Boat speed in still water

[roach711x]

Boat takes 2 hours more going upstream in a water current of 4 mph a distance of 32 miles. What is the rate of the boat in still water?

Distance	Rate	Time
32	b+4	t
32	b-4	t+2

boat traveling with current:
32 = (b+4)t

boat traveling against current:
32 = (b-4)(t+2)

I set them equal to each other to solve for b since distance is the same:
t(b+4) = (b-4)(t+2)

I end up with b = 8t+8

I can't get a whole number for speed of boat in still water....

 

[soroban]

Hello, roach711x!

Boat takes 2 hours more going upstream in a water current of 4 mph a distance of 32 miles.
What is the rate of the boat in still water?

Distance	Rate	Time
32	b+4	t
32	b-4	t+2

You created two variables; we need only one.

\(\displaystyle \text{The boat went downsteam 32 miles at }b\!+\!4\text{ mph.}\)
. . \(\displaystyle \text{This took }\dfrac{32}{b+4}\text{ hours.}\)

\(\displaystyle \text{The boat went upstream 32 miles at }b\!-\!4\text{ mph.}\)
. . \(\displaystyle \text{This took }\dfrac{32}{b-4}\text{ hours.}\)


The upstream time is 2 hours more than the downstream time.

. . \(\displaystyle \dfrac{32}{b-4} \:=\:\dfrac{32}{b+4} + 2\)

Got it?

 

[roach711x]

Hello, roach711x!


You created two variables; we need only one.

\(\displaystyle \text{The boat went downsteam 32 miles at }b\!+\!4\text{ mph.}\)
. . \(\displaystyle \text{This took }\dfrac{32}{b+4}\text{ hours.}\)

\(\displaystyle \text{The boat went upstream 32 miles at }b\!-\!4\text{ mph.}\)
. . \(\displaystyle \text{This took }\dfrac{32}{b-4}\text{ hours.}\)


The upstream time is 2 hours more than the downstream time.

. . \(\displaystyle \dfrac{32}{b-4} \:=\:\dfrac{32}{b+4} + 2\)

Got it?

I think I got it, thanks!