Let's solve the problem step-by-step to find the rate of the boat in still water and the rate of the current.
Given:
- The motorboat travels 78 miles in 3 hours going upstream.
- The motorboat travels 120 miles in 3 hours going downstream.
Let's denote:
- [tex]\( b \)[/tex] as the rate of the boat in still water (in miles per hour, mph).
- [tex]\( c \)[/tex] as the rate of the current (in miles per hour, mph).
### Step 1: Write the equations for both directions
1. Upstream speed:
When traveling upstream, the speed of the boat is reduced by the speed of the current. Therefore, the effective speed upstream is [tex]\( b - c \)[/tex].
According to the problem, the boat travels 78 miles in 3 hours upstream. Hence:
[tex]\[
b - c = \frac{78}{3} = 26 \quad \text{(Equation 1)}
\][/tex]
2. Downstream speed:
When traveling downstream, the speed of the boat is increased by the speed of the current. Therefore, the effective speed downstream is [tex]\( b + c \)[/tex].
According to the problem, the boat travels 120 miles in 3 hours downstream. Hence:
[tex]\[
b + c = \frac{120}{3} = 40 \quad \text{(Equation 2)}
\][/tex]
### Step 2: Solve the system of equations
We now have the following system of linear equations:
[tex]\[
\begin{cases}
b - c = 26 \quad \text{(Equation 1)} \\
b + c = 40 \quad \text{(Equation 2)}
\end{cases}
\][/tex]
To find [tex]\( b \)[/tex] and [tex]\( c \)[/tex], we can solve these equations simultaneously.
1. Add the two equations to eliminate [tex]\( c \)[/tex]:
[tex]\[
(b - c) + (b + c) = 26 + 40
\][/tex]
Simplifying this, we get:
[tex]\[
2b = 66 \implies b = 33
\][/tex]
2. Substitute [tex]\( b = 33 \)[/tex] back into Equation 2:
[tex]\[
b + c = 40 \implies 33 + c = 40 \implies c = 7
\][/tex]
### Step 3: State the final answers
Thus, the rate of the boat in still water is:
[tex]\[
\boxed{33} \frac{\text{mi}}{\text{h}}
\][/tex]
And the rate of the current is:
[tex]\[
\boxed{7} \frac{\text{mi}}{\text{h}}
\][/tex]
