Main content

## Algebra (all content)

### 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

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Solving absolute value inequalities: fractions

Sal solves the inequality |2r-3 1/4| < 2 1/2. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- I have trouble with solving combined absolute value inequalities. For example how would you solve and graph: 9< |4x-5| <23(10 votes)
- One way to solve this problem would be to start by rewriting is as 9 < |4x - 5| and |4x - 5| < 23. Then solve each part.

9 < |4x - 5|

|4x - 5| > 9

4x - 5 > 9 or 4x - 5 < -9

4x > 14 or 4x < -4

x > 7/2 or x < -1

|4x - 5| < 23

4x - 5 < 23 and 4x - 5 > -23

4x < 28 and 4x > -18

x < 7 and x > -9/2

-9/2 < x < 7

Finally, the solution is the numbers that satisfy the criteria between -9/2 and 7 for which x > 7/2 or x < -1. So the answer is -9/2 < x < -1 or 7/2 < x < 7. On a number line, this would be represented by shading the numbers between -9/2 and -1 and the numbers between 7/2 and 7.(17 votes)

- i have a really confusing problem related to this and i need some help on it!:

2|4x+1|-5≤ -1(5 votes)- Okay you want the absolute value by itself. Add 5 to both sides to get 2|4x+1| ≤ 4. Next you divide both sides by 2 to get |4x+1| ≤ 2. Ok now you put 4x+1 ≤ 2 AND 4x+1 ≥ -2. The first one (4x+1 ≤ 2) is simplified to x ≤ 1/4. The second one (4x+1 ≥ -2 ) works out to x ≥ -3/4. Thus, the answer is -3/4 ≤ x ≤ 1/4.(10 votes)

- how so you know if its a "or" or an "and"?(3 votes)
- Think GreaTOR and Less thAND(17 votes)

- How is the absolute value used in the real world?(4 votes)
- ok I think you understand David that - is like debt and + is making money(4 votes)

- how do you know when to put "and" or "or"?(3 votes)
- If the absolute value quantity is less than the other value, use "and". If the absolute value quantity is greater than the other value use "or".(5 votes)

- I have done this before and for example 5647= isn't it the same number?(4 votes)
- I don't know double check it ! ;-)(5 votes)

- When you say, "we'll take the absolute value of it." what do you mean? Are you changing anything such as adding, subtracting, dividing, ect?(4 votes)
- You are simply making the value of the integer positive. It is helpful when considering time and distance, because it is difficult to have negative values for distance and time. That is, unless you're a timelord.(3 votes)

- The absolute value of an expression less than some other number is an AND problem by definition?

So |x|<12 is an AND problem and |x|>12 would be an OR problem? You can tell that by the sign?(2 votes)- Yes, you are correct. |x|<12 is the same as x<12 AND x>-12, while |x|>12 is the same as x>12 OR x<-12. So if a problem has a less than sign, it is an AND problem and if a problem has a greater than sign, it is an OR problem(5 votes)

- What software does Sal use for writing down everything?(2 votes)
- Is this an overall fraction??(2 votes)

## Video transcript

We have the absolute value of
2r minus 3 and 1/4 is less than 2 and 1/2, and we
want to solve for r. So right from the get go we
have to deal with this absolute value. And just as a bit of a review,
if I were to say that the absolute value of x is less
than, well let's just say, less than 2 and 1/2, that means
that the distance from x to 0 is less than 2 and 1/2. That means that x would have to
be less than 2 and 1/2, and x would have to be greater
than negative 2 and 1/2. And just think about
it for a second. If I were to draw it on a number
line right here, that is 0, that is 2 and 1/2, and
that is negative 2 and 1/2. These two numbers are exactly 2
and 1/2 away from 0, because both of their absolute
values is 2 and 1/2. Now, if we want all of the
numbers whose absolute value is less than 2 and 1/2, or that
are less than 2 and 1/2 away from 0, it would be all
of the numbers in between. And that's exactly what these
two statements are saying. x has to be less than 2 and 1/2,
and it has to be greater than negative 2 and 1/2. If this absolute value were
the other way, that the absolute value of x has to be
greater than 2 and 1/2, then it would be the numbers
outside of this, and it would be an or. But we're dealing with the
less than situation right there, so let's just do what
we were able to figure out when it was just an x. The distance from this thing to
0 has to be less than 2 and 1/2, so we can write that 2r
minus 3 and 1/4 has to be less than 2 and 1/2 and 2r minus 3
and 1/4 has to be greater than negative 2 and 1/2. Same exact reasoning here. Let me draw a line so we
don't get confused. Same exact reasoning here. This quantity right here
has to be between negative 2 and 1/2. It has to be greater
than negative 2 and 1/2 right there. And it has to be less
than 2 and 1/2, so that's all I wrote there. So let's solve each of
these independently. Well, this first went over here,
you've learned before that I don't like improper
fractions, and I don't like fractions in general. So let's make all of
these fractions. Sorry, I don't like
mixed numbers. I want them to be improper
fractions. So let's turn all of these
into improper fractions. So if I were to rewrite it, we
get 2r minus 3 and 1/4 is the same thing as 3 times 4
is 12, plus 1 is 13. 2r minus 13/4 is less than-- 2
times 2 is 4, plus 1 is five-- is less than 5/2. So that's the first equation. And then the second question--
and do the same thing here-- we have 2r minus 13 over
4 has to be a greater than negative 5/2. All right, now let's solve each
of these independently. To get rid of the fractions, the
easiest thing to do is to multiply both sides of
this equation by 4. That'll eliminate all of the
fractions, so let's do that. Let's multiply-- let me scroll
to the left a little bit-- let's multiply both sides
of this equation by 4. And what do we get? 4 times 2r is 8r, 4 times
negative 13 over 4 is negative 13, is less than-- and I
multiplied by a positive number so I didn't have to
worry about swapping the inequality-- is less than 5/2
times 4 is 10, right? You get a 2 and a 1, it's 10. So you get 8r minus 13
is less than 10. Now we can add 13 to both sides
of this equation so that we get rid of it on the
left-hand side. Add 13 to both sides and we
get 8r-- these guys cancel out-- is less than 23,
and then we divide both sides by 8. And once again, we didn't have
to worry about the inequality because we're dividing
by a positive number. And we get r is less
than 23 over 8. Or, if you want to write that
as a mixed number, r is less than-- what is that--
2 and 7/8. So that's one condition, but we
still have to worry about this other condition. There was an and right here. Let's worry about it. So our other condition tells us
2r minus 13 over 4 has to be greater than negative 5/2. Let's multiply both sides
of this equation by 4. So 4 times 2r is 8r. 4 times negative 13 over 4 is
negative 13, is greater than negative 5/2 times
4 is negative 10. Now we add 13 to both sides
of this equation. The left-hand side-- these guys
cancel out, you're just left with 8r-- is greater than
negative 10 plus 13 is 3. Or divide both sides of this by
8, and you're left with r has to be greater than 3/8. So our two conditions, r has to
be less than 2 and 7/8 and greater than 3/8. Or we can just write it like
this: r is greater than 3/8, so it's greater than-- maybe I
should say 3/8 is less than r, which is less than 2 and 7/8. So if we were to plot the
solution on the number line-- which I'm about to do, so that's
my number line-- this is 0 right here, maybe
this is 1, 2, and 3. We have 2 and 7/8. We have to be less
than 2 and 7/8. Let's say that this is 2
and 7/8 right there. And we have to be greater
than 3/8. Say that is 3/8, so 3/8
will be some place right around there. And everything in between
is a valid solution. And we could try it out. Let's try out something that,
based on what I just drew, should be a valid solution. 1 should be a valid solution. Let's try it out here. 2 times 1 minus 3 and
1/4, what is that? That's 2 minus 3 and 1/4. And so what is that? 2 minus 3 and 1/4 is-- well, 3
and 1/4 minus 2 is 1 and 1/4, so this will be negative
1 and 1/4. But we're taking the absolute
value of it, so we take the absolute value of it, which is
equal to 1 and 1/4, which is indeed less than 2 and 1/2. Now let's try another number. Let's try 0. Based on this, 0 should
not work. So what happens if
we put 0 here? You get 2 times 0, which
is 0, minus 3 and 1/4. If you take the absolute value
of negative 3 and 1/4, you'll get positive 3 and 1/4,
which won't work. 3 and 1/4 is greater than
2 and 1/2, so that's true, that works out. And same thing for 3. 2 times 3 is 6, minus 3
and 1/4 is 2 and 3/4. Take the absolute value, it's 2
and 3/4, still bigger than 2 and 1/2, so it won't work. So at least the points that we
tried out seem to validate this solution that we got.