document.write( "Then the distance the 2nd one has to go will be \"d+%2B+120\"
\n" ); document.write( "The stopwatch will show the same time for both when they meet
\n" ); document.write( "\"d%2Ft+=+60\" for 1st friend
\n" ); document.write( "\"%28d+%2B+120%29%2Ft+=+80\" for 2nd friend
\n" ); document.write( "-------------------------------------
\n" ); document.write( "\"d%2Ft+=+60\"
\n" ); document.write( "\"d+=+60t\"
\n" ); document.write( "\"%2860t+%2B+120%29%2Ft+=+80\"
\n" ); document.write( "\"60t+%2B+120+=+80t\"
\n" ); document.write( "\"20t+=+120\"
\n" ); document.write( "\"t+=+6\"hrs
\n" ); document.write( "They will meet in 6 hours
\n" ); document.write( "check:
\n" ); document.write( "\"d%2Ft+=+60\"
\n" ); document.write( "\"d%2F6+=+120\"
\n" ); document.write( "\"d+=+720\"
\n" ); document.write( "\"%2860t+%2B+120%29%2Ft+=+80\"
\n" ); document.write( "\"%2860%2A6+%2B+120%29%2F6+=+80\"
\n" ); document.write( "\"360+%2B+120+=+480\"
\n" ); document.write( "\"480+=+480\"
\n" ); document.write( "OK
\n" ); document.write( "
\n" );