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With the current a motorboat can travel 84 miles in two hours.
Against the current the same trip takes three hours.
Find the average rate of the boat in still water and the average rate of the current.
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As usual, the setup in the post by @josgarithmetic is TOTALLY WRONG.
I came to bring a correct solution.
Let r be the motorboat rate in still water and let c be the rate of the current.
Then you have these two equations
r + c = 42 (1) (for the effective rate moving with the current; 42 = 84/2)
r - c = 28 (2) (for the effective rate moving against the current; 28 = 84/3)
Adding equations, you get
2r = 42 + 28 = 70; ---> r = 70/2 = 35.
Then from equation (1), c = 42 - r = 42 - 35 = 7.
ANSWER. The rate of the motorboat in still water is 35 miles per hour;
The current rate is 7 miles per hour.
Solved.
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For your info: still water is not the same as Stillwater.
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After seeing my post, @josgarithmetic rewrote his exactly as my, so now his post is safe (although useless).
At least, I forced him to make it safe . . .