00:02
Here we've got a system of three equations and three unknowns, and we're going to solve these using the method of elimination.
00:10
We can start.
00:11
We're going to eliminate the zs from a and b by taking equation a and multiplying it by two.
00:19
So two times four is eight y minus two times z equals two times negative 13, which is negative 26.
00:29
And we'll take b just the way it is.
00:32
We're going to add that.
00:32
Them together.
00:33
So 3y plus 2 z equals 4.
00:41
And when we add these, we'll get that 11y plus 0 equals negative 22.
00:49
So if we divide both sides by 11, we find that y equals negative 2.
00:55
There's one of our solutions.
00:58
Now, let's take equation b and c and add them together.
01:04
So we can eliminate the zs, and we'll get that 6x minus 2y plus 0 equals 4 plus negative 18 is negative 14.
01:17
And then let's take our y value, y equals negative 2...