SOLUTION: you started on a bike trip, riding at a rate of 9 miles per hour. your friend left one hour later and rode at rate of 12 miles per hour. how long did it take your friend to catch u

Algebra ->  Graphs -> SOLUTION: you started on a bike trip, riding at a rate of 9 miles per hour. your friend left one hour later and rode at rate of 12 miles per hour. how long did it take your friend to catch u      Log On


   

Question 659250: you started on a bike trip, riding at a rate of 9 miles per hour. your friend left one hour later and rode at rate of 12 miles per hour. how long did it take your friend to catch up with you? your distance = your friend's distance. solve for t. t= your time in hours. t+1= friend time in hours
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Your equation:
+d+=+9t%5B1%5D+
Your friend's equation:
+d+=+12t%5B2%5D+
--------------
+9t%5B1%5D+=+12t%5B2%5D+
+t%5B2%5D+=+%283%2F4%29%2At%5B1%5D+
------------------
This doesn't take into account the extra hour
which the friend takes, so I add 1 hr to his time
--------------
Time to catch up occurs when
+%283%2F4%29%2At%5B1%5D+%2B+1+=+t%5B1%5D+
+%281%2F4%29%2At%5B1%5D+=+1+
+t%5B1%5D+=+4+ hrs
--------------
The friend catches up in 4 hrs
check:
+d+=+9t%5B1%5D+
+d+=+9%2A4+
+d+=+36+ mi
and
+d+=+12t%5B2%5D+
+d+=+12%2A%283%2F4%29%2At%5B1%5D+
+d+=+9%2At%5B1%5D+
+d+=+9%2A4+
+d+=+36+
------------
This is done a little differently than the
problem expects, but I think it's right.