Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. System of equation and three variables. When we have 3 equations with three unknowns, we want to look to eliminate 1 variable throughout the entire equation. So when we look at this, there are some key concepts. We want to look for variables that have some terms that are positive and some that are negative. We also want one coefficients when possible. So looking at these three equations, I'm going to eliminate the YS. I'm going to call these equations 1-2 and three. If we combine equation one and two exactly the way they are and we add them, we're going to come up with the fact that the YS will cancel. So if I have one equation and equation 2 and add them together 3X, the YS go away +2 Z equal 14. We're going to call that a new equation. Let's call it equation 4. Now I need to use equation 3 with something, so I'm going to combine one with three, and I'm going to need to take equation 1 and multiply by two. If I take equation 1 and multiply it by two, we get 2X plus two y + 2 Z equal 10. We're going to leave the third equation exactly like it is, so X -, 2 Y plus three Z = 16. When we add these, we're going to end up with 3X. The YS are going to cancel plus 5Z equaling 26. This is going to be equation 5. Now we're going to combine four and five together. If I combine four and five together, we have three X + 2 Z equal 14 and three X + 5 Z equal 26. We want to multiply 1 or the other of the equations by a -1. So let's say multiply equation 4 by -1. If I multiply equation 4 by -1, we're going to get -3 X -2 Z equaling -14. When I combine that with equation 5, my XS are going to cancel and I'm going to get 3Z equaling 12. Well, to get Z by itself, we divide each side by three and we get Z equal 4. Once I know Z, we're going to substitute back in to equation four or five. We can't substitute it into one, two, or three yet because we have 3 variables in 1-2 and three, and we only know one variable. So if I substitute this back into equation four or five, it doesn't matter which three X + 2 instead of Z, I'm going to put in the number four. So three X + 8 equal 14, three X = 6 X equal 2. Now that we know X&Z, we can choose any of the original equations to substitute. So now substitute into one of the originals. It doesn't matter which one. If we substitute into #1, we'd have 2 + y + 4 equal five, 2 + y + 4 equal 5, or 6 + y equal 5 Y equal -1. We give our answer as an ordered triplet, and it always goes in alphabetical order. So in this case X then Y, then Z, or two negative 1/4. So that's an example of three variables with three unknowns, 3 variables, 3 equations, 3 unknowns, 3 unknowns, and three variables. I'm going to look to eliminate the same variable throughout the entire process. I want to look for some that have positive and negative coefficients, and perhaps some that have additive inverses already or have ones as a coefficient. So if I combine one and two directly the way they are, we'd get the YS to cancel and so we'd have 3X the YS go away plus 9Z equaling 27. Now we can see that in this equation, those are actually all divisible by three. So we could come up with an equivalent equation that says X + 3 Z equal 9. Let's call that number four. Now we need to go back to the originals and I have to combine equation 3. So let's do 2 with three because two already had a positive coefficient. If we combine two with three, we're going to take equation 3 and multiply it three by a three so that we get additive inverses. When we add that ten X + 14 Z equal 58. Now those are all even. We could divide by two. Five X + 7 Z is going to equal 29. Let's call that equation 5. When we combine equation 4 and equation 5, we're going to have to multiply to get additive inverses as one of the coefficients. So let's take equation 4 and multiply it by -5 negative five X -, 15, Z equal -45 five X + 7, Z equal 29. When I add those, I get -8 Z equaling -16 or Z equal 2. Once I find the Z, I'm going to substitute it back into equation four or five and it doesn't matter which I'm going to get X + 6 equaling 9 or X equal 3. Once I have X&Z, I'm going to substitute it in to one of the originals to find the Y. So 3 * X - y + 2 * Z equal twelve, 9 - y + 4 equal 1213 - y equal 12 negative Y equal -1 Y equal 1. We always put this as an order triplet in alphabetical order X, then Y, then Z, so 3, then one, then 2-3 variables and three equations. We're going to start by labeling them 1-2 and three. We're going to think about which variable we want to get rid of. We're looking for some that are positive and some that are negative with one coefficients. In this case, it's probably just as easy to do Y as it is ZI think I'll do Y just because the numbers are a little smaller. I'm going to combine equation one and two. When I combine one and two, I need to take equation 1 and multiply it all the way through by A2. When I take equation 2, I don't need to multiply it by anything because the Y was already A2Y. When I add these together, I get seven X -, C equal 10. That's my new equation. Let's call it 4. Now I need to go back to the originals and I'm going to combine one and three. This time. If I combine one and three, I need to multiply 1 by a three. So I get three X -, 3 Y plus 3Z equal 12. I'm going to combine that with the four X + 3 Y -4 Z equal -2. I get seven X -, Z equal 10. This is equation five. Well, the next thing to do is to combine four and five. If I combine four and five, we have seven X -, C equal 10 and seven X -, C equal 10. Well, we need to get rid of one or the other variables. Let's say we're going to get rid of the Z's. So I'm going to take one of those equations and multiply by a -1 negative seven X + Z equal -10 seven X -, Z equal 10. When I add these, I get 0 = 0. Well, 0 = 0 is a true statement, and so it means that these three equations are all going to coincide. They're all going to be the same along a line, or possibly they're all three the same plane. This is called a dependent equation because we got a true statement such as 0 equal 0, 3 variables. With three equations. We're going to start by labeling 1-2 and three. We're going to look to see what variable we might want to get rid of. We want to look for some that have positive variable coefficients and some with negatives, along with A1 if possible. So in this case we're going to eliminate the Y. I'm going to combine one and two. If I combine one and two, I need to take equation 1 and multiply it by three. So 12X plus three y + 3 Z equal 51. I'm going to leave the equation 2 the way it is so that when I add the two together, the -3 YS will cancel and the +3 Y. So thirteen X + 5 Z equal 43. That's going to be our equation 4. Now I'm going to combine equation one with equation 3. If I combine one with three, I'm going to have to multiply equation 1 by A2 8X plus two y + 2 Z equal 34. Going to leave equation 3 exactly like it is so that when we add them, the YS cancel and I get thirteen X + 5 Z equaling 39. This is going to be my equation five. Well, next step is going to be to combine four and five together. If I combine four and five together, I have thirteen X + 5 Z equal 43 and thirteen X + 5 Z equaling 39. Let's say I want to get rid of the X's. So I'm going to multiply 1 or the other equation through by a -1 negative 13 X -5 Z equal -4313 X plus 5Z equal 39. When I add these together, I get 0 equaling -4. Does 0 equal -4? No. So that actually tells me that there are no solutions to this problem. There is no XY and Z that can make all three of these equations true at the same time. It's inconsistent because there is no solution. 3 variables and three equations. When we want to solve for XY and Z, it's important that we pay attention to which variable we're going to choose to eliminate all the way throughout. Some key things to look for. I want some of the coefficients to be positive and others to be negative, and it's always nice if we have a one and somewhere in there if possible. In this example, we're going to eliminate the Z's because it's a +1 Z so I can multiply it by three and by 4 to eliminate the respective equations variables. So let's start by naming my equations 1-2 and three. I'm going to combine one and two, and when I combine one and two I need to multiply equation 2 by three, so two X -, y -, 3 Z equal -1. Now I'm going to take the entire equation 2 and multiply it by three six X -, 3 Y plus 3Z equal -27. When I add those together, I'm going to get eight X - 4 Y, the positive 3Z and -3 Z cancel equaling -28. Now we have 8 X -4 Y equal -28 for my new equation 4. But I can actually simplify that. I could take that entire equation and divide it by 4, and then it would still be an equivalent equation, but with smaller coefficients. So two X -, y equal -7 let's call this our new four. So now I'm going to go back to the original two, and I need to use equation 3. I'm going to combine two with three, and this time I'm going to multiply equation 2 by a +4. So eight X -, 4 Y plus 4Z equaling -36 The third equation I'm going to leave exactly like it is, remembering I need to eliminate the same variable. When I add here, I get nine X -, 2 Y. The Z's are going to cancel, and that's going to equal -19. That's my new equation 5. So now we're going to combine four and five together. If I combine 4:00 and 5:00, I've got two X -, y equal -7, and I've got nine X -, 2 Y equal -19. Let's take equation 4 and multiply it by a -2. If I multiply it by a -2, remember to change all the signs all the way through the equation. So two X -, y equal -7 turns into -4 X +2, Y equal +14. Leave the other equation exactly the same so that when I add them, the YS are going to cancel and I'm going to get 5X equaling -5 or X equal -1. Once I get my X, then I'm going to substitute it into four or five. I need to substitute it into an equation that only had two variables. Substitute into equation four or five. It doesn't matter which one. Once I do this, I'm going to be able to find my Y. So 2 * -1 -, y equal -7 negative 2 -, y equal -7 add 2 to each side, so negative Y equal -5 or Y equal five. Once I know two of my variables, now we're going to substitute into an original equation to get the third unknown. It doesn't matter which one anyone you want. So let's say 2 * -1 - 5 - 3 Z equal -1 negative, 2 - 5 - 3 Z equal -1 negative 7 - 3 Z equal -1. We're going to add 7 to each side and Z is going to equal -2. We're going to give our answer is an order triplet always in alphabetical order, so X, then Y, then Z. To be able to determine if we really did it correctly and didn't make any mathematical errors, we can always check it. To check it, we would actually put it into all three equations. Does 2 * -1 - 5 -3 * -2 really equal -1 so -2 - 5 + 6? And that does indeed check. We need to check it in all three equations. So 2 * -1 - 5 + -2 does that equal -9 negative 2 - 5 - 2 does indeed equal -9. And to check the third equation -1 + 2 * 5 - 4 * -2 = 17 negative 1 + 10 + 8 does indeed equal 17. So we know we absolutely had the right equation just by doing a quick check into all three equations. When having three variables with three unknowns, we want to label our equations and then we want to determine which variable to eliminate. Well, some of these equations don't have all three variables, so I'm going to write them with spaces. For the ones that don't have a variable in existence, we want to figure out which variable to eliminate all the way through. There is an easy method to do this. We want to choose one of the two variables that already has one missing. We also want to think about is 1 positive and one negative or similar coefficients. So we're going to eliminate the Z's in this problem. I'm going to combine equation one and three. So if I take one and three and combine them, we're going to end up with the Z's canceling. So we're going to have two X + y equal 63. This is going to be our new equation, let's call it four. Now if you look, equation 4 has the exact same variables as equation 2, so we can immediately go and combine two with four. So equation two and four, I have negative two X + y = 3 and equation 2 and I have positive two X + y = 63. If I add them just like they are, the X's will cancel. We get two y = 66 or Y is 33. Once I find the Y, I'm going to substitute it back into equation two or four. It doesn't matter. So negative two X + 33 equal 3, subtract the 33 negative 2X equal -30 X equal 15. Once I get X&YI can I can substitute it back into any of the originals? Well if we look at equation 3, equation three had X's and Z's. So now we could solve just by knowing our X and our Z. We get Z equaling 9. So the ordered triplet always given in alphabetical order X then Y, then Z is going to be 1533 and nine. We could always check it if we were concerned whether our math was correct or not. Does 15 + 33 + 9 equal 57? 33 and 15 is 4848 and nine is 57. That one works. Does -2 * 15 + 33 equal three? Well -30 + 33 does indeed equal 3. And finally does 15 - 9 = 6? And the answer to that one is also yes. Thank you and have a wonderful.