# Absolute value inequalities | Linear equations | Algebra I | Khan Academy

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.

They should make khan text books

THIS ISN’T HELPING AHHH

I LOVE YOU, I HAVE AN INTERM EXAM TOMORROW

1:46 helps

I have a massive algebra test tomorrow and I am rewatching these vids to help prepare. This really clarifies everything for me

Thanks for the help 🙂

my kpop brain be like f(x) oh that's a girl group.. AHAHA

Nice 😘

jazaka allah khayran

get your Closed Captioning but out of here

When you don’t pay attention in class

Thank you so much! I understand now:)

how do you know if its and or or

You people are lifesavers

hello sir I I'm shashi and see your videos which is very important for me sir now I have to faced problem in inequalities of modules with fraction so please help me and to making so useful video

#YouCanLearnAnything

Wow u should be a teacher

explicit. Thanks sal.

you saved me from my algebra class thankyou

math exam tomorrow, and u taught this so well that after the first number line i could do the most complicated absolute values. Thanks man

Horrible

Than youuuuuu

You were the only person that brought tears of joy on my cheeks in the middle of tears of unhappiness (FINALS!)

Still wondering about All Real No.s and no solution tho? Can someone teach me

this is very helpful

thanks

what about union and intersection?

i love khan academy sm

HEEEEEY

MEKENI MEKENI DUGDUG DOREMI !!!

i love khan academy so much but

you're road, you know?

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My problem in math is nothing like this it's. ➡️➡️ |x-2| < 10. Does anyone know how to do this 🤔

First 9th grade then 10th and now current 11th.. text books changed ..teachers changed..school changed ..but sal is still there ..thanks a lot..

Who else has the khan academy up but is still watching it on youtube

Khan academy is waste of time!

Very well explained, guys if you don't get this first study the concept of graphical solutions of inequality and intervals then u will surely understand this

Thank you. This is helpful

Thank you so much for this video 🙂

Awesome teaching

You're so helpful. Thanks, Khan!

Omg I've gone my whole life without understanding absolute value correctly. I'm studying for the GRE and was really struggling with these problems. Great video!

man…. wich set works? dose both of them work? can somedody anser this repliy to me pleaze

this was literally no help at all.

thank god for sal

What do I do if the problem is something like: |x-3| = 1

thank you SO MUCH!! this was extremely helpful

as a 7th grader taking math 1 i can confidently say…

i’m gonna die on my quiz tom… yay

if there is an x value on the right side of the greater to or equal than sign, do i carry it over first before I consider the 2 or options and make sure Xs are on one side and it's greater to equal to 0?

This is actually better than my math teacher

Plz tell why did u write lxl=12 as x less than -12

dude thanks

Thank you a lot ,I could understand everything as the way you teach is excellent.

This is why Khan Academy is the boss!

thanks

Thank you 😭♥️🙏🏻

i put 10 of this same video in different tabs and its great. THANKS KHAN ACADEMY!!!!!

Thanks for the clarification Sal! Now I understand why the intervals point where!

In class we went over this for two weeks and never understood it until I watched this video and finally got the hang of it

CONFUSING

So then what is the greatest value of absolute value x+3 less than or equal to 2…

1. -1

2. -2

3. -3

4. -4

5. -5

Wow nice video

#Thanks

You're an awesome educator! thank you!!

im only grade 7 but this is our lesson right now.

Thank you so much, your explanation was easy to understand. Thxxxx!

wow. this is so easy after watching the first 1 minute. thanks!

thanks partner

what if there is no solution how does the graph look?

Oh yeah yeah

Who is actualy a 1st grader?

Okay, but like the first problem: X is less than 12. You write the solution: (-12,12) Yes, -12 is less than 12. But -13 is also less than 12. -14 is also less than 12. -15 is also less than 12, ect. So technically, shouldn't the the solution be [negative infinity, 12)? I just don't understand why you choose negative 12 and stop there. This is not the way the way I was taught.

i can’t understand my teachers teaching but this video actually made me get the hang of it

One primary complaint is incorrect wording – using the word "equation" when should be saying "inequality". This verbiage error occurs multiple times throughout the video.

What I do in my free time

Yall pray for me I have a science test (forces) and a algebra test (absolute values) tommorow !!

lol while is everyone studying for finals, I'm just tryna do ok in 7th-grade maths

wow I understood it now. no memorizing now

Thumbs up if your summer semester Calc II

Why can't you be my math teacher?? I have a math teacher who dont know what i know in math. Education is not equally distributed among the world.

Doing and AP stats review packet and I don’t get the first question so now I’m here 🤪

Thank you sir

Thanq so much!!

your way better then my teacher

Anyone from 9th

this was super helpful im in algebra 2 and we are learning this and it made no sense with the teacher but now it makes so much more sense thank you

This didn’t help

Omg I wish all school teachers taught like this. Sal makes everything easier. I love him.

and he writes this all with a mouse

I think

god bless you.

how did he get the 5???!!!!

Khan i love you

How do you graph 7|5p-7|=-21

I think this is great. I had all the rules and stuff but it really helps you when you understand what is happening. Im so much better at this.

10 years later still saving lives

My teacher doesn’t teach me. Sal does.

This couldn't have being explained any clearer.

oof I have a test in 10 minutes and I didn’t study

8:56 should be "and", not "or"

I’m doing this in the beginning of eight grade all you high school nerds

Its ironic that if school teachers were good and explained the topic good this channel wouldn't be as popular or even exist.