00:01
We're being asked to solve the given nonlinear system of equations.
00:04
Well, to do this, i'm going to use the substitution method, because our first equation is already solved for y.
00:10
It's x squared minus 3.
00:11
So we simply need to substitute this in for y in the other equation.
00:15
So i'm going to come over to the right hand side.
00:17
So what we're going to have is 4x minus the quantity of x squared minus 3, and it's all equal to 6.
00:25
And now we just have to solve this equation for x.
00:28
So first, i must distribute this negative sign to both terms.
00:32
So i'm going to have 4x minus x squared plus 3 is equal to 6.
00:38
Well, we have a quadratic equation here.
00:41
So we must set it equal to 0.
00:42
So i'm going to subtract 6 from both sides.
00:46
And what i find is i have 4x minus x squared minus 3 is equal to 0.
00:51
Now i'm going to rearrange my terms.
00:53
So that way goes my quadratic term, then the linear term, and then our constant.
00:57
So when i do that, i'm going to have negative x squared plus 4x minus 3 is equal to 0.
01:04
Well, now we need to factor.
01:05
However, we don't ever want to factor when that a term is a negative.
01:09
So what i'm going to do is i'm going to multiply the whole equation by negative 1.
01:14
So i'm now going to have x squared minus 4x plus 3 is equal to 0...