Respuesta :
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.
[tex]\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[/tex]
and surely you know how much that is.
what's the boat's speed? well, 36 + c = b.
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.
[tex]\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[/tex]
and surely you know how much that is.
what's the boat's speed? well, 36 + c = b.
