00:01
For this problem, we're given a system of three equations with three unknowns, x, y, and z, and we want to solve by elimination.
00:09
So we're going to start by multiplying the first equation by three.
00:13
So that gives us 3x minus 3y plus 9 z is equal to 24.
00:18
And we're going to take that and subtract it from our second equation.
00:24
So 3x plus y minus 2z is equal to negative 2.
00:31
And so the x is cancel out, and we are left with 4y minus 11z is equal to negative 26.
00:48
And then we want to cancel out the x from one of these other equations as well.
00:54
And so we can multiply the top equation by 2.
00:59
And so we'll end up with 2x minus 2y plus 6z is equal to 6.
01:05
And we can subtract that from our third equation because that one also has 2x.
01:15
So 2x minus y plus z is equal to 0.
01:19
And we end up with our x is canceling out.
01:23
6y minus 5 z is equal to negative 16.
01:29
And so now that we have these two equations we can solve for, or we can eliminate our y term.
01:37
By multiplying both this top one by 3 and this bottom one by 4...