# Absolute Value Inequalities: solving and graphing problems with one variable.

### INEQUALITIES AND ABSOLUTE VALUE

Absolute value inequalities video covers solving and graphing problems with one variable. Starting off with the simplest example, we explore how to remove the absolute value bars, so we can solve and graph the problem. We then solve and graph more complex problems, and show how to check the solutions.

Transcript

## Introduction to Absolute value Inequalities

We have looked at the absolute value of a number equal to 5. Now let’s look at the absolute value of a number that is less than 5.

So, the question is, what are some numbers that are less than 5 spots away from zero, since the absolute value is the distance from zero.

Let’s make a list of numbers that are less than 5 spaces away from zero: $0$, $1$, $-1$, -2, -4, $-\frac{1}{2}, 2\frac{1}{3}$, 2.8, 4.9. But did we write all the possible solutions? No. While 5 and -5 are not solutions, 4.9999 is. So, to graph all the possible solutions we need to put an open circle around 5 and -5 and shade in-between them.

We can describe these numbers as less than 5, but also greater that -5.
So just like when we solved absolute value equations, before we can start we need to write the equation twice. Once with the inequality being positive and once with it being negative. Only with inequalities we have to be careful, because when we write it as a negative we also need to flip the inequality sign. Therefore, to write the absolute value of x is less than 5 we would write $x < 5$ and $x>-5$.

## Let’s look at the inequality the absolute value of x is greater that 5.

This time we are looking for some numbers that are greater than 5 spaces away from zero. Again, we can make a list: 6, 7 8, $9\frac{1}{2}$, 7.5, 6.2 … and a simple way to describe all those numbers would be to say $x > 5$. Are those the only answers, well no, because -6, -7.5, -5.3, $9\frac{1}{2}$ are also numbers that are greater that 5 away from zero and a simple way to describe all those numbers is $x < -5$. To rewrite absolute value of x is greater that 5 without the absolute value bars, we would write $x > 5$ and $x < -5$. ( $|x|>a = -a>x>a = x<-a$ or $x>a$)

So to graph all the possible solutions, we need to put an open circle around 5 and -5 and shade the
area to the left of -5 and to the right of 5. Remember that we need to darken the arrows at the end, to show that all the numbers after the arrow are also part of the solution.

Now let’s do some problems:

## First we will look at |x+5| < 3

To find all the solutions, we first need to write this inequality without the absolute value bars.
We can write that as x+5<3 and x+5>-3, then we can solve both of these inequalities by subtract 5 from both sides.
For x+5<3, we get x<-2 and for x+5>-3, we get x> -8
So, the final solution is x<-2 and x>-8. In other words, x is in-between -8 and -2, and
we can write that as a compound inequality as well: -8<x<-2
Let’s quickly graph all possible solutions.
Now let’s pick a value from the graph to verify our solution. We can try -4 .
-4+5 is 1, so the absolute value is 1. And 1 is less than 3.

## Now, let’s try the problem $|x-3| \geq$ 4.

First, we need to write that inequality without the absolute value bars. We would write it as two inequalities: x-3 is greater than or equal to 4 and x-3 is less than or equal to -4
Now to solve these, we will add 3 to each side of both of the inequalities, getting x is greater than or equal to 7 or x is less than or equal to -1.
Let’s quickly graph all possible solutions. Because the original problem was greater than or equal to, we will need to place a shaded circle on -1 and 7.
Now, let’s pick some test values to verify our solution. We can try -5 and positive 10.
-5 – 3 is -8, and the absolute value of -8 is 8. And 8 is greater than 4.
10 – 3 is 7 and the absolute value of 7 is 7. And 7 is greater than 4. So that seems to work.

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