Solving a distance, rate, time, problem using a system of linear

Question

A motorboat travels 308 kilometers in 7 hours going upstream. It travels 490 kilometers going downstream in the same amount of time. What is the rate of the boat in still water and what is the rate of the current?

Answer 1

Let b be the boat's still water rate and w be the water current rate.

b-w = 308

b+w = 490

Add the two equations

2b + 0 = 798

b = 798/w = 399 kph

b+w = 399+w = 490

w = 490-399 = 91 kph