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### Course: Algebra (all content) > Unit 8

Lesson 3: Solving absolute value inequalities- Intro to absolute value inequalities
- Solving absolute value inequalities 1
- Solving absolute value inequalities 2
- Solving absolute value inequalities: fractions
- Solving absolute value inequalities: no solution
- Absolute value inequalities word problem

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# Intro to absolute value inequalities

To solve inequalities with absolute values, use a number line to see how far the absolute value is from zero. Split into two cases: when it is positive or negative. Solve each case with algebra. The answer is both cases together, in intervals or words. Created by Sal Khan and CK-12 Foundation.

## Want to join the conversation?

- Please help me solve this step by step

5|8-y|=30

Thanks!(38 votes)- 5I8-yI=30

divide by 5 on each side so you get...

I8-yI=6 or I8-yI=-6

Subtract 8 on each each equation to get...

I-yI=-2 or I-yI=-14

Divide by a negative on each side of the equation...

y=2 or y=14(21 votes)

- quick question. When he breaks down the inequality from Abs.val. 5x+3 less than seven, why does he reverse the inequality when saying 5x+3>-7?(21 votes)
- I think your question seems like this I5x+3I<7

so 5x+3 is always less than 7.

so consider few cases

If 5x+3=2; the above given condition is satisfied.

now if 5x+2 = -8 then abs value of 5x+3 is greater than 7 so inequality fails.

so if you break the above mod value then its value must lie in between (-7,7)

and the inequality looks like -7< 5x+3<7.(4 votes)

- For absolute value inequalities, when the constant on one side is negative, is there then no answer, like absolute value equations, or do you solve it normally?(8 votes)
- Great question!

It is a little different with absolute value inequalities. Sounds like you already are aware that with absolute value equations if you have a negative constant on one side after the absolute value is isolated, then the equation has no solution. The reason this happens is the absolute value always creates a positive number. And, a positive number will never = a negative number.

So, let's extend that concept to inequalities.

1) An absolute value < a negative: In this situation, we have a positive value < a negative value. That would always be false. This is a contradiction... therefore, this inequality has "No Solution".

2) An absolute value > a negative: In this situation, we have a positive value > a negative value. That would always be true. It is an identity... the solution to the inequality becomes "all real numbers".

Hope this helps.(13 votes)

- how would you solve a question like |x| + |x-1| <5 ?(10 votes)
- Please help I am so dumbfounded. Do you know how to determine an interval when you have two variables? I have a problem:

"Write your answer as an interval, or as a number if applicable. Find all real number of x and y such that: |7x-iy| = y-i6x "

So I know that you can only add the real terms with the real and the imaginary terms with the imaginary. I tried ignoring the imaginary values since those seem to usually cancel out in some absolute value situations. I was confirmed to be correct that y = [0,inf). However I have been spending way too long trying to figure out what x is I tried a lot of things and I'm just stuck.(6 votes)- |7𝑥 – 𝑦𝑖| = 𝑦 – 6𝑥𝑖

Immediately, I can tell that the only value 𝑥 can be is 0. Why? Well the magnitude of a complex number is always a real number. In this equation, we have equated the magnitude of a complex number to another complex number, and so we must make the imaginary part of the second complex number 0. Hence, 𝑥 = 0. Making this substitution:

|-𝑦𝑖| = 𝑦

√𝑦² = 𝑦

But since magnitude is always nonnegative, we have 𝑦 ∈ [0, ∞). Therefore, the solution is:

𝑥 ∈ {0}

𝑦 ∈ [0, ∞)

Comment if you have questions.(6 votes)

- what is the big difference between Equations and Inequalities? he corrects it at8:29. it's always being corrected.(4 votes)
- An equation means that it has an equal sign. When this happens, it shows that there is the answer, whereas with inequalities, it says that it is greater than this.

For instance, say you had: |x-4| = 7

This would work out to be: x = 11 OR x = -3

If we made this problem an inequality, it would work out the same but the signs would be different: |x - 4| < 7

x < 11 AND x > -3(3 votes)

- How do you solve -7|2X-12|<70?(3 votes)
- -7|2X-12|<70

First, divide both sides of the inequality by -7. Don't forget to switch the sign around for dividing by a negative number.

|2X-12| > -10

And then the answer is all real numbers. Think about it, no matter what X is, after you plug in the numbers, the absolute value sign will make the left hand side be at least 0. It is impossible to get an answer less than 0, let alone -10. So all values of X will provide an answer greater than -10, so all real numbers will work for this inequality.(4 votes)

- If what was is in the absolute value sign in the original inequality is negative, does it change to positive because it's in an absolute value sign? For example if the problem is |-5x| < 3

Does it change to

-3<5x<3 or does it stay as -3<-5x<3?(2 votes)- It really does not matter, because both give the same solution set at the end. Both -3<5x<3 and -3<-5x<3 give -3/5<x<3/5 as the solution.(1 vote)

- How would I solve 4-|8+10r|≥ -28?

Please show me step by step. I'm not sure if this is a special case or not.(2 votes)- Start by isolating the absolute value

1) Subtract 4: -|8+10r| >= -32

2) Multiply both sides by -1 (or divide by -1): |8+10r|<=32

3) This is the point were you would determine if it is a special case. Since the 32 is positive, it is not a special case.

4) So, continue solving by splitting the inequality:

8+10r<=32 AND -(8+10r)<=32

Can you finish from here? Give it a try. Comment back if you have questions.(2 votes)

- To simplify √ (4-x^2) > 0 do we take it to be |4-x^2| > 0 on squaring both sides or

4-x^2 > 0 which will then give x^2 < 4 and |x| < 2 ?(2 votes)- if you are not working with complex numbers, so to speak imaginary units, you are right, yes

as a control you can try inserting some values for x like:

x = 0 => √ (4-0²) = √ (4) = 2 > 0 correct

x = -1 => √ (4- (-1)²) = √ (4 + 1) = √ (5) = 2.236... > 0 correct

x = 5 => √ (4 - 5²) = √ (4 - 25) = √ (-21) ... impossible for real numbers!

x = 2 => √ (4 - 2²) = √ (0) ... is not smaller than 0

you see that you're right... ;)(2 votes)

## Video transcript

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.