How wide is the river?

[Eugenia Gabrielov]

Edit: SORRY! I tried, it seems it's not followable. Maybe I just think funny.

So then, before I add the 10 minute thingy...Let's look at what happens with Boat #1 again.

To review, it goes 720 yards at velocity 1 or v1. it goes 720 + e yards (the whole river) in v1 (TA + TC).  Then it stops for 10 minutes. Then it makes it another 400 yards before it meets up with Boat #2. So the equation goes as such. But what would happen if we kinda...ruled out those 10 minutes for one second? Let's say TF is the time it takes to make those 400 yards

Distance = velocity * time

1120 + E = v1 (TA + TC + TF), and then we'll have to factor in those 10 minutes in a moment.

So what happens to Boat #2?

It makes it E yards. Then it makes it 720 yards. Then it makes it G yards. Forgetting the 10 extra loadage minutes for a second, let's say...

720 + E + G = v2 (TA + TB + TG)

So what exactly is G? G is R - 400, cause it's the distance traveled discluding 400 yards of river..  G is 720 + E - 400 which is equal to 320 + E. Therefore, 

720 + E + 320 + E = 1040 + 2E

1040 + 2E = v2 ( TA + TB + TG), later factoring in 10 minutes of passage time, and...
1120 + E = v1 (TA + TC + TF), again, I'm still not sure precisely how to factor those next 10 minutes in.

[Denman]

it is a perfectly acceptable method.
originally we would use it to see if something is true. say you have 2 equations with the same variables in both, dividing through allows you to see if its true,, as you should get 1=1 or something similar