 One of the interesting features about mathematics is that sometimes we can solve a problem by making it harder. So in general, any system of equations can be solved by reducing it to a single equation in one variable. But this also means we can rewrite a single equation in one variable as a system of equations. And what this does is this gives us a new strategy substitute to simplify. For example, we solved this system of equations earlier, and we solved it by noting that if e to the 2x is y, then e to the 4x is e to the 2x squared, that's y squared. And so we replaced and got a quadratic equation, which we solved. And then once we knew our value for y, since y was e to the power 2x, we could then solve for x. And it gave us our solution y equals something, x equals something. The thing to recognize here is that if you only saw the first equation, you wouldn't realize that this was actually part of a system of equations. So what if you only saw the first equation? So the crucial observation here is that even though these are exponential expressions, they are related to each other. e to the power 4x is e to the power 2x squared. So if we make the substitution y equals e to the power 2x, then we have and when we do that, solving this single equation is the same as solving the system of equations, which we've already solved. And the only real difference is that the problem we're trying to solve only has one variable x, so we only need to state one variable x as our solution. It's worth pointing out that a substitution led to a quadratic equation. This situation is common enough that it actually has a name. We say that the equation is reducible to quadratic. And having introduced the name, we'll probably never specifically refer to it again. What's important is a process. A substitution leads us to a different type of equation that we can solve. The important thing to recognize here is that because these are nonlinear systems, there is no general procedure that always works. We always have to try things out. And if they don't work, we have to try something else or give up. Something we can do is look for any relationships we might find between the variable expressions. And in this case, our variable expressions are y to the fourth and y squared. And we might make the following observation. If we let y squared be x, then y to the fourth is y squared squared or x squared. And so our equation can be rewritten as... And what's not important is that this is a quadratic equation. What is important is that this is an equation that we can solve. So we solve it and find solutions x equal to... But equals means replaceable. If x is equal to 2 or negative 1, x is equal to y squared. So, and again, these are equations we can solve and we get our solutions. Now we can actually do this for equations that we could solve otherwise. So here's a nice tame radical equation that we could solve with some difficulty. But we might also look at our variable expressions x and square root of x. And the observation we might make is that if we let y equals square root of x, then x by itself is y squared. And that means our equation can be rewritten as... And who cares that this is quadratic? What's important is that it's an equation that we can solve. So when we solve it, we get... And equals means replaceable. Since y equals square root of x, we can rewrite. And we can solve the resulting equation. Now sometimes we may have to be a little bit creative. So here we might observe, ah, 4 to the power x. Well that's 2 to the power x squared. So if we let y equals 2 to the x... Well, wait a minute. We don't have a 2 to the power x in our equation. Or do we? So if we apply the rules of exponents, this 2 to the power x plus 2 is the same as 2 to the second times 2 to the x. And so there is a 2 to the power x in our equation. So now making that replacement is worthwhile. Which we can solve. Equals means replaceable. So if y equals 3 or y equals 1, then 2 to the power x is 3 or 1. And we can solve these equations. And let's show off a little bit. Log of 1 is equal to 0. So this solution is 0. And our equation has two solutions.