In the following exercises, translate to a system of equations and solve.

The Jones family took a 12 mile canoe ride down the Indian River in two hours. After lunch, the return trip back up the river took three hours. Find the rate of the canoe in still water and the rate of the current.

The given data is that the Jones family took a 12 mile canoe ride down the Indian River in two hours. After lunch, the return trip back up the river took three hours.

Let x be the rate of the canoe in still water and y be the rate of current.

The family took a canoe ride down for 2 hours and the distance covered is 12 miles and canoe goes down so the current helps canoe so the actual rate of the canoe is $x+y$and the distance covered can be expressed as $2(x+y)=12--\left(1\right)$

The canoe return trip took 3 hours. Since the upstream downs the canoe, the actual rate of the canoe is $x-y$and the distance covered is $3(x-y)=12--\left(2\right)$

Multiply equation (1) with 3 and equation (2) with 2. Thus, the revised equations are,

$$\begin{gathered} 3(2x+2y)=3\left(12\right) \\ 6x+6y=36--\left(3\right) \\ And \\ 2(3x-3y)=2\left(12\right) \\ 6x-6y=24--\left(4\right) \end{gathered}$$

Solve the equations (3) and (4) by subtracting equation (4) from equation (3),

$$\begin{gathered} 6x+6y-6x+6y=36-24 \\ 12y=12 \\ y=1 \end{gathered}$$

Substitute the value of y in equation (1) to get the value of x.

$$\begin{gathered} 2x+2\left(1\right)=12 \\ 2x=10 \\ x=5 \end{gathered}$$