 In this video, we're going to talk about how we can solve a system with three equations and three unknowns using the elimination method. We've learned previously how we can eliminate a 2 by 2 system, and the process is basically the same thing. We have to decide on a variable we want to eliminate, and I'm going to eliminate the first variable x. Now, to do that, you have to pick some pairs, right? So we combine two equations together. So take the first and second equation, for example. We have to adjust the coefficients so that x cancels out, and I'll do that by times the first equation by negative 2. That would look like negative 2x plus 4y plus 2z. You'll notice that my z's are actually crossed so that they don't look identical to my 2. Sorry about my pimmonship there. And then you get negative 2 times 8, which is a negative 16. Then the second equation, we don't have to adjust it at all. We get a 2x minus a 3y plus a 1z, you know, what babies wear, and a 23 here. And we're going to add these together. So because of our choice of coefficients, the 2x's are going to cancel out. The y's, we get 4y minus 3y, which is a y. 2z plus z is a 3z, and then negative 16 plus 23 is a 7. So if we combine together the first and second equations, we can get that by, so we adjust it, we adjust the first equation, combine those together, right? So what we're going to do next is we have to then find a second equation, a second pair of equations, I should say, so that we can eliminate x again. So what if we eliminate x by taking the first equation and the third equation together? And what we're going to do is we're going to adjust things by times the first equation by negative 4. This is going to give us negative 4x plus 8y plus 4z, and then negative 4 times 8 is going to be negative 32 right here. Now we're going to just take the third equation as it was, 4x minus 5y plus 5z is equal to 53, and we want to add these together. Because of the adjustments of the coefficients, the x's are going to cancel out yet again. We're going to get an 8y minus a 5y, which is a 3y. We're going to get a 4z plus a 5z, which is a 9z, and then 53 take away 32 as gives us a 21. And so by eliminating x, you have to do two different pairs of equations. We did the first and the second, we did the first and the third. You can also do the second and the third if you want to, but we only need two pairs to make this work. And so when we put these equations together now, we have a new system of equations that only involves y and z. So we get y plus 3z is equal to 7, and we're going to get 3y plus 9z is equal to 21. And so then what we're going to do is in order to solve this system, it's a 2 by 2 system, we would solve it by any method we feel comfortable with. We could continue to eliminate this system right here. You know, we could eliminate y, we could eliminate z, whatever we want. We could solve it by substitution. We could switch it up and switch to substitution. It doesn't matter. I'm just going to stick with elimination in this one. Let's try to eliminate the y-coordinate. I can accomplish that by timesing the first equation by negative 3. And so what I need to consider here, what I need to consider here is you're going to get negative 3y minus 9z is equal to negative 21. That's just the second equation just times by negative 1. And so when we add this together, you see that everything's going to cancel out. You're going to end up with 0 over 0. This is the dependent case. We're going to have infinitely many solutions, and we should write the solution in general form. So now that we've done this, we need to go back and actually solve for the dependent variables, right? So we're going to treat z here as our free variable. Moving into the other side of the equation, we see that y is equal to 7 minus 3z. So like we talked about before, our variable z here is a free variable. We can choose z to be whatever we want, and that will be a solution to this system of equations. Now, y, on the other hand, is going to be a dependent variable. It will depend on our choice of z. So for example, if we choose z to be 0, then y has to be 7. If we choose z to be 1, then y has to be 4. There's that dependence relationship that's going to be between them. Well, what about the variable x? Can we back up here, right? If we have to sort of make a dependence relationship going on there as well, pick one of the equations you want, you know, maybe like the first one, right? So we get x is equal to 2y plus z plus 8. Now, z is a free variable. It can be whatever it wants to be, right? It's like queen whatever here from the Lego movie part 2. She can be whatever she wants to be, z plus 8. And then y, y is a dependent variable. It depends on our choice of z. And like we saw before, y, remember, is just 7 minus 3z. So we can combine these terms together and get x equals 14 minus 6z plus z plus 8. And so combining like terms here, we're going to get 8 and 14, which is a 22. Negative 6z plus z is a negative 5z. And so we see that x is likewise going to be a dependent variable. It depends on the choice of z. And so when we record this all together, our general solution is going to look something like the following. The solution to the system of equations looks like, well, x is going to be 22 minus 5z, y is going to be 7 minus 3z, and then z is whatever she wants to be there. And so this would give us the general solution to the system of linear equations, which we found using the elimination method, which could be adapted for a 3 by 3 system, a 4 by 4, or however many variables you want.