Let c be the speed of the current and b the speed of the boat in still water (I assume is constant rate)D=RT, distance = rate x timethe distance is 210 in each directionthe rate downstream (current pushing the boat forward) is b+cthe rate upstream (current dragging against the boat) is b-c210 = (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 have21 = b+c3 = b-cadding the left & right sides of the top equation to L&R sides of bottom equation, we have24 = 2b (eliminating c)b = 12 andc=9, since 3 = 12-cSo, the speed of the boat in still water is 12 mph and the speed of the current is 9 mph