salmon's rate of swimming --- x mph
let rate of current be y mph
15/(x+y) = 3
3x+3y=15 ---> x+y=5
15/(x-y)=5
5x - 5y = 15 ----> x - y = 3
add those two equations:
2x = 8
x = 4 , then y = 1
state your conclusion
let rate of current be y mph
15/(x+y) = 3
3x+3y=15 ---> x+y=5
15/(x-y)=5
5x - 5y = 15 ----> x - y = 3
add those two equations:
2x = 8
x = 4 , then y = 1
state your conclusion
Let's assume the speed of the current (stream) is "C" and the speed of the salmon in still water is "S".
During the downstream trip, the salmon's effective speed is the sum of its speed in still water and the speed of the current, which is S + C. We know that the salmon covers a distance of 15 miles in 3 hours, so the equation becomes:
(S + C) * 3 = 15 --- Equation 1
During the upstream trip, the salmon's effective speed is the difference between its speed in still water and the speed of the current, which is S - C. We know that the salmon covers the same distance of 15 miles on the return trip, but it takes 5 hours, so the equation becomes:
(S - C) * 5 = 15 --- Equation 2
Now, we have a system of equations (Equations 1 and 2) that we can solve to find the values of S (speed in still water) and C (speed of the current).
Let's solve the system of equations:
From Equation 1: 3S + 3C = 15
From Equation 2: 5S - 5C = 15
Multiplying Equation 1 by 5 and Equation 2 by 3 to eliminate the coefficient of C, we get:
15S + 15C = 75
15S - 15C = 45
Combining these equations, we have:
30S = 120
Dividing both sides by 30, we find:
S = 4
Therefore, the salmon swims at a speed of 4 miles per hour in still water.