 So let's take a trip back to ancient China about the 1st or 2nd century AD, and look at something called the Fang Cheng Xu, the Fang Cheng method. And so this concerns the problem of solving a system of linear equations. And you learned how to solve linear equations at some point in the ancient past, and if we lived in a kinder, gentler universe, then our solutions to linear equations would always be integers, and as we worked our way through linear equations, we'd always end up with integer coefficients. We don't live in that universe, and so when we have this type of system, we almost always have to deal with fractional coefficients. Now fractions are tremendous amounts of fun, but sometimes they are difficult to work with, and so one of the questions we might ask is can we avoid them for as long as possible? And this is where the Fang Cheng method shows up. So this is a method used in ancient China about the 1st or 2nd century AD, and it's known as the Fang Cheng Xu, which translates as a rectangular tabulation method. And it's based on the following idea. If I have two equations and some ordering of the unknowns, I can multiply the first equation by the leading coefficient of the second, and the second equation by the negative leading coefficient of the first. When I add, that eliminates the leading coefficient. So for example, let's say I have the system of equations or the two equations at least, a11, x, and some other stuff, and then a21x and some other stuff. And if I multiply the first equation by a21, that's the coefficient of the second equation, and I multiply the second equation by negative a11, that's the negative of the coefficient of the first. If I then add the two equations, these coefficients are equal but opposite, so they drop out, and then I get an equation in y and z. And so I get a much nicer equation because I go from equations with three variables to equations with two variables. Well, let's do that in actual practice. So let's take a look at our system of equation, and it's helpful to do something I call keep and clone. What I'm going to do is I'm going to keep all of the working equations, but then I'll make a copy, and I'll use those copies to eliminate the leading variables in the other equations. So let's go ahead and work with this first and second equation. So I'll multiply the first equation by the coefficient of the leading variable as a second. So that's going to be, I'm going to multiply everything here by eight. That's my leading coefficients. That gets me 24x plus 40y plus 16z equals 40. And then I'll multiply the second equation by negative three, the negative coefficient of the leading variable in the other equation. So that gets me minus 24x plus 9y and so on. So this equation here is being multiplied by negative three. And now I have two equations, and the x coefficients are equal, but opposite. So when I add, they drop out. So I add those two equations. I get this new equation here. Likewise, I can work with the first and third equation. So I'll multiply the first equation by two, the leading coefficient of the third equation. So that gets me that. I'll multiply the third equation by negative three. That's the leading coefficient of my first. And again, what that does is that gets my x coefficient to be equal, but opposite. So when I add them, that x coefficient drops out. Now I have three variables. So I actually only need three equations. And since two of the equations only have two variables, this one and this one, I'm going to keep those as being somewhat simpler. And so I'll keep my first equation, which I haven't changed at all. My equation, my second equation, I'll pick this one and my third equation, I'll pick this one. And I don't need the others. So now I have one, two, three simpler equations in the same unknowns. And because everything I did is allowable under the rules of algebra, the solutions of my original system and the solutions to the new system are the same. So now I can continue using the simpler system of equations. And I'll do the same thing as before. So I have my second equation and my third equation. These have two variables. So I'll multiply my second equation by the coefficient, the leading coefficient of the third. And my third equation by the negative of the leading coefficient of the second. So I'll multiply the second equation by 22. That gets me this equation. I'll multiply my third equation by negative 49. That gets me this equation. My coefficients of y are equal but opposite. So when I add, the y variable drops out and I'm left with just the z variable. And once again, I only need three of these equations. So I'll choose the three simplest, these three. And what I've done is I've transformed my original system of equations into a new system of equations that looks like this. And I can't avoid fractions forever, but I can limit their impact to just the last step. So I can solve this system using back substitution. I can start with the simplest equation here. 1266z equals 3177. And so that gets me z equals 3177 over 1266. And if we do take the time to reduce it, that gets me 1059 over 422. Well, I have my second equation, 49y plus 13z equals 13. I know what z is. I'll drop that in. There's my substitution and I solve for y. And again, now I have my third equation, my first equation actually. And so I know what y is. I know what z is. So I can substitute those back into the equation and solve for x. And there's my solution.