A motorboat travels 78 kilometers in 2 hours going upstream. It travels 102 kilometers going downstream in the same amount of time. What is the rate of the boat in still water and what is the rate of the current? Note that the ALEKS graphing calculator can be used to make computations easier. Rate of the boat in still water: Rate of the current: \[ \begin{array}{l} \square \frac{\mathrm{km}}{\mathrm{h}} \\ \square \frac{\mathrm{km}}{\mathrm{h}} \end{array} \]

Step 1 ：Let's denote the rate of the boat in still water as \(b\) and the rate of the current as \(c\).

Step 2 ：When the boat is going upstream, it is going against the current, so the effective speed of the boat is \(b - c\).

Step 3 ：When the boat is going downstream, it is going with the current, so the effective speed of the boat is \(b + c\).

Step 4 ：We know that the boat travels 78 kilometers in 2 hours going upstream, so \(b - c = \frac{78}{2} = 39 \, \text{km/h}\).

Step 5 ：We also know that the boat travels 102 kilometers in 2 hours going downstream, so \(b + c = \frac{102}{2} = 51 \, \text{km/h}\).

Step 6 ：We can solve these two equations to find the values of \(b\) and \(c\).

Step 7 ：Adding the two equations, we get \(2b = 90\), so \(b = \frac{90}{2} = 45 \, \text{km/h}\).

Step 8 ：Subtracting the first equation from the second, we get \(2c = 12\), so \(c = \frac{12}{2} = 6 \, \text{km/h}\).

Step 9 ：Final Answer: The rate of the boat in still water is \(\boxed{45 \, \text{km/h}}\) and the rate of the current is \(\boxed{6 \, \text{km/h}}\).