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## 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

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# Solving absolute value inequalities 1

Sal solves the inequality |h|-19.5 < -12. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- How is absolute value used in the world?(62 votes)
- absolute value is used in real world to find magnitude or length of particular things (scalar value).

Examples:

How far San Fransisco from New York City?

Answer : x km (It will never be negative value)

How long is your pencil?

Answer : 10 cm , not -10 cm(69 votes)

- what exactly is a SOLUTION SET? does it include the number line graphing, or is it the
*>*>_ , or the (_,_) , or all of them together?(13 votes)- The solution set of a given equation is the set of all the elements that satisfy the equation. This means that all elements in the set make the equation a true proposition.

For example, consider the equation x+3=4

The solution set is the set {1} because when you substitute x in the equation you get 1+3=4 which is a true proposition.(19 votes)

- In equations, we can shift the numbers/variables from one side to another But, in inequalities can we shift the numbers/variables from one side to another?(10 votes)
- Of course, as long as you divide and multiply by positive numbers it is fine. But as soon as you multiply or divide by negative numbers, you have to swap the inequality. The addition and subtraction can be handled with easily, and the inequality can retain its direction.(18 votes)

- What if we have | x | = - 2 ? ... Is this even solvable?(7 votes)
- Absolute values are always positive, because the distance from zero to any number is never going to be negative. Since the absolute value of no number is -2, there is no solution.(12 votes)

- Are there different ways to do absolute values??

cause my teacher taught me a different way and I don't exactly understand the reasoning behind what she does??(6 votes)- There is technically only one way to solve absolute values, which is to make the non variable side both negative and positive, but if you are talking about simplification, there are many ways to simplify an absolute value expression to solve it.(5 votes)

- Okay, I answered this question on my test the other day, and I know what the answer is, but I can't figure out how to get there. I get half of the inequality right, but I can't get the other half. Here's the question:
`|(1/2)x-1|+(3/2)x<10`

I got the absolute value isolated:`|(1/2)x-1|<10-(3/2)x`

And rewrote the equation to get rid of the absolute value signs:`-10+(3/2)x<(1/2)x-1<10-(3/2)x`

Add 1 to both sides:`-9+(3/2)x<(1/2)x<11-(3/2)x`

Multiply both sides by 2:`-18+(6/2)x<x<22-(6/2)x`

Split the inequality into two parts:`-18+3x<x`

and`x<22-3x`

Subtract 3x and add 3x:`-18<-2x`

and`4x<22`

Divide by -2 and 4:`9>x`

and`x<11/2`

So I get the`x<11/2`

right, but the other half is supposed to be`0<x`

not`9>x`

. Where'd I go wrong!?(2 votes) - I understand this example for the -7 1/2 is less than h, which is less than 7 1/2. But the answer to a problem such as: n-3 (is less than) 6 would be -3(is less than) n (is less than) 9. How does that work if it's not just a negative version of the answer you first got? (Sorry if this question is unclear)(2 votes)
- how do you do the problems with...

{x|x|<5(2 votes) - When Sal Adds 19 1/2 to - 12 how come he does not have to make the denominator on -12 equal to 19 1/2? Isn't -12 the same thing as -12/1 so if we are adding/subtracting fractions dont the denominators have to be equal?(1 vote)
- If we're subtracting what are called "common fractions", like 37/4 or 20/3, where the only value is a fraction and not a mixed number, then we'd probably make the fractions have equal denominators. But in this case, we don't need to, though we certainly
**could**- it's easier to leave them as improper fractions and subtract 12 from 19 to get 7 and then add 1/2, since (19+1/2)-12 = (19-12)+1/2.(2 votes)

- Is the sign supposed to flip?(2 votes)
- I am not sure I understand...(1 vote)

## Video transcript

We're told to graph all possible
values for h on the number line. And this is a especially
interesting inequality because we also have an absolute
value here. So the way we're going to do it,
we're going to solve this inequality in terms of the
absolute value of h, and from there we can solve it for h. So let's just get the absolute
value of h on one side of the equation. So the easiest way to do this
is to add 19 and 1/2 to both sides of this equation. I often like putting that as an
improper fraction, but 1/2 is pretty easy to deal with,
so let's add 19 and 1/2 to both sides of this inequality. Did I just say equation? It's an inequality, not an
equation, it's an inequality sign, not an equal sign. So plus 19 and 1/2. On the left-hand side, these
guys obviously cancel out, that was the whole point, and we
are left with the absolute value of h on the left-hand
side is less than. And then if we have 19 and 1/2,
essentially minus 12, 19 minus 12 is 7, so it's going
to be 7 and 1/2. So now we have that the absolute
value of h is less than 7 and 1/2. So what does this tell us? This means that the distance,
another way to interpret this-- remember, absolute value
is the same thing as distance from 0-- so another
way to interpret this statement is that the distance
from h to 0 has to be less than 7 and 1/2. So what values of h are going
to have less than a distance from 7 and 1/2? Well, it could be less than 7
and 1/2 and greater than 0, or equal to 0. So let me put it this way. So h could be less
than 7 and 1/2. But if it gets too far negative,
if it goes to negative 3, we're cool, negative
4, negative 5, negative 6, negative 7, we're
still cool, but then at negative 8, all of a sudden the
absolute value isn't going to be less than this. So it also has to be greater
than negative 7 and 1/2. If you give me any number in
this interval, its absolute value is going to be less than 7
and 1/2 because all of these numbers are less than 7
and 1/2 away from 0. Let me draw it on the number
line, which they want us to do anyway. So if this is the number line
right there, that is 0, and we draw some points, let's say
that this is 7, that is 8, that is negative 7, that
is negative 8. What numbers are less than
7 and 1/2 away from 0? Well, you have everything all
the way up to-- 7 and 1/2 is exactly 7 and 1/2 away, so you
can't count that, so 7 and 1/2, you'll put a circle
around it. Same thing true for negative 7
and 1/2, the absolute value, it's exactly 7 and 1/2 away. We have to be less than 7 and
1/2 away, so neither of those points are going to be included,
positive 7 and 1/2 or negative 7 and 1/2. Now, everything in between is
less than 7 and 1/2 away from 0, so everything else counts. Everything outside of it is
clearly more than 7 and 1/2 away from 0. And we're done.