Nathan is out rafting. he rafts 16 miles with the river current. at the end of 16 miles, he turns around and rafts the same distance against the river current. the journey takes him 4 hours overall. if he can raft at a speed of 9 mph in still water, what is the speed of the current of the river he is in? a 5 mph b 4 mph c 6 mph d 3 mph

S = d/t

st = d

t = d/s

The time going is t1.
The time returning is t2.
The total time is 4 hours, so we have t1 + t2 = 4

The speed of the current is c.
The speed going is 9 + c.
The speed returning is 9 - c.

t1 = 16/(9 + c)

t2 = 16/(9 - c)

t1 + t2 = 16/(9 + c) + 16/(9 - c)

4 = 16/(9 - c) + 16/(9 + c)

1 = 4/(9 - c) + 4/(9 + c)

(9 + c)(9 - c) = 4(9 - c) + 4(9 + c)

81 - c^2 = 36 - 4c + 36 + 4c

81 - c^2 = 72

c^2 = 9

c^2 - 9 = 0

(c + 3)(c - 3) = 0

c + 3 = 0   or   c - 3 = 0

c = -3   or   c = 3

We discard the negative answer, and we get c = 3.

The speed of the current is 3 mph.

20y5 times 2 is 1010 times -2 is -20, plus y

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