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A woman paddling a canoe downstream with the river current travels 20 miles in 2 hours.
On the return trip, she paddles the same distance upstream against the river current in 5 hours.
Write and solve a system of two linear equations to determine the rate of the boat in calm water and the rate of the current,
both in miles per hour.
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Your first equation is
= u + v, or
10 = u + v. (1)
The term in the left side is the speed of the canoe (relative to the bank of the river) on the trip downstream with the river current.
Here u is the canoe speed in the still water and v is the current speed.
Your second equation is
= u - v, or
4 = u - v. (2)
The term in the left side is the speed of the canoe (relative to the bank of the river) on the trip upstream against the river current.
Again, here u is the canoe speed in the still water and v is the current speed.
Rewrite the equations (1) and (2) in this way:
u + v = 10, (3)
u - v = 4. (4)
To solve them, add equations (3) and (4) (both sides). You ill get
2u = 10 + 4, or 2u = 14. Hence, u = = 7.
You just found the speed of the canoe in still water. It is 7 mph.
Now from (3) v = 10 - u = 10 - 7 = 3.
Thus the current rate is 3 mph.
Answer. The speed of the canoe in still water is 7 mph. The current rate is 3 mph.
For many other solved problems on upstream and downstream trips see the lessons
- Wind and Current problems
- More problems on upstream and downstream round trips
- Selected problems from the archive on the boat floating Upstream and Downstream
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.
Please ignore everything that the other tutor, "josgarithmetic", wrote in his post.
He neither knows how to solve it nor how to explain.