Intro to absolute value inequalities

I now want to solve some inequalities that also have absolute values in them. And if there's any topic in algebra that probably confuses people the most, it's this. But if we kind of keep our head on straight about what absolute value really means, I think you will find that it's not that bad. So let's start with a nice, fairly simple warm-up problem. Let's start with the absolute value of x is less than 12. So remember what I told you about the meaning of absolute value. It means how far away you are from 0. So one way to say this is, what are all of the x's that are less than 12 away from 0? Let's draw a number line. So if we have 0 here, and we want all the numbers that are less than 12 away from 0, well, you could go all the way to positive 12, and you could go all the way to negative 12. Anything that's in between these two numbers is going to have an absolute value of less than 12. It's going to be less than 12 away from 0. So this, you could say, this could be all of the numbers where x is greater than negative 12. Those are definitely going to have an absolute value less than 12, as long as they're also-- and, x has to be less than 12. So if an x meets both of these constraints, its absolute value is definitely going to be less than 12. You know, you take the absolute value of negative 6, that's only 6 away from 0. The absolute value of negative 11, only 11 away from 0. So something that meets both of these constraints will satisfy the equation. And actually, we've solved it, because this is only a one-step equation there. But I think it lays a good foundation for the next few problems. And I could actually write it like this. In interval notation, it would be everything between negative 12 and positive 12, and not including those numbers. Or we could write it like this, x is less than 12, and is greater than negative 12. That's the solution set right there. Now let's do one that's a little bit more complicated, that allows us to think a little bit harder. So let's say we have the absolute value of 7x is greater than or equal to 21. So let's not even think about what's inside of the absolute value sign right now. In order for the absolute value of anything to be greater than or equal to 21, what does it mean? It means that whatever's inside of this absolute value sign, whatever that is inside of our absolute value sign, it must be 21 or more away from 0. Let's draw our number line. And you really should visualize a number line when you do this, and you'll never get confused then. You shouldn't be memorizing any rules. So let's draw 0 here. Let's do positive 21, and let's do a negative 21 here. So we want all of the numbers, so whatever this thing is, that are greater than or equal to 21. They're more than 21 away from 0. Their absolute value is more than 21. Well, all of these negative numbers that are less than negative 21, when you take their absolute value, when you get rid of the negative sign, or when you find their distance from 0, they're all going to be greater than 21. If you take the absolute value of negative 30, it's going to be greater than 21. Likewise, up here, anything greater than positive 21 will also have an absolute value greater than 21. So what we could say is 7x needs to be equal to one of these numbers, or 7x needs to be equal to one of these numbers out here. So we could write 7x needs to be one of these numbers. Well, what are these numbers? These are all of the numbers that are less than or equal to negative 21, or 7x-- let me do a different color here-- or 7x has to be one of these numbers. And that means that 7x has to be greater than or equal to positive 21. I really want you to kind of internalize what's going on here. If our absolute value is greater than or equal to 21, that means that what's inside the absolute value has to be either just straight up greater than the positive 21, or less than negative 21. Because if it's less than negative 21, when you take its absolute value, it's going to be more than 21 away from 0. Hopefully that make sense. We'll do several of these practice problems, so it really gets ingrained in your brain. But once you have this set up, and this just becomes a compound inequality, divide both sides of this equation by 7, you get x is less than or equal to negative 3. Or you divide both sides of this by 7, you get x is greater than or equal to 3. So I want to be very clear. This, what I drew here, was not the solution set. This is what 7x had to be equal to. I just wanted you to visualize what it means to have the absolute value be greater than 21, to be more than 21 away from 0. This is the solution set. x has to be greater than or equal to 3, or less than or equal to negative 3. So the actual solution set to this equation-- let me draw a number line-- let's say that's 0, that's 3, that is negative 3. x has to be either greater than or equal to 3. That's the equal sign. Or less than or equal to negative 3. And we're done. Let's do a couple more of these. Because they are, I think, confusing, but if you really start to get the gist of what absolute value is saying, they become, I think, intuitive. So let's say that we have the absolute value-- let me get a good one. Let's say the absolute value of 5x plus 3 is less than 7. So that's telling us that whatever's inside of our absolute value sign has to be less than 7 away from 0. So the ways that we can be less than 7 away from 0-- let me draw a number line-- so the ways that you can be less than 7 away from 0, you could be less than 7, and greater than negative 7. Right? You have to be in this range. So in order to satisfy this thing in this absolute value sign, it has to be-- so the thing in the absolute value sign, which is 5x plus 3-- it has to be greater than negative 7 and it has to be less than 7, in order for its absolute value to be less than 7. If this thing, this 5x plus 3, evaluates anywhere over here, its absolute value, its distance from 0, will be less than 7. And then we can just solve these. You subtract 3 from both sides. 5x is greater than negative 10. Divide both sides by 5. x is greater than negative 2. Now over here, subtract 3 from both sides. 5x is less than 4. Divide both sides by 5, you get x is less than 4/5. And then we can draw the solution set. We have to be greater than negative 2, not greater than or equal to, and less than 4/5. So this might look like a coordinate, but this is also interval notation, if we're saying all of the x's between negative 2 and 4/5. Or you could write it all of the x's that are greater than negative 2 and less than 4/5. These are the x's that satisfy this equation. And I really want you to internalize this visualization here. Now, you might already be seeing a bit of a rule here. And I don't want you to just memorize it, but I'll give it to you just in case you want it. If you have something like f of x, the absolute value of f of x is less than, let's say, some number a. Right? So this was the situation. We have some f of x less than a. That means that the absolute value of f of x, or f of x has to be less than a away from 0. So that means that f of x has to be less than positive a or greater than negative a. That translates to that, which translates to f of x greater than negative a and f of x less than a. But it comes from the same logic. This has to evaluate to something that is less than a away from 0. Now, if we go to the other side, if you have something of the form f of x is greater than a. That means that this thing has to evaluate to something that is further than a away from 0. So that means that f of x is either just straight up greater than positive a, or f of x is less than negative a. Right? If it's less than negative a, maybe it's negative a minus another 1, or negative 5 plus negative a. Then, when you take its absolute value, it'll become a plus 5. So its absolute value is going to be greater than a. So I just want to-- you could memorize this if you want, but I really want you to think about this is just saying, OK, this has to evaluate, be less than a away from 0, this has to be more than a away from 0. Let's do one more, because I know this can be a little bit confusing. And I encourage you to watch this video over and over and over again, if it helps. Let's say we have the absolute value of 2x-- let me do another one over here. Let's do a harder one. Let's say the absolute value of 2x over 7 plus 9 is greater than 5/7. So this thing has to evaluate to something that's more than 5/7 away from 0. So this thing, 2x over 7 plus 9, it could just be straight up greater than 5/7. Or it could be less than negative 5/7, because if it's less than negative 5/7, its absolute value is going to be greater than 5/7. Or 2x over 7 plus 9 will be less than negative 5/7. We're doing this case right here. And then we just solve both of these equations. See if we subtract-- let's just multiply everything by 7, just to get these denominators out of the way. So if you multiply both sides by 7, you get 2x plus 9 times 7 is 63, is greater than 5. Let's do it over here, too. You'll get 2x plus 63 is less than negative 5. Let's subtract 63 from both sides of this equation, and you get 2x-- let's see. 5 minus 63 is 58, 2x is greater than 58. If you subtract 63 from both sides of this equation, you get 2x is less than negative 68. Oh, I just realized I made a mistake here. You subtract 63 from both sides of this, 5 minus 63 is negative 58. I don't want to make a careless mistake there. And then divide both sides by 2. You get, in this case, x is greater than-- you don't have to swap the inequality, because we're dividing by a positive number-- negative 58 over 2 is negative 29, or, here, if you divide both sides by 2, or, x is less than negative 34. 68 divided by 2 is 34. And so, on the number line, the solution set to that equation will look like this. That's my number line. I have negative 29. I have negative 34. So the solution is, I can either be greater than 29, not greater than or equal to, so greater than 29, that is that right there, or I could be less than negative 34. So any of those are going to satisfy this absolute value inequality.