.
Let x be the rate of the canoe in still water.
Then the effective rate upstream is (x-2) miles per hour, and the time upstream is hours.
The effective rate downstream is (x+2) miles per hour, and the time downstream is hours.
So, your time equation is
+ = 4.
Multiply both sides by (x-2)*(x+2). You will get
6*(x+2) + 6*(x-2) = 4*(x-2)*(x+2)
6x + 12 + 6x - 12 = 4(x^2 - 4)
12x = 4x^2 - 16
4x^2 - 12x - 16 = 0
x^2 - 3x - 4 = 0
(x-4)*(x+1) = 0.
Only positive root x= 4 is meaningful.
Answer. The rate of the canoe in still water is 4 miles per hour.
Solved.
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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.