00:02
What we're looking at here is a system of three equations and three unknowns, and our goal is to find the value of z.
00:09
So it would be nice if all of these equations were in the same format, kind of like the format we see in the second and the third equation.
00:17
So let's make some changes to equation one.
00:20
I'm going to distribute the two.
00:22
So we have x minus 2z is equal to 2y minus 2z.
00:29
We'll notice that there's a minus 2 z on both sides.
00:32
So if we add 2 z to both sides, we get x equals 2y.
00:36
So that wasn't exactly what i was going for, but that simplifies things a lot because what we can do now is substitution, and we can put 2y in for x in our other two equations.
00:49
So that makes the second equation 2 times 2y minus 6y plus z equals 1, and the third equation 3 times 2y plus y minus 2z equals 4.
01:03
And that's good because now these equations only have two variables in them.
01:09
So in the second equation here, we have 4y minus 6y.
01:13
So we have negative 2y plus z equals 1...