# Compound inequalities examples

CCSS Math: HSA.REI.B.3

## Video transcript

Let's do some compound inequality problems, and these are just inequality problems that have more than one set of constraints. You're going to see what I'm talking about in a second. So the first problem I have is negative 5 is less than or equal to x minus 4, which is also less than or equal to 13. So we have two sets of constraints on the set of x's that satisfy these equations. x minus 4 has to be greater than or equal to negative 5 and x minus 4 has to be less than or equal to 13. So we could rewrite this compound inequality as negative 5 has to be less than or equal to x minus 4, and x minus 4 needs to be less than or equal to 13. And then we could solve each of these separately, and then we have to remember this "and" there to think about the solution set because it has to be things that satisfy this equation and this equation. So let's solve each of them individually. So this one over here, we can add 4 to both sides of the equation. The left-hand side, negative 5 plus 4, is negative 1. Negative 1 is less than or equal to x, right? These 4's just cancel out here and you're just left with an x on this right-hand side. So the left, this part right here, simplifies to x needs to be greater than or equal to negative 1 or negative 1 is less than or equal to x. So we can also write it like this. X needs to be greater than or equal to negative 1. These are equivalent. I just swapped the sides. Now let's do this other condition here in green. Let's add 4 to both sides of this equation. The left-hand side, we just get an x. And then the right-hand side, we get 13 plus 14, which is 17. So we get x is less than or equal to 17. So our two conditions, x has to be greater than or equal to negative 1 and less than or equal to 17. So we could write this again as a compound inequality if we want. We can say that the solution set, that x has to be less than or equal to 17 and greater than or equal to negative 1. It has to satisfy both of these conditions. So what would that look like on a number line? So let's put our number line right there. Let's say that this is 17. Maybe that's 18. You keep going down. Maybe this is 0. I'm obviously skipping a bunch of stuff in between. Then we would have a negative 1 right there, maybe a negative 2. So x is greater than or equal to negative 1, so we would start at negative 1. We're going to circle it in because we have a greater than or equal to. And then x is greater than that, but it has to be less than or equal to 17. So it could be equal to 17 or less than 17. So this right here is a solution set, everything that I've shaded in orange. And if we wanted to write it in interval notation, it would be x is between negative 1 and 17, and it can also equal negative 1, so we put a bracket, and it can also equal 17. So this is the interval notation for this compound inequality right there. Let's do another one. Let me get a good problem here. Let's say that we have negative 12. I'm going to change the problem a little bit from the one that I've found here. Negative 12 is less than 2 minus 5x, which is less than or equal to 7. I want to do a problem that has just the less than and a less than or equal to. The problem in the book that I'm looking at has an equal sign here, but I want to remove that intentionally because I want to show you when you have a hybrid situation, when you have a little bit of both. So first we can separate this into two normal inequalities. You have this inequality right there. We know that negative 12 needs to be less than 2 minus 5x. That has to be satisfied, and-- let me do it in another color-- this inequality also needs to be satisfied. 2 minus 5x has to be less than 7 and greater than 12, less than or equal to 7 and greater than negative 12, so and 2 minus 5x has to be less than or equal to 7. So let's just solve this the way we solve everything. Let's get this 2 onto the left-hand side here. So let's subtract 2 from both sides of this equation. So if you subtract 2 from both sides of this equation, the left-hand side becomes negative 14, is less than-- these cancel out-- less than negative 5x. Now let's divide both sides by negative 5. And remember, when you multiply or divide by a negative number, the inequality swaps around. So if you divide both sides by negative 5, you get a negative 14 over negative 5, and you have an x on the right-hand side, if you divide that by negative 5, and this swaps from a less than sign to a greater than sign. The negatives cancel out, so you get 14/5 is greater than x, or x is less than 14/5, which is-- what is this? This is 2 and 4/5. x is less than 2 and 4/5. I just wrote this improper fraction as a mixed number. Now let's do the other constraint over here in magenta. So let's subtract 2 from both sides of this equation, just like we did before. And actually, you can do these simultaneously, but it becomes kind of confusing. So to avoid careless mistakes, I encourage you to separate it out like this. So if you subtract 2 from both sides of the equation, the left-hand side becomes negative 5x. The right-hand side, you have less than or equal to. The right-hand side becomes 7 minus 2, becomes 5. Now, you divide both sides by negative 5. On the left-hand side, you get an x. On the right-hand side, 5 divided by negative 5 is negative 1. And since we divided by a negative number, we swap the inequality. It goes from less than or equal to, to greater than or equal to. So we have our two constraints. x has to be less than 2 and 4/5, and it has to be greater than or equal to negative 1. So we could write it like this. x has to be greater than or equal to negative 1, so that would be the lower bound on our interval, and it has to be less than 2 and 4/5. And notice, not less than or equal to. That's why I wanted to show you, you have the parentheses there because it can't be equal to 2 and 4/5. x has to be less than 2 and 4/5. Or we could write this way. x has to be less than 2 and 4/5, that's just this inequality, swapping the sides, and it has to be greater than or equal to negative 1. So these two statements are equivalent. And if I were to draw it on a number line, it would look like this. So you have a negative 1, you have 2 and 4/5 over here. Obviously, you'll have stuff in between. Maybe, you know, 0 sitting there. We have to be greater than or equal to negative 1, so we can be equal to negative 1. And we're going to be greater than negative 1, but we also have to be less than 2 and 4/5. So we can't include 2 and 4/5 there. We can't be equal to 2 and 4/5, so we can only be less than, so we put a empty circle around 2 and 4/5 and then we fill in everything below that, all the way down to negative 1, and we include negative 1 because we have this less than or equal sign. So the last two problems I did are kind of "and" problems. You have to meet both of these constraints. Now, let's do an "or" problem. So let's say I have these inequalities. Let's say I'm given-- let's say that 4x minus 1 needs to be greater than or equal to 7, or 9x over 2 needs to be less than 3. So now when we're saying "or," an x that would satisfy these are x's that satisfy either of these equations. In the last few videos or in the last few problems, we had to find x's that satisfied both of these equations. Here, this is much more lenient. We just have to satisfy one of these two. So let's figure out the solution sets for both of these and then we figure out essentially their union, their combination, all of the things that'll satisfy either of these. So on this one, on the one on the left, we can add 1 to both sides. You add 1 to both sides. The left-hand side just becomes 4x is greater than or equal to 7 plus 1 is 8. Divide both sides by 4. You get x is greater than or equal to 2. Or let's do this one. Let's see, if we multiply both sides of this equation by 2/9, what do we get? If you multiply both sides by 2/9, it's a positive number, so we don't have to do anything to the inequality. These cancel out, and you get x is less than 3 times 2/9. 3/9 is the same thing as 1/3, so x needs to be less than 2/3. So or x is less than 2/3. So that's our solution set. x needs to be greater than or equal to 2, or less than 2/3. So this is interesting. Let me plot the solution set on the number line. So that is our number line. Maybe this is 0, this is 1, this is 2, 3, maybe that is negative 1. So x can be greater than or equal to 2. So we could start-- let me do it in another color. We can start at 2 here and it would be greater than or equal to 2, so include everything greater than or equal to 2. That's that condition right there. Or x could be less than 2/3. So 2/3 is going to be right around here, right? That is 2/3. x could be less than 2/3. And this is interesting. Because if we pick one of these numbers, it's going to satisfy this inequality. If we pick one of these numbers, it's going to satisfy that inequality. If we had an "and" here, there would have been no numbers that satisfy it because you can't be both greater than 2 and less than 2/3. So the only way that there's any solution set here is because it's "or." You can satisfy one of the two inequalities. Anyway, hopefully you, found that fun.