# A boat goes 50 miles upstream in 3 hours and later returns to the starting point in 2 hours. What is the speed of the boat in still water and what is the speed of the current?

The boat's speed is 20 5/6 mph, and the river's speed is 4 1/6 mph. Here's how we solve it: In going upstream, the boat moves against the river's current, so the speed of the current is subtracted from the speed of the boat. The opposite it true for it travel downstream, because in that instance, the speed of the river is added to the speed of the boat. If we us B as the speed of the boat and R as the speed of the river, then in moving upstream 50 miles in 3 hours, 3 times the boat's speed minus 3 times the river's speed is 50 miles. Downstream, the 50 mile trip took 2 hours, so 2 times the boat's speed plus 2 times the river's speed is 50 miles. Here are the equations. 3B - 3R = 50 2B + 2R = 50 When we have two unique equations, which we have, that contain two variables, which we also have (the speed of the boat and the speed of the river), we can solve by the method of simultaneous equations. In this method, we "add" the two equations together algebraically. Our goal in this, however, must be to make one of the variable "drop out" or "cancel out" so that only one variable remains. If we multiply the first equation by 2 and the second equation by 3, we'll end up with a "-6R" term in the first equation, and a "+6R" term in the second equation. When these two terms are added algebraically, they'll "disappear" and leave a single variable in the resulting equation. Let's multiply as we suggested. 2 (3B - 3R = 50) = 6B - 6R = 100 3 (2B + 2R = 50) = 6B + 6R = 150 Adding the two equations together algebraically will now make one term "go away" and we can solve for the other variable. Here's how it looks when we add: (6B + 6B) + (-6R + 6R) = (100 + 150) 12B = 250 [Notice that the R term has "cancelled out" and we are left with one variable.] Solving for B, the speed of the boar, we have this: 12B = 250 B = 20 5/6 miles per hour Returning with that to one equation, we can solve for the other variable. Let's plug it into the second equation. 2B + 2R = 50 2(20 5/6) + 2R = 50 41 2/3 + 2R = 50 2R = 50 - 41 2/3 2R = 8 1/3 R = 4 1/6 mph [This is the speed of the river's current.] Let's check by substituting both answer back into the first equation. 3B - 3R = 50 3(20 5/6) - 3(4 1/6) = 50 (62 1/2) - (12 1/2) = 50 50 = 50 [Our work checks, and the speeds of the boat and current are correct. Note that we could have arrived at the same answer by multiplying the first equation through by -2 and getting -6B + 6R = -100, and then multiplying the second equation through by 3 and getting 6B + 6R = 150 for our two equations. Then adding them algebraically, we get 12R = 50 and our B term has dropped out. Solving for this to find the speed of the current we get R = 4 1/6 mph, which is identical to what we got when we worked it through the first time.