A boat traveled 180 miles downstream and back. The trip downstream took 10 hours. The trip back took 30 hours. Find the speed of the boat in still water and the speed of the current.

Solve each system using any method

d=v*t

180=(vb-vs)*30
180=(vb+Vs)*10

6=vb-vs
18=vb+Vs

24=2Vb
vb=12 mil/hr
vs=you do that.

uk some kind of elaboration would be greatly appreciated by many.

other than that its cool

To solve this problem, let's define variables to represent the unknowns:

Let's call the speed of the boat in still water "b" and the speed of the current "c".

Now, let's analyze the given information:

1. On the downstream trip, the boat traveled 180 miles in 10 hours. In this case, the boat's speed is increased by the speed of the current, so the effective speed is (b + c).

2. On the upstream trip, the boat traveled the same 180 miles but took 30 hours. In this case, the boat's speed is decreased by the speed of the current, so the effective speed is (b - c).

Now, we can set up a system of equations:

1. For the downstream trip: distance = speed × time
180 = (b + c) × 10

2. For the upstream trip: distance = speed × time
180 = (b - c) × 30

Now, we can solve this system of equations using any method like substitution or elimination.

Let's use the method of substitution:

From equation 1, we have:
b + c = 18 [Equation A]

From equation 2, we have:
b - c = 6 [Equation B]

Now, we can solve this system of equations by adding equation B to equation A:

(b + c) + (b - c) = 18 + 6
2b = 24

Dividing both sides by 2, we get:
b = 12

Now, substitute the value of b (12) into Equation A to solve for c:

12 + c = 18
c = 18 - 12
c = 6

Therefore, the speed of the boat in still water is 12 mph, and the speed of the current is 6 mph.