00:01
In this problem, we have a system of three equations with three unknowns, x, y, and z, and we want to solve for these three unknowns.
00:09
And so we're going to use elimination here, and we're going to start by taking our first two equations.
00:15
And you can see that we have one x term and one y in each equation.
00:21
And so if we take our second equation, x plus y plus 3z is equal to 14, and we subtract our first equation from it, so x plus y plus 2 z is equal to 11, then we will end up with x minus x is 0 plus y minus y is 0 plus 3 z minus 2 z is just equal to z, 14 minus 11 is equal to 3.
00:57
And so from this equation we know that z is equal to 3.
01:00
Then we can take our second and third equations and plug in this z value that we already know.
01:07
So our second equation becomes x plus y plus three times z.
01:14
So three times three is equal to 14.
01:19
And that is if we move the three times three to the right side of the equation, we get x plus y is equal to 14 minus 9.
01:30
So is equal to 5.
01:33
If we do the same thing for our third equation, x minus 2y minus z, which is 3, is equal to 5, then we end up with x minus 2y...