00:01
So we're going to solve this system of two equations using the process of elimination.
00:06
So when we do elimination, we need to have one variable sort of matching up with the same variable and the other equation, so that if we were to add or subtract these two equations from each other, one variable would be eliminated in that process.
00:21
Now, we don't have that in the current state.
00:23
We do have x squared and y in both of them, but the coefficients aren't the same.
00:28
What we can do, though, is multiply, both equations by a constant in order to get the coefficients to equal the same thing.
00:36
There's a few ways to do this, but what will work really smoothly is to multiply this first equation by 2 and the second equation by negative 3.
00:45
So that will have 6x squared and negative 6x squared, and that will allow us to eliminate the x.
00:51
So let's do that.
00:52
We'll multiply this first equation by 2 and we have to multiply everything by 2.
00:57
So we get 6x squared plus 8 y equals 34.
01:04
And then we'll do this one by negative 3 to get negative 6 x squared minus 15 y equals negative 6.
01:17
And so then we're ready to add these two together so that we can eliminate.
01:22
So the x is get eliminated.
01:24
And then we have 8y minus 15 y.
01:27
So that's negative 7y.
01:30
And then we have 34 minus 6.
01:33
Which is 28...