HOW DO SOLVE A BOAT TRAVELED 336 MILES DOWNSTREAM AND BACK. THE TRIP DOWNSTREAM TOOK 12 HOURS. THE TRIP BACK TOOK 14 HOURS. WHAT IS THE SPEED OF THE BOAT IN STILL WATER? WHAT IS THE SPEE OF THE CURRENT

To solve this problem, we need to use the formula for the speed of an object in still water, which is the sum of the speed of the object relative to the current and the speed of the current itself.

Let's denote the speed of the boat in still water as 'b' and the speed of the current as 'c'. According to the problem, the boat traveled 336 miles downstream and back.

We know that the trip downstream took 12 hours, and the boat's speed relative to the current would be the boat's speed in still water (b) plus the speed of the current (c). Therefore, the equation for the downstream trip is:

336 = (b + c) * 12 -- equation 1

Similarly, the trip back took 14 hours, and the boat's speed relative to the current would be the boat's speed in still water (b) minus the speed of the current (c). Thus, the equation for the upstream trip is:

336 = (b - c) * 14 -- equation 2

We now have a system of two equations with two variables. Let's solve it:

First, simplify equation 1:

336 = 12b + 12c

Divide both sides of the equation by 12:

28 = b + c -- equation 3

Next, simplify equation 2:

336 = 14b - 14c

Divide both sides of the equation by 14:

24 = b - c -- equation 4

Now we have a system of equations 3 and 4, which we can solve simultaneously to find the values of 'b' and 'c'.

Add equations 3 and 4 together:

28 + 24 = (b + c) + (b - c)

52 = 2b

Divide both sides of the equation by 2:

b = 26

Now substitute the value of 'b' back into equation 3 or 4 to find the value of 'c'. Let's use equation 3:

28 = 26 + c

Subtract 26 from both sides of the equation:

c = 2

Therefore, the speed of the boat in still water is 26 mph, and the speed of the current is 2 mph.