 In this video, I'm going to talk about solving linear systems with three variables. So if you notice the system that you see here, we have three equations and three variables x's, y's, and z's. So I'm going to show you kind of some, hopefully show you a couple of strategies to use when solving these systems. One of the things that you need to do to effectively solve this system is you need to be able to organize everything that you're going to be using. So as I go through this, I will show you how to organize all of your equations so that you could solve for all three of these variables. OK. First thing we're going to do is you want all your variables on one side and numbers on the other. So that's how everything is set up here. What I want to do is I want to make sure and label these three equations. So one, two, and three simply is what I'm going to label these equations as. Now what I want to do is I'm going to use elimination to solve this system. Now when using elimination, just like with the system of two variables, you need to eliminate one of your variables. So as I look at this, I think right here, these z's would be the easiest to eliminate. So I'm going to take my second equation and I'm going to take my third equation. But I'm going to take the third equation and multiply it times a negative 1 so that I have a negative z here. And then when I add these two equations together, the z's will be eliminated. So here this looks like 2x plus 3y plus z is equal to 1. And the second, so 2 is rewritten as you see it. But equation 3, I'm going to multiply everything times a negative 1. So that's a positive 3x, a positive 4y, and a negative z is equal to a negative 4. There we go. So now when I add everything together, now when I add these two equations together, I get a 5x plus 7y and then plus 0 equals negative 3. The z's get eliminated in this case. So all I'm left with is now an x and a y. Now we're going to use this equation again, so we've got to keep track of it. I'm going to call this one equation 4. Now what I'm going to do is I'm going to do this process all over again, except for I'm going to use two different equations. And I'm going to eliminate the same variable that I did this first time. So in this case, I eliminated the z. So in the next two that I choose, I'm also going to eliminate the z. So this time, I'm going to use equations 1 and equation 2. Now what I have to do is this second equation, I need to multiply times a negative 2, so I have a positive 2z and a negative 2z. So that's what I'm going to do over here. Take my first equation and then the second equation, I'm going to multiply times negative 2. So the first equation, I'm going to rewrite as is. Negative x plus y plus 2z is equal to 7. And then I'm going to take my second equation and I'm going to multiply times negative 2. So I get a negative 4x minus 3 times 3 minus 9. Oh, no, I'm not multiplying times 3. This is why you need to organize everything so you remember what you're multiplying by. So negative 2 times 3 is a negative 6y. There we go, negative 6y. And negative 2z equals negative 2. All right, just double check, make sure I got those right. Yep. All right, so now, again, the idea is to add these two equations together. This is going to be a negative 5x plus 5y. The z's go away and I get 7 minus 2 is going to be 5. Okay, now again, this is going to be, I need to keep track of this, so this one's going to be my fifth equation, this will be my fifth one. Okay, now essentially what we've done, this fourth equation and this fifth equation, basically what I've done is I've turned this from a system of three variables into a system of two variables. Okay, notice the system of two variables, xy, xy. So now I'm going to take these two equations and I'm going to eliminate one more variable. And once I eliminate that one variable, I'll have only one left over and I will be able to solve for that one. Now notice here, I have a negative 5x and I have a positive 5x. So coincidentally, this will be really nice because now all I have to do is just add these two together. So I'm going to take equation four and equation five and I'm going to add them together, 5x plus 7y is equal to negative three. Negative 5x plus 5y is equal to five. And, and, and, now that's, you know what? As I look back through my work, I realized I did make a mistake. So you know what? I'm going to go back and fix it. Now, normally you think, well, why don't you just redo the video? Well, as you go through this, you need to always be double checking yourself. One thing I realized as I look back to my equation up here, this is a positive 5y, that's not supposed to be positive. It's supposed to be negative. Okay, so that's, I got to take that off there, take that off there. So that makes this a negative, negative 5y, okay. All right, now, and again, you always want to be going back and double checking that kind of stuff because if you make one tiny little mistake like that, it throws off everything from that point on. Okay, and as you, and we're, we're about halfway done with this problem. And, and if I make a mistake right here, the rest of the problem is going to be completely wrong. But anyway, it's a good thing, I do a good thing, I double check myself to make sure that that was right. Okay, so now, same thing, I'm going to add down my columns here to eliminate a variable. So eliminate the x's, I get 0, 7 minus 5 is a 2y, and then 3, negative 3 and a positive 5 make a positive 2. So in this case, y is equal to 1. Okay, so now, I have effectively, after all that, I've effectively solved for 1. I have solved for 1 of my variables. Now, what I need to do is I need to take this, and just like in a system of two equations, I need to take this and plug it back into one of my equations here. So in this case, I'm actually going to plug it back into 4. I'm not going to plug it back into 5 because I got a lot of negatives here. So I'm going to plug this back into 4, and I'm going to solve for x. Since if I take y and plug it back in, if I plug it in right here, x is going to be the only variable left. So I'm going to take y equals 1, and I'm going to plug it into equation number 4. Okay, so that's going to give you 5x plus 7 times 1 equals negative 3. Okay, and now solve this. This is a positive 7, so subtract that over. So 5x is equal to negative 10. 5x is equal to negative 10, so that means x is equal to negative 2. All right, so now what I've done is I've solved for another one of my variables. Now what I'm going to do is I'm going to take both of these. I'm going to take my y and my x, and I'm going to go back up to my original equations up here and try and solve for the z variable, which is the first one that I eliminated. Now in this case, what I'm going to do is I'm actually going to use the second equation here. Notice 2, 3, 1, and 1. I don't have any negatives or anything like that. So when I plug these two into number 2, I'm going to solve for z. I think this would be the easiest one to solve for. So if x is equal to negative 2 and y is equal to 1, I'm going to plug that into equation number 2. Plug that into equation number 2, and so that's going to be 2 times negative 2 plus 3 times 1 plus z is equal to 1. So this is a negative 4 and a positive 3, which makes a negative 1 plus z equals 1. So then z is equal to 2 after you add 1 over to the other side. All right, so there are my three results. So x is equal to negative 2, y is equal to positive 1, and z is equal to 2. OK, so that's just one example of solving a linear system in three variables. Now notice all the organization that we had to have here. You really need to be able to write down these 1s, 2s, and 3s and keep track of all the equations that you're using. And this is a really good strategy to use to simply labeling everything. And then also, right here, when you plug numbers back in to whatever equation, I get a keeping track of that. And then back up here for my final step to find my final variable, both of these I'm plugging into equation 2. It's really easy notation to read, and it's really easy to follow. The organization is what's going to get a lot of students on this. You need to be able to organize this so that you can effectively and efficiently and accurately solve these systems of equations. OK.