Question 206439
Let c be the speed of the current and b the speed of the boat in still water (we assume this speed is constant)
D=RT, distance = rate x time

the distance is 210 in each direction
the rate downstream (current pushing the boat forward) is b+c
the rate upstream (current dragging against the boat) is b-c
210 = (b+c) x 10   (downstream)
210 = (b-c) x 70   (upstream)    
dividing both sides of top equation by 10 and bottom equation by 70 we have
21 = b+c
3 = b-c     
adding the left & right sides of the top equation to L&R sides of bottom equation, we have
24 = 2b  (eliminating c)
b = 12 and
c=9, since 3 = 12-c
So, the speed of the boat in still water is 12 mph and the speed of the current is 9 mph