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## Algebra (all content)

### Course: Algebra (all content) > Unit 2

Lesson 14: One-step inequalities# Inequalities using addition and subtraction

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

## Want to join the conversation?

- What does inequality mean? I know it is dumb to ask but I don't understand this kind of stuff. Thanks to Khan Academy for all the help they have given me this year. Me, my friends and my family have learned a lot. THANKS!(20 votes)
- An inequality is an expression that includes < > ≤ ≥ and sets up a scenario where the two terms on either side of the expression are not equal (hence inequality).(20 votes)

- so the "<" and ">" when the pointy parts points to the number it means smaller than the other number?(6 votes)
- That's a good way to remember.

The way I remembered is that the big opening points towards the bigger number.(7 votes)

- At4:05why the sign didn't switch?(8 votes)
- at4:59he says it could be a googol how many zeros/digits is that?(3 votes)
- And googolplex is a 1 with google zeros after it!(4 votes)

- This is so confusing? When will we use this in the real would, I am genuinely asking.(5 votes)
- I still don't understand how to graph it on a number line. Can anyone help me>?(3 votes)
- There are some videos on that topic. You can find them at this link: https://www.khanacademy.org/math/algebra-home/alg-basic-eq-ineq#alg-inequalities

Hope this helps.(5 votes)

- but if it says it must be less than 35, wouldn't the limit be 39?(5 votes)
- I know how to do this, but my question is how to add or subtract inequalities themselves. Ex) 8>2x + 6x-5<14(3 votes)
- Your inequality make complete sense. Double inequalities should have 2 less than symbols like 8<2x+6x-5<14, or 2 greater than symbols. Was the problem given to you in this form? Or, did you change something. Please clarify.

There is a section on compound inequalities that might help you:

https://www.khanacademy.org/math/algebra-home/alg-basic-eq-ineq#alg-compound-inequalities(2 votes)

- when do you switch the inequality sign when adding and subtracting(2 votes)
- You don't. You only reverse the inequality if you multiply or divide the inequality by a negative number.(2 votes)

- so the open circle means its not equall to?(1 vote)
- Yes. You are correct. A filled in circle included equal to.(3 votes)

## 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.