00:01
In this situation, we're solving a system by a substitution.
00:03
We want to find the easiest equation that's given to us to work at x and y by itself.
00:08
That seems to be the first one here.
00:10
So we chose that because we have coefficients of one for both x and y.
00:14
You can solve for either x or y.
00:15
I would just solve for x in this case.
00:17
So x plus y equals zero.
00:20
I want to take away y from both sides.
00:22
So we know therefore that x is equal to negative y, right? basically, that they're opposites.
00:28
Now you can take that value of x negative y and replace x in the other original equation and substitute in to figure out the y position of the intersection between these two lines.
00:41
So let's do that.
00:42
And we get four times whatever x is, and we said x was negative y, plus two times y, is equal to three.
00:52
Now when you evaluate this, we're going to get four times negative y, which is negative 4y, then plus 2y is equal to 3.
01:00
Negative 4y and plus 2y combine to be negative 2y.
01:04
So negative 2y is equal to 3...