 What can we do with algebraic expressions that involve subtraction? One thing mathematicians like to do is to reduce problems to previously solved problems. So we might remember that we can add like terms by adding their coefficients. And so what can we do about subtraction? We can remember this theorem that when we add the additive inverse, it's the same as subtracting, or equivalently, to subtract, we'll add the additive inverse. And this is especially useful because addition is commutative and associative. It's better if we add the additive inverse than worry about what we mean by a subtraction. The only problem is what is the additive inverse of a term? So let's tackle that problem. We want to find the additive inverse of 8x. Well, by our definition, if we take 8x and we add the additive inverse of 8x, we get 0. Since we're able to add these terms, they must be like terms. Their variable portion has to be the same. So both of these must be something times x, which means we can apply the distributive property. And that'll give us 8 plus something times x is going to be 0. But if I want something times x to be 0, my something has to be 0. And that means whatever I'm adding to 8 must give me 0. And that means the thing I'm adding must be the additive inverse of 8. But wait, if I have this equation 8 plus the additive inverse of 8 times x equals 0, then I can apply the distributive property, which gives me 8x plus the additive inverse of 8 times x equals 0. So now let's compare these two equations. Both of these are 8x plus something equals 0. And what this suggests is that our somethings have to be the same thing. And so that means this additive inverse of 8x must be the same thing as the additive inverse of 8 times x. And so this suggests the following theorem. To produce the additive inverse of a term, replace the coefficient with its additive inverse. So for example, let's find the additive inverse of the quantity 8x squared minus 3x plus 7. Now, first of all, we have to take care of the thing inside the parentheses. And because there's a subtraction here, let's change that to adding the additive inverse. So instead of minus 3x, we'll write plus the additive inverse of 3 times x, and everything else remains the same. Now I have the additive inverse of a sum. Well, my theorem says the additive inverse of a sum is the sum of the additive inverses. So we'll have the additive inverse of 8x squared, which we'll produce by replacing the coefficient 8 with the additive inverse, plus the additive inverse of the additive inverse of 3x, plus the additive inverse of 7. But we can clean this up a little bit. This additive inverse of the additive inverse of 3 is just going to be 3 by itself. And so our final answer, additive inverse of 8x squared plus 3x plus additive inverse of 7. So let's take an expression like 8x plus 5y minus 3x plus 2x minus 7y. Now, because I have the subtractions minus 3x minus 7y, what I'll do is I'll first rewrite this as an addition of the additive inverse. And again, the reason that this is useful is that once this is an addition, I can rearrange the terms in any order that I want. And I want to put the like terms close to each other, mostly for convenience so I can work with them more easily. So we'll rearrange things a little bit. So now I can add these like terms by adding the coefficients. 8x plus additive inverse 3x plus 2x is going to be 7x, and 5y plus additive inverse of 7y is going to be additive inverse 2y. What about 8x minus quantity 3x minus 5? So again, the first thing we're going to want to do is to change the subtractions into adding the additive inverse. And order of operation says we have to take care of the stuff inside the parentheses first. So this 3x minus 5 becomes 3x plus the additive inverse of 5. Now we have a subtraction. We want to change this to adding the additive inverse. But notice this is the additive inverse of a sum, which is going to be the sum of the additive inverses. And now that we have a sum, then we don't need all of these extra parentheses. We're adding the terms 8x plus additive inverse of 3x plus additive inverse of additive inverse of 5. Well, the additive inverse of the additive inverse of 5 is just 5, and the others are like terms so we can combine them to get our final answer. Or take a look at something like this. Now in this particular case, we see that there's no written coefficient of x squared, but that's okay. We have an implied coefficient of 1, and it's convenient to write that down. So we're subtracting 4x plus 1x squared. So that's the same as adding the additive inverse of 4x plus 1x squared. So the additive inverse of a sum is the sum of the additive inverses. I have just an addition at this point, so we can rearrange this any way we want to. We'll combine the like terms. And we have our final answer, 11x squared, plus the additive inverse of 2x. And this is a perfectly good final answer. However, there's one more thing we might want to do, which is mostly a matter of style. And that's this. We can write plus the additive inverse, but we prefer to write our final answers without this notation plus an additive inverse. Well, that's okay because I know that when I'm adding an additive inverse, it's the same as a subtraction. So instead of 11x squared plus the additive inverse of 2x, I can write this as 11x squared minus 2x. Again, this form is a perfectly good final answer. This form is stylistically better.