Let's assume the speed of the boat in still water is B and the speed of the current is C.

When the boat travels downstream, it gets an additional boost from the current, so its effective speed is B + C. We can use the formula distance = speed × time to write the equation 20 = (B + C) × 2.

When the boat travels upstream, it has to fight against the current, so its effective speed is B - C. We can again use the formula distance = speed × time to write the equation 20 = (B - C) × 5.

Now we have a system of two equations:

2(B + C) = 20
5(B - C) = 20

Let's simplify equation 1 by dividing both sides by 2:

B + C = 10

Now we can solve the system of equations by either substitution or elimination. Let's use substitution:

From equation 1, we know that B = 10 - C.

Substituting this into equation 2, we get:
5(10 - C - C) = 20
50 - 10C - 5C = 20
50 - 15C = 20
-15C = 20 - 50
-15C = -30
C = -30 / -15
C = 2

Now that we know the speed of the current is 2 mph, we can substitute this back into equation 1 to solve for B:

B + 2 = 10
B = 10 - 2
B = 8

Therefore, the speed of the boat in still water is 8 mph and the speed of the current is 2 mph.

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