 Alright, so how can we solve nonlinear systems of equations? And so what do we mean by a nonlinear system? Well, let's take an expression like 12x plus 17 over square root 37 pi e to the fifth y, and this is a linear expression because it's a polynomial of degree 1. We have a whole bunch of messy terms, but these are all real numbers and our variables, x and y, are only raised to the first power, so this is a degree 1 polynomial. And degree 1 polynomials aren't too bad. A nonlinear equation is one where one, we have more one or more nonlinear expressions of the variables. So for example, this horrible thing, well, this is still a linear equation because this is a linear expression of x and y. On the other hand, x to the power one-half equals 2. Well, here's a nonlinear equation. x is raised to the power one-half. This is not a degree 1 polynomial. This is not even a polynomial, so we have a nonlinear equation. In general, if you get a nonlinear system of equations, these are very, very, very, very hard to solve. They're extremely difficult to solve, most nonlinear systems, and there are no general approaches to solving nonlinear systems of equations. There's only some vague guidelines that we can use and we try them out and see what works. We tend to try a combination of two things. We might try to solve for one variable, and that might not actually be possible. Our nonlinear system might not be solvable for one of the variables. The other thing we could also try is we might try our addition method, and again, this might not work. This might just make a horrible mess of things. And the key to nonlinear systems is there is no general approach. We try what we can. So let's take a nice simple nonlinear system, y equals x squared minus through x minus 5, y equals 2x plus 3. Again, nonlinear because we have an x raised to the second power. And, well, conveniently, we already have one variable solved in terms of the other. And I have one variable solved in terms of the other, and so that means substitution makes a lot of sense as a way of trying to solve this system. So I have y equals x squared minus through x minus 5. Equality says whenever I see the one, I can replace it with the other. So here I see y equals 2x plus 3. Any place I see y, I can replace it with 2x plus 3. Or any place I see y, I can replace it with x squared minus 3x minus 5. So I have this y here. I don't want to replace it with x squared minus 3x minus 5 because then I get x squared minus 3x minus 5 equals x squared minus 3x minus 5. That's not useful. But I can replace y with 2x plus 3, which is the same thing. So I'll do that substitution and I end up with a quadratic equation. And so now I can solve this quadratic equation like I've solved every other quadratic equation. And I'll get everything over to one side. I have this nice quadratic equation, x squared minus 5x minus 8. And I'll drop that into the quadratic formula. So I drop that into the quadratic formula. I do a little bit of arithmetic and simplify. Now this gives us our x values, but if I want to solve the system, what I need is also the y values. So I'll solve this system for y. Now, because I already know the x values, what I can do to find y is I can drop them into either equation. So again, the solution is going to make both equations true, so I can use either equation to find the y value. So I can drop this thing into this first equation y equals x squared minus 3x minus 5. I can find y equals x squared minus 3x minus 5. I can evaluate this or I can drop this x value into this equation and find y equals 2 times x plus 3, your choice. Now, because we want to do things the hard way, we'll pick the...no, we don't want to do that. We want to pick the easier expression, much easier to evaluate this than it is to evaluate that. So I'll evaluate this expression here and I'll substitute our x values into our second equation and after all the dust clears, I end up with 8 plus or minus square root of 57. Now, it's important to remember that this plus or minus actually tells us we have two different solutions for x. And depending on which solution we drop into here, we'll get two different values for y. And so we do want to express our solution to the system as two different solutions. So the solutions x, y... well, if I use 5 plus root 57 over 2 as my x value, my y value is 8 plus root 57. So there's my first solution. If I use 5 minus square root of 57 over 2, my y value is going to be 8 minus square root of 57. And so there's my second solution and I have my 1, 2 solutions to this system of equations.