00:01
To solve a system of equations where we have 3a plus 2b equals 27.
00:08
We also have 5a minus 7b plus c is equal to 5.
00:20
And we have negative 2a plus 10b and plus 5c equals negative 29.
00:33
So in order to solve this, we need to create a, we have our first equation up here has only a and b in it.
00:40
So we need to create another equation that's not equivalent that only has a and b.
00:47
So to do that, i'm going to multiply my middle equation by negative 5.
00:52
You can see how that will change my coefficient in front of c to negative 5, and i'm going to add it to my bottom equation.
01:00
Okay, and we'll see where that gets us.
01:03
So negative 5 times 5 is negative 25, negative 25 plus negative 2 is negative 27a, negative 5 times negative 7 is 35, 35 plus 10 is 45b, and then negative 5 times 5, our cs will cancel out, and when we add them, and then that was the whole point of multiplied by negative 5.
01:32
Negative 5 times 5 is negative 25, and then negative 25, and negative 29 is negative 54.
01:42
So now we have two new equations.
01:44
We have the same one from before, so we have 3a plus 2b equals 27...