SOLUTION: I have a bad time with word problems. Please help me with this: Junior's boat will go 15 mph in still water. If he can go 12 miles downstream in the same amount of time as it

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I have a bad time with word problems. Please help me with this: Junior's boat will go 15 mph in still water. If he can go 12 miles downstream in the same amount of time as it       Log On


   



Question 253209: I have a bad time with word problems. Please help me with this:
Junior's boat will go 15 mph in still water. If he can go 12 miles downstream in the same amount of time as it takes to go 9 miles upstream, then what is the speed of the current?
I know that we have to find the current so do I subtract from the original 15 mph? downstream = x-3 and upstream = x-6? Pleas help me.

Found 2 solutions by Greenfinch, richwmiller:
Answer by Greenfinch(383) About Me  (Show Source):
You can put this solution on YOUR website!
In a unit of time T
speed is 15 + current
so distance is T(15 + c) = 12
and, T(15 - c) = 9
so (15 - c) x 12 = (15 + c) x 9
180 - 12c = 135 + 9c
45 = 21c
c = 45/21 = 2 1/7 mph

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
downstream (with the current) you add x the speed of the current
upstream you subtract x the speed of the current
rt=d
we have the times equal but distance and speed different
since the times are the same let's solve for t
t=d/r
downstream
t=12/(15+x)
upstream
t=9/(15-x)
since t=t we will set them equal
12/(15+x)=9/(15-x)
multiply both sides by (15+x)
12=9*(15+x)/(15-x)
multiply both sides by (15-x)
12(15-x)=9*(15+x)
divide both sides by 3
4*(15-x)=3(15+x)
60-4x=45+3x
subtract 45 from both sides
15-4x=3x
add 4x to both sides
15=7x
15/7=x
2 1/7 mph=x speed of current