00:02
Here we have a system of three equations with three unknowns, and we are going to use the method of elimination to solve.
00:10
Looks like we can either eliminate our x's or eliminate our zs.
00:16
Both of them have twos and fours for the coefficient, so it shouldn't be too difficult.
00:22
Let's go for the x's.
00:24
So we'll start with equation a, and we'll multiply that by 2.
00:30
So negative 4x, negative 6y, negative 4z equals negative.
00:42
And we'll take b as is and add it.
00:45
So 4x plus 5y plus 2z equals 3.
00:52
And then we add those together, we get negative y minus 2z equals negative 7.
01:02
If we multiply everything by negative 1, we will get that y plus 2z equals positive 7.
01:14
And we'll call that equation d.
01:16
Now let's try to eliminate x another way.
01:19
Let's take equation c and we'll add negative a to it.
01:25
So if we take a and multiply it by negative 1, we'll get negative 2x minus 3y.
01:33
Minus 2z equals negative 5.
01:39
Yes.
01:41
And when we add these together, again, the x is eliminate.
01:44
4y minus 3y equals y.
01:47
4z minus 2z equals positive 2z, and 12 minus 5 equals 7.
01:53
Well, look at that.
01:55
We just found equation d again...