If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Inequalities using addition and subtraction

Inequalities Using Addition and Subtraction. Created by Sal Khan and CK-12 Foundation.

Want to join the conversation?

Video transcript

what i want to do in this video is a handful of fairly simple inequality videos. But the real value of it, I think, will be just to get you warmed up in the notation of inequality. So, let's just start with one. we have x minus 5 is less than 35. So let's see if we can find all of the x's that will satisfy this equation. And that's one of the distinctions of an inequality. In an equation, you typically have one solution, or at least the ones we've solved so far. In the future, we'll see equations where they have more than one solution. But in the ones we've solved so far, you solved for a particular x. In the inequalities, there's a whole set of x's that will satisfy this inequality. So they're saying, what are all the x's, that when you subtract 5 from them, it's going to be less than 35? And we can already think about it. I mean 0 minus 5. That's less than 35. Minus 100 minus 5. That's less than 35. 5 minus 5. That's less than 35. So there's clearly a lot of x's that will satisfy that. But what we want to do is come up with a solution that essentially encompasses all of the x's. So the way we do that is essentially the same way that we solved any equations. We want to get just the x terms, in this case, on the left-hand side. So I want to get rid of this negative 5, and I can do that by adding 5 to both sides of this equation. So I can add 5 to both sides of this equation. That won't change the inequality. It won't change the less than sign. If something is less than something else, something plus 5 is still going to be less than the other thing plus 5. So on the left-hand side, we just have an x. This negative 5 and this positive 5 cancel out. x is less than 35 plus 5, which is 40. And that's our solution. And to just visualize the set of all numbers that represents, let me draw a number line here. And I'll do it around-- let's say that's 40, this is 40, 41, 42. And then we could go below 40. 39, 38. You can just keep going below 40. It just keeps going on in both directions. And any x that is less than 40 will satisfy this. So it can't be equal to 40, because if x is equal to 40, 40 minus 5 is 35. It's not less than 35. So x has to be less than 40. And to show this on the number line, we do a circle around 40 to show that we're not including 40. But then we can shade in everything below 40. So everything that's just exactly below 40 is included in our solution set. So everything I've shaded in yellow is included in our solution set. So 39, 39.999999, repeating, which is about as close as you can get to 40 as possible, that's in our solution set. But 40 is not. That's why we put that open circle around it. Let's do another one. Let me do it in another color as well. So let's say we have x-- let me do it over on this top right corner. Say we have x plus 15 is greater than or equal to negative 60. Notice, now we have greater than or equal. So let's solve this the same way we solved that one over there. We can subtract 15 from both sides. And I like to switch up my notation. Here I added the 5, kind of, on the same line. You could also do your adding or subtracting below the line, like this. So if I subtract 15 from both sides, so I do a minus 15 there, and I do a minus 15 there. Then the left-hand side just becomes an x. Because obviously you have 15 minus 15. That just cancels out. And you get x is greater than or equal to negative 60 minus 15 is negative 75. If something is greater than or equal to something else, if I take 15 away from this and from that, the greater than or equal sign will still apply. So our solution is x is greater than or equal to negative 75. Let's graph it on the number line. So let me draw a number line here. I'll have-- let's say that's negative 75, that's negative 74, that's negative 73, that's negative 76. And so on and so forth. I could keep plotting things. Now, x has to be greater than or equal to negative 75. So x can be equal to negative 75. So we can include the point, because we have this greater than or equal sign. Notice we're not making it hollow like we did there, we're making it filled in because it can equal negative 75, or it needs to be greater than. So greater than or equal. We'll shade in everything above negative 75 as well. So in orange is the solution set. And this obviously, we could keep going to the right. x could be a million, it could be a billion, it could be a googol. It can be an arbitrarily large number as long as it's greater than or equal to negative 75. Let's do a couple more. Let's do x minus 2 is less than or equal to 1. Once again, we want to get just our x on the left-hand side. Get rid of this negative 2. Let's add 2 to both sides of this equation. Plus 2. The left-hand side just becomes an x. You have a less than or equal sign. That won't change by adding or subtracting the same thing to both sides of the inequality. And then 1 plus 2 is 3. So x needs to be less than or equal to 3. Any x that is less than or equal to 3 will satisfy this equation. So let's plot it. And I'd try out any x that's less than or equal to 3 and verify for yourself that it does indeed satisfy this inequality. I shouldn't call it an equation. This inequality. So let me graph the solution set. So let's say this is 0, 1, 2, 3, 4. That's negative 1, negative 2. So x has to be less than or equal to 3. It can be equal to 3, so we fill in the dot, or less than 3. So the solution set for over here is in pink. Anything less than or equal to 3. And verify it for yourself. If x is equal to 3, you get 3 minus 2, which is equal to 1. And that is valid because it could be less than or equal. If you do 2.999999 minus 2, you get 0.999999, which is less than 1. And you can keep trying for any of these numbers in this pink solution set here. Let's do one more. Let's say we have x minus 32 is less than or equal to 0. Same drill as before. Let's add 32 to both sides of this equation. The left-hand side just becomes x, and then the right-hand side is less than or equal to 32. Pretty straightforward. Same drill when we graph this equation. We draw the number line. If this is 32, this is 33, this is 31. I could keep adding things above and below 32. For the solution set is everything less than or equal, so we can-- it could be equal to 32, or less than. So we fill in everything below that. Remember, the reason why we're filling in this solid, the reason why 32 is an acceptable solution to this original inequality, is because of this less than or equal sign. Over here, you didn't have less than or equal, and that's why 40 wasn't part of the solution set.