 All right, well, what about some other equations that involve absolute values? For example, x plus 3 absolute value is equal to 5. And one problem we may run into is I know how to interpret the absolute value of a minus b. And I don't have a way of interpreting the absolute value of a plus b. So one of the things I might do is I might use the rules of integer arithmetic to rewrite this. Standard thing in mathematics. If you're ever confronted with something you don't know how to handle, turn it into something you do know how to handle. That's the basic rule of algebra, actually. So I have x plus 3, and I can change that x minus negative 3 is equal to 5. And now I have something I can understand. This is telling me the difference between whatever x is and negative 3 is going to be 5. So I'm looking for a number that's 5 away from negative 3. So I'll set down my negative 3, and it's possible that x is 5 more than 3, negative 3. It tells me that x has to be 2. The other possibility is that x is 5 less than 3, and that tells me that x has to be negative 8. And so there are my two solutions. Well, we might try and find a way of solving this problem directly without having to convert the plus an amount into minus the additive inverse. And a convenient way of thinking about this is to remember that the absolute value of a, well, a itself is the same as a minus 0. So the absolute value of a is a difference. Is the difference between a and 0, wherever a is and the origin? So let's think about that. So this x plus 3 absolute value equal to 5. Well, the idea is I have x. I know how to interpret x plus 3, so I'm going to add 3 to get some place. And the difference between where I am and the origin is equal to 5. So where I am to the origin is going to be 5 units. Now there's two possibilities. Either the origin is over here someplace to the right, or it's over here someplace to the left. Well, let's take those two possibilities one at a time. If the origin is off to the right, then someplace up here it's the origin, and then x is way back here someplace. And how do I get there? Well, that's back 5, back 3 more. So that tells me that x is located at negative 8. The other possibility is the origin might be off to the left. So again, x plus 3, the origin is 5 units away, but this time it's 5 units back this way. So here's my origin, and the question I want to find out is what x is, and what I know is that this distance is 5, this distance is 3. That means this left over distance here has to be 2, and so I know that x is equal to 2. And so I get my two solutions, x equals negative 8, and x equals 2.