00:01
For this problem, we're going to be solving the system of equations using the elimination method.
00:07
So looking at these two equations, there's nothing we can immediately eliminate, nor can we only multiply one of the equations by a coefficient in order to eliminate one of the variables.
00:20
So that means we're going to have to find the lowest common multiple for the coefficients of one of the variables.
00:26
It can be either of them, but i'm just going to choose x.
00:29
So for that, it's going to probably be six because three times two is six, and there's nothing lower that both of them have multiples of.
00:39
So i'm going to multiply the first equation by two, and then the second equation by negative three, because it needs to be negative in order for them to cancel each other out.
00:51
So that is going to give us 6x plus 4y equals 20, and negative 6x minus 15y equals negative 9.
01:12
So i'm going to add those together and we're going to end up with 0 minus 11y equals 11...