Let xx be the width of the river. It is given that, when they have met the first time, boat11 has travelled 14001400 m and boat22, (1400−x)(1400−x) metres.
When they meet for the second time, boat11 has travelled 600+(x−1400)=(x−800)600+(x−1400)=(x−800) m, and boat22, 1400+(x−600)=(x+800)1400+(x−600)=(x+800) m.
Note that, the relationship of the distances each traveled, is the same to both meetings. Thus,
14001400−x=x−800x+80014001400−x=x−800x+800⇒x=?⇒x=?Hope you can take it from here.