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Their rate canoeing downstream was = 6 miles per hour,
and this rate is the SUM of the rate of the canoe in still water and the rate of the current:
u + v = 6 (1)
Their rate upstream was = 2 miles per hour,
and this rate is the DIFFERENCE of the rate of the canoe in still water and the rate of the current:
u - v = 2 (2)
Add equations (1) and (2). You will get
2u = 6 + 2 = 8, which implies u = 8/2 = 4 miles per hour is the rate of the canoe in still water.
Now substitute it into equation (1) to get
v = 6 - u = 6 - 4 = 2 miles per hour is the rate of the current.
ANSWER. The rate of the canoe in still water is 4 miles per hour.
The rate of the current is 2 miles per hour.
Solved.
Check my answer on your own by substituting the found values into equations (1) and (2).
By doing it, you will understand the solution much better - you will understand it IN FULL.
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It is a typical and standard Upstream and Downstream round trip word problem.
You can find many similar fully solved problems on upstream and downstream round trips with detailed solutions in lessons
- Wind and Current problems
- More problems on upstream and downstream round trips
- Wind and Current problems solvable by quadratic equations
- Unpowered raft floating downstream along a river
- Selected problems from the archive on the boat floating Upstream and Downstream
in this site, where you will find other similar solved problems with detailed explanations.
Read them attentively and learn how to solve this type of problems once and for all.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this textbook under the section "Word problems", the topic "Travel and Distance problems".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.