1.5 Solving Inequalities

All right, we have 1.5 solving linear inequalities. We're going to have to write and graph our solutions in interval notation. And that may be new to some of you right here. So first of all, let's review with some simple ones. I have. Now first of all, in inequality means where we have greater than and less than or greater than and equal to or less than equal to. In other words, it's not equal to. So remember that word in right here means that not. So it's not equal to. So here I have X is less than three. So if we have a number line and I do simple number line graphs here, I'm not going to be too picky on this. But if this were my zero and let's say this is my three value, I want everything that is less than three. Now notice it's not equal to three. So what we do is we do an open circle because everything I shade. See how I'm coloring this right here. Everything I shade is implying that that is a solution. It's an answer. So there's many answers here because negative three is less than three, zero is less than three. There's many answers. In fact, infinite amount of answer goes going this way. But three itself is not less than three and therefore I have an open circle signifying that I'm not shading that in. Now, what's new to you is interval notation. It's how do I write that answer as one big solution? And what I have here is if I have any time I have the arrow, it's going to be a parenthesis. I'm going to read left to right. And what I'm basically doing is all of my solutions are from negative infinity to three. Notice it's an open parentheses. When I have it not equal to, I use parentheses. If I have something that's actually including that number or equal to it, I use a bracket. So right now, I'm going to use both parentheses in my answer to my solutions for this problem are negative infinity to three. Notice I went from left to right. Next, X is greater than or equal to negative two. So let's make a graph, get another color here. You don't have to do other colors, but I like to. And here is my value negative two, and I just put a zero here just to kind of for people to see. And it says all my solutions are larger than or equal to negative two. So that means a closed circle and everything to the right. Now I don't do any memorization like, oh, if the arrow is going this way, go right. I just know logically, it's larger, so it would go to the right. So what would be my interval notation for this? Okay. So my left hand point would be a bracket negative two, and then it goes to the right forever for infinity. Now infinity is never going to be equal because it's not an actual number. It's a concept. So you're always going to buy the infinities use a parenthesis. All right, next one, set. Solving inequalities with variables on one side, note when you multiply and divide both sides by a negative number, you reverse the inequality. So what we want to know is that we're going to solve just like we would do an equation, but we reverse the sign when we actually have multiplying or dividing by a negative. So let's take a look at this. Here I would go ahead and add 7 to both sides, just like I would do with a regular problem. That would be an equation. So I get three halves X is greater than two. Now to undo this, I could do this to a different ways. I could first multiply both sides by two, then divide both sides by three, or I could do real quickly, multiply by two thirds. Because this gives me one and then I would get X is greater than four thirds. So that's my solution. Now what would be my graph? Well, here is zero. And I'm just going to put four thirds here. I'm not going to go ahead and like, oh, it's partially in between here. I'm just going to put four thirds here. We know that that is greater than zero. And it says that our solutions are greater than four thirds. It's not equal to four thirds, so be careful. It's going to be an open circle. Greater than four thirds. All right, now what would be for this one? What would be my interval notation start from left? So here's left to right, okay? So I would have parentheses four thirds infinity because this is my left mouse, my rightmost is infinity, both would be parentheses because it is open. All right. 5 minus two X is greater than or equal to 27. So again, we're undoing it. I would subtract 5 on both sides. And I get negative two X is greater than or equal to 27. Now here's what we're talking about. I now have to divide both sides by negative two. So when I do this, I switch the sign. Because I'm dividing by a negative, okay? So now, because what's happening that's flipping over on the other side of your number line and where your X is, it's going to change where it's located. So the solution. That's the solution. So what is my graph? Well, in my head, what does that equal? 27. 13.5, right? So just showing you it's a negative 13.5. So let's have here zero, and I'm just going to write out negative 13.5. It says it's less than or equal to. So close it, less than or equal to. So notice it's close. So left to right, parenthesis, negative infinity. Notice it's negative infinity once going left. And I'd have negative 13.5 brackets. Solving inequalities with the variable on both sides. All right, so solve each inequality right and grab the solution set in interval notation. I suggest you always bring variables to the left side. So let's see what this would look like. I have 9 N plus 6 is less than or equal to 7 N plus four. I like to bring it to the left side because then how you graph it is relative to that variable, not relative to the number. So that sign will be relative to that variable. So make it a habit just go to the left hand side. And I can show you example of why that's best. Subtract 7 in on both sides. So I get two N plus 6 as less than or equal to four. Now, I go ahead and subtract 6 on both sides. And now I divide by two. So it's okay to divide by a positive. So I get my solution as this. Now what is my graph to this? Well, my graph to the is okay, here's my negative one. It's less than or equal to, so closed circle. Go in this way. And so my interval notation would be negative infinity, that's the leftmost. Comma negative one, bracket. Why is it a bracket? 'cause it's closed circle. It's equal to negative one. Here, we've got to do some distributing first. All right. Let's get everything over the letters to the left hand side and numbers to the right. So I'm going to add four X. Because when I added four X to a negative 5, I just get a negative X so now I have negative X as the last thing or equal to 7. Now I don't want to leave it like this. So what am I really dividing by? Negative one. So X change your sign because you're dividing by a negative one. My graph to this. Last thing are equal to negative 7. Interval notation. All right. Compound inequalities. These are dealing with and or or a compound inequality is too simple inequalities joined by and or and let's consider the difference. So let's first take a look at and. And is restrictive. Think of like the sandwich type, okay? See how little cute little memory aid there. Right here, these two conditions are going to be true. So when you see this right here, this isn't going to be like a normal equation, but I'm going to deal with all three parts. So when I have to undo this and get my variable by itself, I'm subtracting 8 on all three parts. So I get negative ten. As less than or equal to three T is less than or equal to two. Now I divide by three on all three parts. So negative ten thirds. Okay. So my graph, what this is saying. So let's put my negative ten thirds down here in my two thirds over here. What it's saying is T is everything in between that. It's larger than negative ten thirds and it's equal. But it's also less two thirds. So it's going to be all in here. Think of the sandwich that both are true at the same time. Now my interval notation would look like this. Because you have that equal to, and so therefore it would be a bracket. Number 7, negative 12 is less than or equal to three X minus three is less than 15. So I'd have first of all add three to all three parts. So I get negative 9 as less than or equal to three X 18. Now I want to divide by three on all three parts. So negative three. And that would be 6. To graph this, I have my negative three, 6, put them in order. Now this one is closed, but this is open. And so I would get negative three, 6. So what about or think of oars on a boat? What are about? You have two of them, okay? And so you're going to have on the side where of the oars on the boat, they're on both on the outside of the boat, okay? So think of that for oars. They're on two solutions. Each side. Here I'm going to solve this inequality in this inequality. So here I have two X is less than two. X is less than one. Over here, I have four X is greater than 16. X is greater than four. How do we graph this? So this will make you see what I mean by oars on a boat. So here's my one. Here's my four. Everything less than one. Not equal to one, but everything less than one. So see how it's going that way. And here is everything greater than four. Can you imagine your little boat here? Here's your boat. And you have the two oars there. They're going on opposite sides. So think of that for oars on a boat. So what's my interval notation? Now, wait a second. That's a lot. So first I got to do for interval notation. I have to do this part and then I have to do that part. So for interoperable location I'd have negative infinity to one, union with four to infinity. So I think combining two different interval notations is what you're doing. So let's try another one. So again, solve for this, watch out, has a little star there because we've got a really watch out for that negative. Negative two X is less than or equal to negative four. I am dividing by a negative, so therefore we change our inequality sign. Here, two X is less than ten. X is less than 5. Let's graph this. I have a two, and I have a 5. All right, now because this is an or. It's saying I have everything greater than two. And everything less than 5. What is my solution? Now, if this was an and we would only color in what they both had in common, but this is an or. So therefore my solution is all solutions. And there you have it. So let's get started and practice our compound. And regular inequalities.