We assume that we need to find the speed of the boat in still water.

Let

t1 = the time of the upstream trip

t2 = the time of the downstream trip

d = 9 km the distance traveled in each direction

v = the speed of the boat in still water

c = the speed of the current

The speed upstream is: v - c = 5 follows c = 5 + v

The speed downstream is: v + c

Since speed = distance/time => distance = speed*time we have:

(v - c)*t1 = d

(v - c)*t1 = 9

(v + c)*t2 = d

5*t2 = 9

The time for the round trip is 9 hrs:

t1 + t2 = 9

For (v - c)*t1 = 9 we solve for t1:

t1 = 9/(v - c)

plug c = 5 + v and we get

t1 = 9/(v - 5 - v)

t1 = - 9/5

We got negative result for the time, and since the time can not be negative the problem has no solution.

We are assuming that there is a typo in the text of the problem. Now we will give solution to the following problem:

'A boat travels 9 km upstream and 9 km back. The time for the round trip is 6 hrs. The speed of the stream is 5 km/hr. What is the speed of the boat in still water?'

Let

t1 = the time of the upstream trip

t2 = the time of the downstream trip

d = 9 km the distance traveled in each direction

v = the speed of the boat in still water

c = 5 km/h the speed of the current

The speed upstream is: v - c

The speed downstream is: v + c

Since speed = distance/time => distance = speed*time we have:

(v - c)*t1 = d

(v - 5)*t1 = 9

(v + c)*t2 = d

(v + 5)*t2 = 9

The time for the round trip is 6 hrs:

t1 + t2 = 6

For (v - 5)*t1 = 9 we solve for t1: t1 = 9/(v - 5)

For (v + 5)*t2 = 9 we solve for t2: t2 = 9/(v + 5)

Plug into t1 + t2 = 9:

9/(v - 5) + 9/(v + 5) = 6

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click here to see the equation solved for v

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The speed of the boat in still water is 6.72 km/h.