Systems of Three Variables Systems of Three Variables
Systems of Three Variables
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- 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 can imagine if I were to draw a 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 may it might actually look.
- It 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.
- It 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
- and then it goes like that. I'm just doing this for visualization purposes.
- And so the intersection of this plane - the x, y 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. And a lot of times these three equations with three unknown systems
- 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 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 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 and a negative y and then another positive y
- it seems like we can eliminate the Ys.
- 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 will have all of the x and z constraint information embedded in it because
- 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 three z is equal to negative ten.
- And x minus y plus two z is equal to three. So over here if we want to eliminate y, we can literally
- just add these two equations. So on the left hand side, x plus x is two x. Y plus negative y cancels out
- And then negative three z plus two z - that gives us just a negative z
- and then we have negative ten plus three, which is negative seven.
- So using these two equations we got
- two x minus z is equal to negative seven - just adding these two equations.
- Now let's do these two equatons. 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 two z is equal to three.
- And we have two x plus y minus z is equal to negative six.
- If we want to eliminate the Ys, we can just add these two equations.
- So x plus two x is three x. Negative y plus y cancels out. Two z minus z - well that is just z.
- And that is going to be equal to three plus negative six, which is negative three.
- So if I add these two equations, I get three x plus z is equal to negative three. 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 two x 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, the Zs cancel out. Otherwise if it didn't happen so naturally, we'd
- have to multiply one of these equations, or maybe both of them, by some scaling factor.
- But we can just add these two equations up.
- On the left hand side, two x plus three x is five x. Negative z plus z cancels out.
- Negative seven plus negative three - that is equal to negative ten.
- Divide both sides of this equation by five and
- we get x is equal to negative two. Now we can substitute back to find the other variables.
- Maybe we can substitute back into this equation to figure out what z must be equal to.
- So we have two times x. Two times negative two minus z is equal to negative seven.
- Or negative four minus z is equal to negative seven.
- We can add four to both sides of this equation and then we get
- negative z is equal to negative seven plus four, which is negative three.
- Multiply or divide both sides by negative one and you get z is equal to three. And now we can go and
- substitute back into one of these original equations. So we have x. We know x is negative two.
- So we have negative two plus y, minus three times z.
- Well, we know z is three (so minus three times three)
- should all be equal to negative ten. And now we just solve for y.
- So we get negative two plus y minus nine is equal to negative ten. And so negative two minus nine,
- that's negative eleven. So we have
- y minus eleven is equal to negative ten. And then we can add eleven to both
- sides of this equation. And we get y is equal to negative ten plus eleven, which is one.
- So we're done!
- We've got x is equal to negative two. Z is equal to three and y is equal to one.
- Now I can actually go back and check it.
- Verify that this x, y and z works for all three constraints
- that this three dimensional coordinate lies on all three planes.
- So let's try it out. We've got x is negative two, z is three, y is one.
- So if we substituted - let me do it into each of them - so in this first equation
- that means that we have negative two plus one (remember y was equal to one).
- Let me write it over here - y is equal to one, x is equal to negative two, z is equal to three.
- That was the result we got. Yup, that's the result we got.
- So when we test it into this first one, you have negative two plus one minus three times three.
- So minus nine. This should be equal to negative ten. And it is.
- Negative two plus one is negative one, minus nine is negative ten.
- So it works for the first one. Let's try it for the second equation right over here.
- So we have negative two minus y (so, minus one) plus two times z (so, z is three, so two times three)
- So, plus six needs to be equal to three.
- So this is negative three plus six, which is indeed equal to three.
- So this satisifies the second equation. And then we have the last one right over here!
- We have two times x, so two times negative two, which is negative four. Negative four.
- Plus y, so plus one. Minus z, so minus three. Minus three.
- Needs to be equal to negative six. Negative four plus one is negative three,
- and then you subtract three again. It equals negative six.
- So it satisfies all three equations, so we can feel pretty good about our answer.
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