SOLUTION: Solve using a system: Roy and Chuck are two miles apart riding toward each other on their bicycles. They will meet in 20 minutes. If Roy were to ride twice as fast and Chuck were t

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Question 941929: Solve using a system: Roy and Chuck are two miles apart riding toward each other on their bicycles. They will meet in 20 minutes. If Roy were to ride twice as fast and Chuck were to ride three times as fast, the could meet in 8 minutes. How fast is each boy riding?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
In both cases, I can think of this as one
of them ( either one ) standing still, and
the other moving at the sum of their speeds
Let +s%5Br%5D+ = Roy's speed in mi / min
Let +s%5Bc%5D+ = Chuck's speed in mi / min
---------------------------------------
1st case:
(1) +2+=+%28+s%5Br%5D+%2B+s%5Bc%5D+%29%2A20+
2nd case:
(2) +2+=+%28+2s%5Br%5D+%2B+3s%5Bc%5D+%29%2A8+
---------------------------
(1) +s%5Br%5D+%2B+s%5Bc%5D+=+1%2F10+
(2) +2s%5Br%5D+%2B+3s%5Bc%5D+=+1%2F4+
-------------------------
Multiply both sides of (1) by +2+
and subtract (1) from (2)
(2) +2s%5Br%5D+%2B+3s%5Bc%5D+=+1%2F4+
(1) +-2s%5Br%5D+-+2s%5Bc%5D+=+-2%2F10+
----------------------------
+s%5Bc%5D+=+5%2F20+-+4%2F20+
+s%5Bc%5D+=+1%2F20+
and
(1) +s%5Br%5D+%2B+s%5Bc%5D+=+1%2F10+
(1) +s%5Br%5D+%2B+1%2F20+=+2%2F20+
(1) +s%5Br%5D+=+1%2F20+
---------------------
Both of them travel at 1 mi in 20 min
+%283%2F3%29%2A%281%2F20%29+=+3%2F60+
Both travel at 3 mi/hr
----------------------
check:
(1) +2+=+%28+s%5Br%5D+%2B+s%5Bc%5D+%29%2A20+
(1) +2+=+%28+1%2F20+%2B+1%2F20+%29%2A20+
(1) +2+=+%28+2%2F20+%29%2A20+
(1) +2+=+2+
OK
(2) +2+=+%28+2%2A%281%2F20%29+%2B+3%2A%281%2F20%29+%29%2A8+
(2) +2+=+%28+2%2F20+%2B+3%2F20+%29%2A8+
(2) +2+=+%28+5%2F20%29%2A8+
(2) +2+=+40%2F20+
(2) +2=+2+
OK