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Solve this system. And here we have three equations with three unknowns.

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And just so you have a way to visualize this,

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each of these equations would actually be a plane in three dimensions.

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And so you're actually trying to figure out where three planes in three dimensions intersect.

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I won't go into the details here, I'll focus more on the mechanics,

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but you can imagine if I were to draw a three dimensional space over here.

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Now all of a sudden we'll have an x, y, and z axes.

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So you can imagine that maybe this first plane

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and I'm not drawing it the may it might actually look.

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It might look something like that. (I'm just drawing part of the plane.)

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And maybe this plane over here.

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It intersects right over there and it comes popping out like this and then it goes behind it like that.

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It keeps going in every direction, I'm just drawing part of the plane.

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And maybe this plane over here, maybe it does something like this.

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Maybe it intersects over here and over here.

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And so it pops out like that and then it goes below it like that

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and then it goes like that. I'm just doing this for visualization purposes.

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And so the intersection of this plane - the x, y and z coordinates that would satisfy

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all three of these constraints the way I drew them - would be right over here.

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So that's what we're looking for. And a lot of times these three equations with three unknown systems

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will be inconsistent. You won't have a solution here, because it's very possible to have three planes

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that all don't intersect in one place. A very simple example of that is

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well, one, they could all be parallel to each other, or they could intersect each other but maybe they

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intersect each other in kind of a triangle, so maybe one plane looks like that, then another plane maybe

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pops out like that, goes underneath. And then maybe the third plane cuts in.

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It does something like this: where it goes into that plane

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and keeps going out like that, but it intersects this plane over here.

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So you see kind of forms a triangle and they don't all intersect in one point

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so in this situation, you would have an inconsistent system. So with that out of the way, let's try to

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actually solve this system. And the trick here is to try to eliminate one variable at a time from all

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of the equations, making sure that you have the information from all three equations here

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so what we're going to do is we could maybe - it looks like the easiest to eliminate

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since we have a positive y and a negative y and then another positive y

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it seems like we can eliminate the Ys.

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We can add these two equations and come up with another equation

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that will only be in terms of x and z. And then we could use these two equations

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to come up with another equation that will only be in terms of x and z.

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But it will have all of the x and z constraint information embedded in it because

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we're using all three equations. So let's do that. So first let's add these two equations right over here.

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So we have x plus y minus three z is equal to negative ten.

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And x minus y plus two z is equal to three. So over here if we want to eliminate y, we can literally

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just add these two equations. So on the left hand side, x plus x is two x. Y plus negative y cancels out

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And then negative three z plus two z - that gives us just a negative z

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and then we have negative ten plus three, which is negative seven.

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So using these two equations we got

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two x minus z is equal to negative seven - just adding these two equations.

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Now let's do these two equatons. And we can reuse this equation as long as

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we're using new information here. Now we're using the extra constraint of this bottom equation.

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So we have x minus y plus two z is equal to three.

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And we have two x plus y minus z is equal to negative six.

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If we want to eliminate the Ys, we can just add these two equations.

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So x plus two x is three x. Negative y plus y cancels out. Two z minus z - well that is just z.

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And that is going to be equal to three plus negative six, which is negative three.

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So if I add these two equations, I get three x plus z is equal to negative three. Now I have a system

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of two equations with two unknowns. This is a little bit more traditional of a problem. So let me write

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them over here. So we have two x minus z is equal to negative seven. And then we have three x plus z

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is equal to negative three and the way this problem is set up, it gets pretty simple pretty fast, because

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if we just add these two equations, the Zs cancel out. Otherwise if it didn't happen so naturally, we'd

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have to multiply one of these equations, or maybe both of them, by some scaling factor.

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But we can just add these two equations up.

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On the left hand side, two x plus three x is five x. Negative z plus z cancels out.

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Negative seven plus negative three - that is equal to negative ten.

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Divide both sides of this equation by five and

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we get x is equal to negative two. Now we can substitute back to find the other variables.

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Maybe we can substitute back into this equation to figure out what z must be equal to.

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So we have two times x. Two times negative two minus z is equal to negative seven.

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Or negative four minus z is equal to negative seven.

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We can add four to both sides of this equation and then we get

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negative z is equal to negative seven plus four, which is negative three.

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Multiply or divide both sides by negative one and you get z is equal to three. And now we can go and

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substitute back into one of these original equations. So we have x. We know x is negative two.

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So we have negative two plus y, minus three times z.

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Well, we know z is three (so minus three times three)

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should all be equal to negative ten. And now we just solve for y.

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So we get negative two plus y minus nine is equal to negative ten. And so negative two minus nine,

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that's negative eleven. So we have

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y minus eleven is equal to negative ten. And then we can add eleven to both

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sides of this equation. And we get y is equal to negative ten plus eleven, which is one.

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So we're done!

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We've got x is equal to negative two. Z is equal to three and y is equal to one.

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Now I can actually go back and check it.

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Verify that this x, y and z works for all three constraints

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that this three dimensional coordinate lies on all three planes.

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So let's try it out. We've got x is negative two, z is three, y is one.

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So if we substituted - let me do it into each of them - so in this first equation

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that means that we have negative two plus one (remember y was equal to one).

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Let me write it over here - y is equal to one, x is equal to negative two, z is equal to three.

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That was the result we got. Yup, that's the result we got.

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So when we test it into this first one, you have negative two plus one minus three times three.

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So minus nine. This should be equal to negative ten. And it is.

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Negative two plus one is negative one, minus nine is negative ten.

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So it works for the first one. Let's try it for the second equation right over here.

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So we have negative two minus y (so, minus one) plus two times z (so, z is three, so two times three)

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So, plus six needs to be equal to three.

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So this is negative three plus six, which is indeed equal to three.

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So this satisifies the second equation. And then we have the last one right over here!

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We have two times x, so two times negative two, which is negative four. Negative four.

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Plus y, so plus one. Minus z, so minus three. Minus three.

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Needs to be equal to negative six. Negative four plus one is negative three,

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and then you subtract three again. It equals negative six.

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So it satisfies all three equations, so we can feel pretty good about our answer.