 What about solving nonlinear systems? In general, solving a linear system of equations is easy if a bit time-consuming. On the other hand, solving a nonlinear system of equations is much, much harder. And, at least one of the major motivations per calculus is turning nonlinear systems into linear systems. And that's because nonlinear systems are so much harder to work with. How much harder? There is no general method for solving nonlinear systems of equations. You have to try things until you solve it or until you give up. For example, let's take a system of equations like this. And the first thing we might notice is that since both equations are solved for why, we might try substitution. So again, most of success in life and in math is bookkeeping, so we might start by writing down y equals y. But equals means replaceable, so I can replace y with 3x plus 5, or I can replace y with x squared minus 4x minus 7. So I'll make those replacements. And now I have a second degree equation. So I'll get all the terms onto one side. That gives me a nice little quadratic equation. The quadratic formula gives me the solutions. Well, actually these are half solutions. These are the values of x. We do need the values of y. Fortunately, we have this nice equation that tells us how to find y if we know the value of x. Even better, we have a simpler equation that tells us how to find y once we know the value of x. So if x equals 7 plus root 97, then y equals, and we won't even bother to simplify, if x equals 7 minus root 97 over 2, then y equals. And again, we don't really gain anything by simplifying. And so the solutions to this system, x equals this, y equals that, or x equals this, y equals that. How about this system of equations? Well, we already have one equation solved for y, so let's replace every occurrence of y with e to power 2x. Except y doesn't appear in this other equation. Let's take a closer look. And the thing that's worth noting here is e to power 2x is y. Well, we're given that as part of our system. But this other expression, e to power 4x, well, that's really e to power 2x squared, and equals means replaceable. So that means this is y squared. So we might make the following observation. If y equals e to power 2x, then y squared is e to power 4x. So we can substitute, and that gives us a quadratic equation. So the quadratic formula gives us solutions. Now, equals means replaceable. So if I know that y is equal to this mess, then I know e to power 2x is equal to this mess. So we'll replace. Now, this actually gives us two equations. e to power 2x is minus 5 plus root 125 over 2, and e to power 2x equals minus 5 minus root 125 over 2. However, because this is an exponential expression, we know that it always has to be positive, and this number is negative. And what that means is that this equation won't lead to a solution, so we can discard it. This equation, on the other hand, will give us a solution, so we'll hit both sides with a log, solve for x, gain street cred among bands of wandering mathematicians by showing off, or not, you don't really need to do this simplification. While the simplification isn't actually necessary, you might want to do it anyway as a reminder that any time you have a number multiplied by a log, it's the same as a log of something raised to a power. Whether you simplify or not, x equals this quantity is only half a solution. Our system had two variables x and y, so we'll need to include the y value. Now, what's important here is to remember that we actually had two solutions for y, but one of them, negative 5 minus root 125 over 2, didn't give us a solution. So the only y value that we want to include is the y value that gave us this solution for x. The important thing to remember about nonlinear systems of equations is there is no general method for solving nonlinear systems of equations. Try things until you solve them or until you give up. So we have our system of equations. Well, we could try to solve one equation for a variable. So if I take that first equation and try to solve it for x, I get a rather scary looking expression. Well, maybe I should take a look at trying to solve the other equation for x, so I'll do that. And that's not any better. Now, we could actually use substitution and solve this system of equations, but the problem is we'd be substituting things that are like this, and maybe we don't want to do that. Well, the other way we have for solving systems of linear equations is to use addition. So we might notice here that the coefficients of x squared are 1 and 2. So if we multiply one equation by 2 and the other by negative 1, we get. And if we add them, we get, which is a quadratic equation, and so we can solve this to find half a solution. Remember, our system of equations had two variables, so we need to give both variables as part of our solutions. So we'll take our y values one at a time. If y equals 2, then we can find x using either equation. Equals means replaceable, so we'll replace y with 2 in, I don't know, the first equation seems a little easier. And we can solve that. So our solution requires both an x and y value, x equals 5, y equals 2 is a solution, so is x equals negative 5, y equals 2. But wait, there's more. We have a second value of y, so we'll replace y with negative 2411's in either equation and solve for x. And the key here is not to get too lost in the arithmetic, this thing is just a real number, so we can play around with it as a real number without worrying about what its actual value is. And we solve finding x equals, so again, if x is positive principal square root of this mess, y equals negative 2411's, and if x equals negative principal square root of this mess, y equals negative 2411's. And so we have our four solutions.