Question

a boat takes 4 hours to go 20 miles upstream. it can go 32 miles downstream in the same time. find the rate of the current and the rate of the boat in still water. (hint: because the current pushes the boat when it is going downstream the rate of the boat downstream is the sum of the rate of the boat and the rate of the current. the current slows down the boat when it is going upstream so the rate of the boat upstream is the difference of the rate of the boat and the rate of the current)
omg what?! please help

Answer 1

D=V*T

Let V be the speed of the boat and W be the speed of the current. Then:

20=(W-V)*4
32=(W+V)*4

Because if he's going upstream, the total speed is his speed minus the speed of the current slowing him down. Vice versa downstream. Now just solve the system of equations for V and W

Answer 2

i really dont know

Answer 3

where do i get the w and the v

Answer 4

those are the variables you have to figure out what they are

Answer 5

how do you solve a system of equations

Answer 6

if i said.... 20 = 4X how would you solve for X?

Answer 7

you would divide

Answer 8

so X = ?

Answer 9

5

Answer 10

right, well in this case the "X" is w+v and w-v

so divide both equations by 4 to start

Answer 11

well 20/4 =5
and 32/4 is8

Answer 12

yes

...
w-v = 5
w+v = 8

what 2 numbers add to 8 but have difference of 5

Answer 13

A boat takes 4 hours to 20 miles upstream.

r = 20miles / 4 hrs

r = 5 mph

It can go 32 miles downstream in the same amount of time as 4 hours.

r -c = 20/4 = 5 mph
r+c = 32/4 = 8

(Adding the equations together to eliminate C)

Then solve for R and C :)

Answer 14

r+c = 32/4 = 8 mph*

Answer 15

1 and 7?

Answer 16

no...7-1=6

@Asylum15 gave you hint: add equations together

you need to learn how to solve system of 2 equations before you can do these word problems
this might help:
http://www.khanacademy.org/math/algebra/systems-of-eq-and-ineq/v/systems-of-equations

Answer 17

that website really helped.