A motorboat that travels with a speed of 20 km/hour in still water has traveled 36 km against the current and 22 km with the current, having spent 3 hours on the entire trip. Find the speed of the current of the river.

To find the speed of the current of the river, we need to set up a system of equations based on the given information.

Let's assume the speed of the current is "c" km/hour.

Against the current:
The speed of the boat relative to the water is reduced by the speed of the current, so the effective speed is (20 - c) km/hour. The time taken to travel 36 km against the current is given by distance/speed, which is 36/(20 - c) hours.

With the current:
The speed of the boat relative to the water is increased by the speed of the current, so the effective speed is (20 + c) km/hour. The time taken to travel 22 km with the current is given by distance/speed, which is 22/(20 + c) hours.

The total time taken for the trip is 3 hours, so we can write the equation:
36/(20 - c) + 22/(20 + c) = 3

Now, let's solve this equation to find the value of "c", the speed of the current.

To simplify the equation, we can multiply both sides by (20 - c)(20 + c) to get rid of the denominators:
36(20 + c) + 22(20 - c) = 3(20 - c)(20 + c)

Expanding and rearranging, we have:
720 + 36c + 440 - 22c = 3(400 - c^2)

Combining like terms:
1160 + 14c = 1200 - 3c^2

Rearranging again:
3c^2 + 14c - 40 = 0

Now we have a quadratic equation. To solve it, we can either factor or use the quadratic formula.

Factoring the equation, we have:
(3c - 4)(c + 10) = 0

Setting each factor to zero, we have two possible solutions:
1) 3c - 4 = 0, which gives c = 4/3
2) c + 10 = 0, which gives c = -10

Since the speed of the current cannot be negative, we discard the solution c = -10.

Therefore, the speed of the current of the river is 4/3 km/hour.

Let's assume the speed of the current is 'c' km/hour.

When the motorboat travels against the current, its effective speed is reduced by the speed of the current. Therefore, the speed of the motorboat relative to the river is (20 - c) km/hour.

When the motorboat travels with the current, its effective speed is increased by the speed of the current. Therefore, the speed of the motorboat relative to the river is (20 + c) km/hour.

We are given that the motorboat traveled 36 km against the current and 22 km with the current, and the total time spent on the trip is 3 hours.

Using the formula: distance = speed × time, we can set up two equations:

36 = (20 - c) × t1 (1)
22 = (20 + c) × t2 (2)

Since we know that the total time spent on the trip is 3 hours, we can also write:

t1 + t2 = 3 (3)

Now, let's solve the system of equations.

From equation (1):

36 = (20 - c) × t1

We can rewrite t1 in terms of c:

t1 = 36 / (20 - c) (4)

From equation (2):

22 = (20 + c) × t2

We can rewrite t2 in terms of c:

t2 = 22 / (20 + c) (5)

Substituting equations (4) and (5) into equation (3):

36 / (20 - c) + 22 / (20 + c) = 3

To simplify the equation, we can multiply through by the denominators:

36(20 + c) + 22(20 - c) = 3(20 - c)(20 + c)

Expanding and simplifying:

720 + 36c + 440 - 22c = 3(400 - c^2)

Combining like terms:

1160 + 14c = 1200 - 3c^2

Rearranging the equation:

3c^2 + 14c - 40 = 0

Now we can solve this quadratic equation. Factorizing, we get:

(3c - 4)(c + 10) = 0

Setting each factor equal to zero:

3c - 4 = 0 or c + 10 = 0

Solving for c:

3c = 4 or c = -10

c = 4/3 km/hour (speed of the current)

Since the speed of the current cannot be negative, we discard the solution c = -10.

Therefore, the speed of the current is 4/3 km/hour.

since time = distance/speed,

36/(20-s) + 22/(20+s) = 3