00:02
We've got a system of three equations with three unknowns, and we want to use the method of elimination to solve the system.
00:10
Since our x's and zs have one or negative one as their coefficients, that's where we're going to start.
00:17
And it looks like x is the candidate.
00:19
Let's eliminate x.
00:21
We can do that two ways.
00:22
The first way, we'll add the equation a to the equation c.
00:25
So a is negative x plus two, y.
00:29
Oh, we could save ourselves some space.
00:33
If we put equation a right underneath c, we'll get negative x plus 2y minus z equals negative 4.
00:43
And if we add c plus a together, x minus x is 0, y plus 2y is 3y, negative z plus negative z is negative 2z, and that will equal negative 8.
00:56
We'll call that equation d.
00:58
We'll also eliminate x a second way, and we'll do that by multiplying equation.
01:04
A times 2.
01:06
So 2 times equation a will get us negative 2x plus 4y minus 2z equals negative 8.
01:19
And we'll add equation b to that.
01:23
So that'll be positive 2x, positive 5y, negative 4z equals negative 16.
01:33
We add those two together.
01:35
Our x is are eliminated again.
01:37
Plus 5 is 9.
01:38
So we have 9y.
01:40
Negative 2 and negative 4 is negative 6.
01:43
Z.
01:45
And negative 8 plus negative 16 is negative 24.
01:49
And that can reduce if we divide every term by 3.
01:53
We'll get 3 times y minus 2 times z equals negative 8.
02:00
And that is what we just found.
02:03
That's d...