 So this is a little bit of speculation, but the use of the counting board might have led to algebra as follows. So remember, arithmetic is bookkeeping. And so we can use the counting board for arithmetic as follows. The columns of the counting board might represent different arithmetic units. Ones, tens, hundreds, and so on. But algebra is generalized arithmetic, and so they could instead represent different algebraic units. A first variable, x, a second variable, y, a third variable, z, and so on. And this allows us to represent a system of equations on the counting board. So for example, we might be able to represent the equation 3x plus 7y plus 2z equals 15 on a counting board. Now the Chinese didn't, but it's convenient to label our different columns with our different variables, x, y, z, and the number. And arithmetic is bookkeeping. Algebra is generalized arithmetic. We want to keep track of how many of which units. So we have three of the axes. We have seven of the y's, two z's, and our number is 15. And for convenience, we're not going to split that up into two different columns. Or let's say I have a counting board representation of a system of equations. So assuming that our first, second, and third columns correspond to x, y, and z, let's interpret these equations. So our first row shows we have 2 of x, 1 of y, 2 of z, and the number 5. So the equation is 2x plus 1y plus 2z equals 5. The second row shows we have no x's, but we do have a y and 2z's and a 7. So our equation is y plus 2z equals 7. And the third row shows a z and a 2, and so the equation is z equals 2. So the Chinese also recorded negative amounts on the counting board. Positive coefficients used red counting rods, and negative coefficients used black counting rods. So we might represent the equation 3x minus 2y plus 12z equals 38. So we'll set up our counting board and label the columns with x, y, z, and the number. We have three axes. We have minus 2y's, so we'll put down two black counting rods. We'll put down a 12, and again we're not going to split the columns, and a 38. To solve systems of equations, the Chinese used an approach known as Fang Cheng Xu. This should be viewed as being done on the counting board, but we'll demonstrate using more familiar numbers. So let's try to solve a system with three variables. So setting up our array, our first equation, 3x plus 1y plus 2z equals 33, gives us the row 3, 1, 2, 33. The second equation, 2x plus 2y plus 3z equals 32, gives us the row 2, 2, 3, 32. And the third equation will give us the row 1, 1, 5, 23. Now, here's a bit of modern terminology that will make the process easier. The non-zero number in the upper left-hand corner is called pivot. So we'll multiply every number in the second and all other rows by the pivot. So now we'll subtract the first row from the second. So 6 minus 3 gives us 3, 6 minus 1 gives us 5, 9 minus 2 is 7, and 96 minus 33 is 63. Since our leading entry in that second row isn't 0, we'll subtract again. 3 minus 3, 5 minus 1, 7 minus 2, 63 minus 33. And now we'll subtract the first row from the third row. So now onto the second row. The second row pivot is 4, so multiplying the third row by 4. Now we'll subtract the second row from the third once, twice, and now let's look at our results. The last row gives us the equation 42z equals 84, which we can solve, telling us z is equal to 2. The second row gives us the equation 4y plus 5z equals 30, but we already know what z is, so we can substitute and solve. And finally, our first row gives us the equation 3x plus y plus 2z equals 33. We know what y and z are, so we'll substitute. So while this seems to be a cumbersome process, remember that this is actually being done on a counting board. So here's a quick view of what that might look like. We'll set up our equations. So our first equation 3x, 1y, 2z, and 33. Our second equation 2x, 2y, 3z, and 32. And our third equation 1x, 1y, 5z, and 23. Now our first row pivot is 3, which means we're going to want three copies of each of the other two rows. So we'll just put down copies of the other rows. Now we want to subtract the first row from the second. So if you look at the first row, it tells us exactly the sorts of pieces we should be removing from the second row. So we'll want to remove three rods from the first column entry of the second row, then one rod from the second column entry, two rods from the third column entry, and three horizontal and three vertical from the fourth column. And since we still have rods in that first column, we'll remove them again. Cell will remove three from the first column, one from the second, two from the third, and three horizontal, three vertical from the fourth. And we'll do the same subtraction with the third row. Now it's not critical, but we could clean this up a bit. So we have a 10 here. So we can replace these with a 10 rod. And there's more than five rods here, so we can replace it with a five rod. Now when we wrote this out, we multiplied the third row by the second row pivot, which is 4. But the purpose of the multiplication is so that we can subtract the second row from the third row. And notice that if we just duplicate the third row, we'll be able to do that. And the second row tells us how much we're going to remove from the third row, so we're moving. Again, we can simplify. There are two fives here. That makes a 10. And here's a nice little thing we can do. This two horizontal, one vertical is actually present twice in the last column. And that means we can actually do the division, because for everything in the third column, we have two in the fourth column. So as before, we can read this last line as saying z is equal to two. But what that also means is that we can remove one z and two numbers at the same time and still have equality. So let's break up some of those 10s so we can do this removal. And we'll remove one z and two number at the same time. We'll remove one z and two number and keep going. I can reduce, make change, reduce again, which gives me y equals five. And again, as before, I can remove one z and two numbers or one y and five numbers. So let's do that and we obtain our solution.