 Let's talk about parameterizing the solutions to a system of equations. And so if I have a relationship between two or more variables, I might be able to parameterize the variables. I might be able to express some of them in terms of new variables, which are called the parameters. So for example, let's take the system of equations 3x plus 4y minus equals 5, 2x minus y equals 0, and a parameterization of the solutions. Again, I have three variables x, y, and z. So what I might do is I might try to set up new variables, and maybe I'll say x equals s, y equals, so I'd say t, z equals r, and there is a parameterization where I've expressed my variables x, y, and z in terms of my parameters, s, t, and r. Well, this is a parameterization. It's actually pretty useless because all I've done is I've changed the names of the variables. Also, we can go a little bit farther. Remember that a parameterization isn't complete until I can tell you what s, t, and r could be, and the problem is that there are certain restrictions on what they can be because not all values of x, y, and z will solve this system of equations. So this is a useless parameterization. So let's see if we can construct one that's actually useful. So here's a nice general rule of life and also of mathematics. The more powerful something is, the more it's going to cost you. And this cost can show up in one of two ways. It may cost you in terms of how much effort you have to put into setting it up. But then the other thing is that it may cost you in that you might not be able to do as much after you've used it. So we'll take a look at both instances as the course goes on. So parameters are extremely powerful. They allow us to express several things in terms of just one thing. And so they should be viewed as being very expensive. And what we want to do is we want to use as few of them as possible. And wherever we can, we want to try and recycle them. So let's think about this. If an equation has two variables, I can make one of them a parameter. But then I can solve and express the other variable in terms of that parameter. And what this suggests is a general strategy. So I might proceed as follows. If I can find an equation with just two unknowns, what I can do is express one of them in terms of a parameter and solve for the other one in terms of that same parameter. And otherwise, if I can't find an equation with just two unknowns, I'll find an equation with the fewest number of unknowns. I'll choose parameters for all but one of those variables and solve for the remaining unknown. And now that I have a parameterized set of variables, if I have other equations, I can substitute in the parameterized versions of the variables wherever I can. And if I parameterized all variables at this point, I'm done. Otherwise, I'll go back to set one and look for an equation with just two unknowns. And importantly, these two unknowns are distinct at this point from the things that we parameterize. So for example, let's say I have this system of variables, system of equations, and I want to use as few parameters as possible. So the equation has two variables, the last equation has two variables, so I can start by parameterizing one of those variables. So let x equals s to avoid fractions, and they'll solve the equation for y. To s minus y equals three, I'll solve that for y. And so I have y. My variable y is now in terms of s. My variable x is also in terms of s. And now I have this variable z left over. Let's see what I can do with that. Well, I have two variables expressed in terms of parameters. I can substitute those into the first equation. That's 3x, 3s, 4y, 2s minus 3. z is the last remaining variable, and I can actually solve this equation for z. And so now I have z expressed also in terms of my parameters. And so now whatever x is, I know what y is, I know what z is, and I now have one, two, all three variables expressed in terms of the parameterization of s. Well, I don't have to use that particular parameterization. What if I like fractions? So I can still start with the equation, which has just two variables and choose one of them to be our parameter. So I'll let y equals s, for example, and my second equation, 2x minus y equals 3. Well, that becomes 2x minus s equals 3, and I'll solve this for x. And well, I do introduce fractions, but we like fractions, or maybe we like fractions. Now I have y equals s, x equals 3, plus s over 2. I can drop those into the first equation. I'll get my new equation. The only variable I have here is z. Again, s is a parameter, and we'll consider that to be not a variable. And that's all for z, 11s minus 1 over 2. And there's my parameterization, same system of equations, radically different parameterization. And again, the reason that we end up with this parameterization is we chose y to be s instead of x. Well, let's take a look at another important example, find a parameterization for the variables in this system. So again, the thing to notice here is that there's no equation that has just two variables. So what can we do? Well, for now, what we can do is we can start off by picking the equation with the fewest number of variables. So that would be the second equation here. And I can make two of these variables parameters and then solve for the third variable. Now, for example, this equation has variables x, z, and w. So maybe I'll pick z and w to be my parameters. If z is s, w is t, I can substitute those in and solve for x. I get x equals minus one half s minus one third t. And it's a perfectly usable parameterization, but maybe I want to avoid having those fractions there because I know, looking forward, I'm going to have to drop them into here and do more algebra with them. So maybe I'll try something else. I know that I'm taking one half z, one third w. So maybe I'll make z and w things I can take half and a third of very easily. So maybe I'll let z equals two s, w equals three t. So I'll drop those into my equation. I get x plus s plus t equals zero. So x equals minus s minus t. Now, I have my other equation here. I have a parameterization of z, w, and x. That takes care of z, w, and x. I still have two variables left, y and v. So I need to let one of them be a parameter. So again, because I want to avoid fractions, I'll let y equal r. So I drop those in 3x. There's my parameterization for x, 3x, for y, y is r. So that's going to become 4r, z is 2s, w, 3t, v is still a variable. So I'll leave that. And I will expand and solve. And after all the dot settles, I get v equals s minus 4r. And so now I have my parameterization. x equals s minus t, y equals r, z equals 2s, w equals 3t, v equals s minus 4r. And so there's my parameterization.