Intro to linear systems with 3 variables

solve this system and here we have three equations with three unknowns and just so you have a way to visualize this each of these equations would actually be a plane in three dimensions and so you're actually trying to figure out where three planes in three dimensions intersect I won't go into the details here I'll focus more on the mechanics but you could imagine if I were to draw three dimensional space over here now all of a sudden we'll have an x y and z-axes so you can imagine that maybe this first plane and I'm not drawing it the way it might actually look might look something like that I'm just drawing part of the plane and maybe this plane over here it intersects right over there and it comes popping out like this and then it goes behind it like that keeps going in every direction I'm just drawing part of the plane and maybe this plane over here maybe it does something like this maybe it intersects over here and over here and so it pops out like that and then it goes below it like that it goes like that I'm just doing this for visualization purposes and so the intersection of this plane the XY and z coordinates that would satisfy all three of these constraints the way I drew them would be right over here so that's what we're looking for in a lot of times these three sister equation with three unknown systems won't they will be inconsistent you won't have a solution here because it's very possible to have three planes that all don't intersect in one place a very simple example of that is well one they could all be parallel to each other or they could intersect each other but maybe they intersect each other in kind of a triangle so maybe one plane looks like that then another plane another plane maybe pops out like that goes underneath and then maybe the third plane cuts in it does something like this where it goes into that plane and keeps going out like that but it intersects this plane over here so you see this kind of forms a triangle and they don't all intersect in one point so in this situation you would have an inconsistent system so with that out of the way let's try to actually solve this system and the trick here is to try to eliminate one variable at a time from all of the equations making sure that you have the information from all three equations here so what we're going to do is we could maybe it looks like the easiest to eliminate since we have a positive Y to negative Y and then another positive Y it seems like we can eliminate the Y's we can add these two equations and come up with another equation that will only be in terms of X and Z and then we could use these two equations to come up with another equation that will only be in terms of X and Z but it'll have all of the all of the X and z constraint information embedded in it because it has all we're using all three equations so let's do that so first let's add these two equations right over here so we have X plus y minus 3z is equal to negative 10 and X minus y plus 2z is equal to 3 so over here we if we want to eliminate Y we can literally just add these two equations so on the left hand side X plus X is 2x y plus negative Y cancels out and then negative 3z plus 2z that gives us just a negative Z and then we have negative 10 plus 3 which is negative 7 so using these two equations we got 2x minus Z is equal to negative 7 just adding these two equations now let's do these two equations we can and we can reuse this equation as long as we're using new information here now we're using the extra constraint of this bottom equation so we have X minus y plus 2z is equal to 3 and we have 2x 2x plus y minus z is equal to negative 6 if we want to eliminate the Y's we can just add these two equations so X plus 2x is 3x negative y plus y cancels out 2 Z minus Z well that is just Z and that is going to be equal to 3 plus negative 6 is negative 3 so if I add these two equations I get 3x plus Z is equal to negative 3 now I have a system of two equations with two unknowns this is a little bit more traditional of a problem so let me write them over here so we have 2x minus Z is equal to negative seven and then we have three X plus Z is equal to negative three and the way this problem is set up it gets pretty simple pretty fast because if we just add these two equations disease cancel out otherwise if it didn't happen so naturally we would have to multiply one of these equations or maybe both of them by some scaling factor but we can just add these two these two equations up on the left-hand side 2x plus 3x is 5x negative Z plus Z cancels out negative 7 plus negative 3 that is equal to negative 10 divide both sides of this equation by 5 and we get X is equal to negative 2 now we can substitute back to find the other variables we will use we can substitute back into this equation to figure out what Z must be equal to so we have 2 times X 2 times negative 2 minus Z is equal to negative 7 or negative 4 minus Z is equal to negative 7 we can add 4 to both sides of this equation and then we get negative Z is equal to negative 7 plus 4 is negative 3 multiply or divide both sides by negative 1 and you get Z is equal to 3 and now we can go and substitute back into one of these original equations so we have X we know X is negative 2 so we have negative 2 plus y minus 3 times Z well we know Z is 3 minus 3 times 3 should be equal to negative 10 and now we just solve for y so we get negative 2 plus y minus 9 is equal to negative 10 and so negative 2 minus 9 that's negative 11 so we have Y minus 11 is equal to negative 10 and then we have so we can add 11 to both sides of this equation and we get Y is equal to negative 10 plus 11 is 1 so we're done we got X is equal to negative 2 Z is equal to 3 and Y is equal to 1 and now we can actually go back and check it verify that this XY and Z works all three constraints that this this three dimensional coordinate lies on all three planes so let's try it out we got X is negative 2 Z is 3 y is 1 so if we substitute it let me do it into each of them so in this first equation that means that negative negative 2 plus 1 remember Y was equal to 1 let me write it over here Y is equal to 1 X is equal to negative 2 Z is equal to 3 that was the result we got that's the result we got so when we tested in this first one you have negative 2 plus 1 minus 3 times 3 so minus 9 this should be equal to negative 10 and it is negative 2 plus 1 is negative 1 minus 9 is negative 10 so it works for the first one let's try it for the second equation right over here we should have so we have negative 2 minus y so minus 1 plus 2 times Z so Z is 3 so 2 times 3 so plus 6 needs to be equal to 3 so this is negative 3 plus 6 which is indeed equal to 3 so it satisfies the second equation and then we have the last one right over here we have 2 times X so 2 times negative 2 which is negative 4 negative 4 plus y so plus 1 minus Z so minus 3 minus three needs to be equal to negative 6 negative 4 plus 1 is negative 4 plus 1 is negative 3 and then you subtract 3 again it equals negative 6 so it satisfies all three equations so we can feel pretty good about our answer