Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. When solving system of equations in three variables, we want to look for coefficients with A1 and things that have positives and negatives. So in this first example, I'm going to look to eliminate the Y's because I have a positive Y and negative Y and a -2 Y. In the next example, I might go, oh, in this case, I might go again with the YS because I have a -3 YA +3 Y and a negative Y. So let's go back and solve this. I'm going to label the equations 1-2 and three because I'm going to need to combine two of them and then two more and take those results to combine together. So if I look at one and two directly, I can see that if I just add 1 and 2 exactly like they are, the YS will go away. So X + y + Z equal 5, two X -, y + Z equal 9. So now I get a new equation 3X plus 2Z equaling 14. Let's call this equation 4. Equation 4 has 2 variables in it. Going back, I now need to use equation 3 and I can combine it with either one or two. I'm going to choose one because one had a positive Y and three has a -2 Y, so I'm going to take one and three, and I'm going to multiply equation 1 by a 2. So I'm going to get two X + 2 I +2 Z equaling 10. Combining that with X -, 2 Y plus three Z = 16. When I add those, I can see the YS are going to cancel three X + 5 Z equal 26. Let's call that equation 5. So when I look at equation 4 and equation 5, if I've done this correctly, I will have the same variables in four as I do in five. If you realize that one has an X and AZ and the other one may be AY and AZ, you didn't keep and solve the same variables when you were doing the elimination the first time. So now we're going to combine four and five just like any 2 variable elimination. So once again, we want additive inverses. So I'm going to take that equation 4 and multiply everything through by a -1. We're going to combine it with equation 5, leaving it just the same because I had a negative 3X and a positive 3X. Those will cancel, leaving ME3Z equaling 12 or Z equal 4. Once I get ZI can substitute it into either four or five, it doesn't matter. I think I'll do four and solve for the X. So three X + 2 * 4 equaling 14. So three X + 8, subtract 8 from each side, divide by three, I get X equaling 2. Once I have X and ZI can substitute it into any of the original equations to find my Y, and it doesn't matter which one I substitute it into, I'm going to use one. So Y is going to equal -1. Now, I recommend that we always, always always check these. So does 2 + -1 + 4 really equal 5? Yes. Does 2 * 2 -, a negative one or plus 1 + 4 equal 9? And the answer's yes to that also. And does two -2 * -1 would be a positive 2, and 3 * 4 would be 12. So does 2 + 2 + 12 = 16, the answer is yes. We're going to give the answer as an ordered triplet in alphabetical order X, then Y, then Z. I'm going to go ahead and let you do #2. We're going to jump down and look at #3 well, here I think maybe we'll solve for the Z's this time, just to show a different variable being solved for. Let's label them 1-2 and three again. So I'm going to take equation 1:00 and 2:00. And if I do one and two, I need to multiply equation 1 by a three. So we get three X -, 3 Y plus 3Z equaling 12, and we have 5X plus two y -, 3 Z equaling 2. When we add these, I get eight X -, y equaling 14. Remember, we're going to call that equation 4. So now I'm going to go back and I'm going to use one and three this time. And I want one because it had a coefficient of 1. So I can multiply that equation through by a four and use it fairly easily. Four X -, 4 Y plus four Z = 16. Remember, I have to keep eliminating the same variable, so I want to get rid of the Z again. When I add these together, I get eight X -, y equaling 14. That's my equation 5. So when I combine equations 4:00 and 5:00, eight X -, y equal 14. Let's multiply equation 5 by a negative. So negative 8X plus Y equal -14 zero equals zero. 0 = 0 is a true statement, so there's infinitely many solutions. Now, we don't know if they're the same plane or if they're a line segment. All we really know at this point is that there's infinitely many solutions. So we could refer to that even as consistent equations. Looking at #4 as the next example, I'm going to go back and solve the why's because I have a one coefficient on this top 1 and I have some positives and some negatives doing the exact same steps as before. So we're going to have 1-2 and three. I'm going to take equation 1 and multiply it all the way through by three. So 12X plus three. Y + 3 Z equaling 51 X -3 Y +2 Z equal -8. When I add this, I get thirteen X + 5 Z equaling 40 3. That's going to be our equation 4. So this was one with two and now we're going to do 1 and 3. So I'm going to multiply equation one now through by a 2. So 8X plus two y + 2 Z equal 34 and five X -, 2 Y plus 3Z equal 5 S 13 X. The YS cancel 5Z equaling 39. That's equation 5. I'm going to combine four and five, and when I do this, remember I need additive inverses. So thirteen X + 5 Z equal 43 and -13 X -5 Z equaling -39. When I add these I get 0 equaling 4. This is no solutions because that's not a true statement. 0 does not equal 4, it never equals 4. So these could be parallel planes, or two of the planes might intersect and two of the planes might intersect, but not all three at the same time. I'm going to go ahead and let you practice #5 I want to look at six because six has actually got only two variables in two of the equations. Now it doesn't matter which one we choose, but we want to consider getting one of the equations that already is missing a variable to be combined with one of the other ones. So sometimes I rewrite them so that I can see them lined up. If I look here, I can see that there is no Z in the second equation. So if I could combine equations one and three and get rid of the Z, I'll have a new equation 4 that only has X's and Y's in it. And then I could combine that directly with equation 2. So let's go ahead and combine one and three. If I add one and three, I'm going to get X + y + C equaling 57 and X -, C equaling 6. So two X + y is going to equal 63 and that's my equation 4. So now I can combine 2 directly with four. Remember we need additive inverses so negative two X + y equal 3 and two X + y = 63 so we get 2Y equaling 66 Y is 33. Once I get the Yi can substitute it back in to get the X. So two X + 33 = 63. Two X equal 30X is 15 and once I get the XI can put it back in to any of the equations I want to get the Z. So 15 -, Z = 6 Z is going to equal 9 so I put that one into equation 3 and back here I substituted this one into equation 4 so make sure to check them. But 1533 and 9 should be correct if we did all the math right. You finish up and do 7 and 8 also. Thank you and have a wonderful day.