Question

A motorboat takes 3 hours to travel 108 km going upstream. The return trip takes 2 hours going downstream. What is the rate of the boat in still water and what is the rate of the current?

Answer 1

Recall your d = rt, distance = rate * time.

let's say the boat has a still water rate of "b", and the current has a a rate of "c", ok.... when the boat is going upstream, is not really going "b" fast, is going  slower at "b - c", due to the current going in the opposite direction.

when the boat is coming downstream, is not going "b" fast either, is going faster, is going "b + c", due to the current adding speed to it.

we know the trip up was 108 kms, thus the return trip is also 108 kms.

$\bf \begin{array}{lccclll} &distance&rate&time\\ &-----&-----&-----\\ Upstream&108&b-c&3\\ Downstream&108&b+c&2 \end{array} \\\\\\ \begin{cases} 108=3(b-c)\implies \cfrac{108}{3}=b-c\implies 36+c=\boxed{b}\\\\ 108=2(b+c)\\ --------------------\\ 108=2\left(\boxed{36+c} +c \right) \end{cases} \\\\\\ \cfrac{108}{2}=36+2c\implies 54=36-2c\implies \cfrac{54-36}{2}=c$

and surely you know how much that is.

what's the boat's speed? well, 36 + c = b.