00:01
Okay, so for this question, we're dealing with this system of linear equations right here.
00:07
Then we're asked to eliminate the x term from the third equation by adding something times the first equation to the third equation.
00:16
And then we'll rewrite the third equation in its new form with no x term.
00:21
So to do this, what we're really doing is we're taking the third equation.
00:26
I'll just call it three for now.
00:28
We're adding some number, we call it a, times the first equation.
00:37
And this will result in something x, we'll call it bx, plus something y, let's call it c, plus something z, plus, sorry, is equal to another something.
00:59
But we don't know yet, we're going to figure all these out.
01:02
And what we do know actually is that the goal is to have this equal to zero.
01:07
So this variable b that i've put here is going to be equal to zero.
01:13
So we can start by multiplying some unknown constant a through this first equation.
01:23
So when we do that, what we've got is ax minus ay plus a z is equal to to a.
01:40
And we're going to add this to the third equation.
01:44
So we can just add the coefficients for each variable together between this equation right here and our third equation.
01:57
So doing that, what we get is 3 plus a times x, minus plus 1 minus a times y plus plus negative 2 plus a said is equal to 2 plus 2a 2a.
02:32
Now i got that just from taking the coefficients of the third equation 3, 1 and negative 2, and adding the coefficients from the first equation when we multiply by some unknown variable which is what we want to solve for here.
02:46
Now the piece of information that we have is that this one here is supposed to be 0.
02:54
So 3 plus a is equal to 0.
02:58
So solving that we get a is equal to negative 3...