Question 1009854: Identify the variables, write a system of equations to solve
(Problem on my exam study guide)
When she rows with the current, Mindy can row 24 miles in 3 hours. Against the same current, Mindy can row only ⅔ of this distance in 4 hours. Find the rate of the current and Mindy's rowing rate in still water. Assume that both are constant.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! rate * time = distance
let a = the rate of the boat
let b = the rate of the current.
with the current, the rates are additive.
against the current, the rate are subtactive.
going with the current, rate * time becomes:
(a + b) * 3 = 24
going agains the current, rate * time becomes:
(a - b) * 4 = 16
simplify these equations to get:
3a + 3b = 24
4a - 4b = 16
multiply both sides of the first equation by 4 and multiply the second equation by 3 to get:
12a + 12b = 96
12a - 12b = 48
add the equations together to get:
24a = 144
solve for a to get a = 6
your original equations are:
(a + b) * 3 = 24
(a - b) * 4 = 16
replace a in the first equation to get:
(6 + b) * 3 = 24
distribute the multiplication to get:
18 + 3b = 24
subtract 18 from both sides to get:
3b = 6
solve for b to get:
b = 2
you have a = 6 and b = 2
this means the rate of the boat is 6 mph and the rate of the current is 2 mph.
go back to your original equations again and replace a with 6 and b with 2 to get:
(a + b) * 3 = 24 becomes (6+2) * 3 = 24 which becomes 8*3 = 24 which becomes 24 = 24.
(a - b) * 4 = 16 becomes (6-2) * 4 = 16 which becomes 4 * 4 = 16 which becomes 16 = 16.
this confirms the solutions are correct.
the rate of the boat is 6 miles per hour.
the rate of the current is 2 miles per hour.
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