We can ignore the denominators.

Both of their denominators is 36.

And now we have different denominators here.

What happens when we have different denominators?

You're taking the difference of those two denominators, and now you're taking the sum of the two denominators.

Any common denominators in cases of spontaneous combustion?

In this example, the denominators are already the same.

A final note on the denominators. In these equations.

We multiply the denominators 3 times 2 is 6.

So, what are the common denominators in these three cultures?

Here you multiplied the numerators, and you multiplied the denominators.

We first check to make sure the denominators are the same.

And you multiply the denominators 8 times 25 is going to be 200

We'll learn in a second what we do when the denominators are different.

So this is equal to 48 over and then you just multiply the denominators.

Although now we have two expressions we want to get out of the denominators.

All I did is I re write both of these fractions with different denominators.

So here we have two fractions we're adding together, but we have different denominators.

If the denominators are the same, the denominator of the answer will be the same.

So LCD of the 2 fractions is going to be the LCM of both of these denominators over here. and if we can make this as common denominators we can add the two fractions

Well the denominators or the bottom numbers, the bottom numbers in the fraction stayed the same.

So whenever you add fractions, the first thing you have to do is check the denominators.

In fact we've done two more Blue Zone expeditions since this and these common denominators hold true.

I'm just multiplying the numerator and then multiplying the denominators 3x plus 2 times x minus 5.

And of course, when we multiply fractions, we're just multiplying all the numerators over all the denominators.

We just add the numerators, three plus two is five, and we keep the denominators the same.

So hopefully you're now satisfied that the focal length of a hyperbola is the sum of these two denominators.

See if we subtract let's just multiply everything by 7, just to get these denominators out of the way.

So what's probably bothering you, because it's bothering me, is these x's that we have in the denominators right over here.

So the denominators are the same, so we know that this, when you add this together, you have to get this.

And then multiplying the denominators, 3 times 5 is 15 but hopefully this now makes conceptual sense as 2 3 of 4 5.

Now, we've written both of them as improper fractions, but we still have different denominators, so we have to have a common denominator.

And I have six plus and the denominators are same here so we're gonna get something over sixteen and twelve minus eight is four.

We have a 4 in the denominator, that'll eventually be in the denominator, so let's divide our eventual numerators and our eventual denominators both by 4.

So we have one fourth plus two fourths, so we know that we just add the numerators, three, and the denominators are the same, three fourths.

And all that is, is taking this expression up here and turning it into two simpler expressions where the denominators are the factors of this lower term.

When you add anything, you find the common denominator, and the common denominator, the easiest common denominator, is to multiply the two denominators, so let me write this here.

Where does this put us, so both of these are positive intergers, so we can just add them, we just have to find the like denominators we have 25 8 3 4.

If we have the same denominator, the denominator in the difference is going to be the same as the denominators in the two numbers that we're subtracting, so it's going to be 18.

And like we've done in every problem so far, the denominators are the same, so we can just set the numerators equal to each other and try to solve for A, B, and C.

So, I personally don't like having x's in my denominators, so we want to get it outside of the denominator into a numerator or at least not in the denominator as soon as possible.

We now bring this expression to the right side, multiply the denominators up, and sort all the expressions containing pi to the left, which gives us pi 3 8, exactly the number we were at before.

Well, that was a pretty tiring problem, and you can see, the partial fraction decomposition becomes a lot more complicated when you have a higher degree denominators, but hopefully you found that a little bit useful.

Now, whenever I like to deal with fractions especially when I have different denominators i like to deal them with improper fractions, it makes the addition, and the subtraction, and actually the multiplication and divison a lot easier.

But we have just successfully decomposed this pretty I don't want to say that we necessarily simplified it, because you could say, oh, we only have one expression here, now I have three but I've reduced the degree of both the numerators and the denominators.