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## 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.

## 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.