I am not good in math so I took your problem to a friend of mines...these are his results...I hope it helps.

Let the width of the river be d.
Let the speed of boat A be a
and the speed of boat B be b.
Let t1 and t2 be the time they meet the first time and the second.
...

I am not good in math so I took your problem to a friend of mines...these are his results...I hope it helps.

Let the width of the river be d.
Let the speed of boat A be a
and the speed of boat B be b.
Let t1 and t2 be the time they meet the first time and the second.

A travels 700 m in t1 and continue to travel d-700 + 300 m in t2 and meet with B.
B travels d-700 m in t1 and continue to travel 700 + (d-300) in t2 and meet with A.

For boat A after t1
t1 = 700/a
For boat B after t1
t1 = (d-700)/b

700/a = (d-700)/b ---------------------------------(1)

For boat A after t2
t2 = (d-700 + 300 )/a = (d-400)/a

For boat B after t2
t2 = [700 + (d-300)]/b = (d+400)/b

(d-400)/a = (d+400)/b ---------------------------(2)


Equation (1) ÷ (2)

[700/a] / [(d-400)/a] = [(d-700)/b] / [(d+400)/b]

canceling a and b you have

[700] / [(d-400)] = [(d-700)] / [(d+400)]
d²-1100d+2800 = 700d+2800
d²-1800d = 0
d = 1800


The river has a width of 1800 m