.
Let u be the motorboat speed in still water and v be the current rate.
The effective speed going upstream is
= = 36 km/h.
It is the DIFFERENCE of the motorboat speed in still water and the rate of the current. It gives you your first equation
u - v = 36. (1)
The effective speed going downstream is
= = 54 km/h.
It is the SUM of the motorboat speed in still water and the rate of the current. It gives you your second equation
u + v = 54. (2)
Thus you have this system of two equations in 2 unknowns
u - v = 36, (1) and
u + v = 54. (2)
Add the two equations. You will get
2u = 36 + 54 = 90 ====> u = = 45 km/h.
So, you just found the speed of the motorboat in still water. It is 45 km/h.
Then from the equation (2) you get v = 54 = 45 = 9 km/h is the current rate.
Answer. The speed of the motorboat in still water is 45 km/h.
The current rate is 9 km/h.
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.
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.