 When you're working with math, you're going to have to learn how to work with formulas. And formulas aren't always going to be so easy as 2 plus 3 equals x. Sometimes you could be running into 2 plus x equals 5 and you have to solve for what x is. That means you're going to have to move that formula around, which is called transposition. And that's what this video is all about. Let's start out super simple. 3 plus x equals 12. What we're going to have to do here is we've got to get x alone, which means that if we want to get x alone, we're going to have to get rid of this 3. Now this 3 plus means that if I want to get rid of that 3, I'm going to have to subtract the 3 out of there. If I subtract the 3 on one side of the equation, I have to subtract the 3 on the other side of the equation. There, I've got x is equal to 12 minus 3. And if you do the math quickly in your head, those of you who need a calculator, perhaps this isn't for you, x is equal to 9. So that's a very simple one. You're just subtracting it out of there. What happens if I need to divide something out of there? In this example, I've got 3 times x is equal to 12. This one's almost easier. Because again, I've got 3 times x, that's what that 3 beside the x means, is equal to 12. I've got to get rid of that 3. But now I'm not subtracting out there because I don't have that plus sign that's in there. I have to divide the 3 out. So I take this side and I divide 3, and I take this side and I divide 3. Which gives me x is equal to 12 divided by 3, which will then give me 4. Now those are some pretty simple ones, but let's take a look at a couple of electrical formulas and how we can solve for different components. Okay, here's a very common formula for inductive reactants. We're not going to go too into the theory now, but we will in a later course. This one is 2 pi times f times l is equal to xl. The f is frequency, the l is inductance. The xl is inductive reactants. Doesn't matter though. We're going to try to solve for a different variable here. We're going to try to solve for l. So we need to get l alone. We need just to get this guy. It's just like we did in that last one, though. We need to get rid of this 2 pi f. It's gots to go. So if I've got to get rid of that, divide it out of there. And again, like every other formula, whatever you do in one side of the equation, you have to do on the other side. So you divide those out, which basically effectively cancels that side out, and you end up with l is equal to xl over 2 pi f. So it's not that difficult. We've just transposed to figure what that is. Let's take another look at another electrical formula. This is the formula for capacitive reactants, which is the opposite of inductive reactants, which is why we have the 1 over. So capacitive reactants is equal to 1 over 2 pi fc. And now we're going to try to solve for c. So that's the first thing we have to do in this case. We're going to have to cross-multiply, because this fraction, it's hard to work with. What I'm going to do is I'm going to go x times 2 pi fc is 1 times and then xc over 1. So it's 1 times 1. So let's move that formula around. We're cross-multiplying. I end up with 2 pi fc times xc is equal to 1. All I did was cross-multiply there. Now we want to get c alone. So if we want to get c alone, I'm going to have to get rid of the 2 pi f and the xc, which means we're going to have to divide those out. And again, whatever I do in one side of the formula, I'm going to have to do to the other side, just like this. So I'm dividing out 2 pi fxc. On one side, I'm dividing out 2 pi fxc on the other side. And there you go, I've solved for c. C is equal to 1 over 2 pi f times xc. Let's take a look at another formula. This is the formula for frequency when you have number of poles on an alternator and its RPM. So its frequency is equal to your rotations per minute times your number of poles divided by 120. I will go to much more detail on this when we get into alternators in a later course. Now what we're going to do is try to solve for a different variable. We're going to try to solve for p. Now just like in that last example there, I've got this fraction here. That fraction is annoying. I need to get rid of it. So I'm going to cross-multiply. f times 120 is equal to np times 1. And I get n times p is equal to 120 f. And I want to get p alone. It becomes very easy now. I just divide the n out on that side and the n out on the other side. np divided by n is equal to 120 f over n. And there you have it. You calculated for the period, or sorry, the number of poles. p is equal to 120 f over n. That's all transposition is. If you get a little hung up on it, just keep watching the video over and over. I've gone through every example that you possibly could use in the electrical course here.