In this video, we will be learning how to solve absolute value equations. After you finish this lesson, view all of our Algebra 1 lessons and practice problems.

***Remember*** Absolute Value is always Positive! (or zero) Therefore, some equations may have No Solution

For Example:

$|x+9|+3=12$

$|x+9|+3-3=12-3\leftarrow$ First subtract 3 to isolate the quantities in the absolute value brackets

$|x+9|=9\leftarrow$ Now remove the absolute value brackets and separate the equation into 2 cases as shown below

$x+9=9$                                    $x+9=-9$

$x+9-9=9-9$                $x+9-9=-9-9$

$x=0$                                          $x=-18$

## Examples of Absolute Value Equations

### Example 1

$|4+5x|+2=16$

First subtract 2 to isolate the quantities in the absolute value brackets
$|4+5x|+2-2=16-2$

Now remove the absolute value brackets and separate the equation into 2 cases

$|4+5x|=14$

$4+5x=14$                                    $4+5x=-14$

$4+5x-4=14-4$                      $4+5x-4=-14-4$

$5x=10$                                          $5x=-18$

$\dfrac{5x}{5}=\dfrac{10}{5}$                                         $\dfrac{5x}{5}=\dfrac{18}{5}$

$x=2$                                             $x=\dfrac{18}{5}$

### Example 2

$|-2x-1|-2=11$

First add 2 to isolate the quantities in the absolute value brackets
$|-2x-1|-2+2=11+2$

Now remove the absolute value brackets and separate the equation into 2 cases

$|-2x-1|=13$

$2x+1=13$                                    $2x+1=-13$

$2x+1-1=13-1$                      $2x+1-1=-13-1$

$2x=12$                                          $2x=-14$

$\dfrac{2x}{2}=\dfrac{12}{2}$                                         $\dfrac{2x}{2}=\dfrac{-14}{2}$

$x=6$                                             $x=7$

## Video-Lesson Transcript

Let’s go over absolute value equations.

The absolute value is the distance of a number from $0$.

The absolute value is the positive of that number.

For example:

$\mid 5 \mid = 5$

Likewise $\mid {-5} \mid = 5$

No matter what number we put in the absolute value, the answer is always positive.

Even if it’s zero. Zero is neither positive nor negative.

$\mid 0 \mid = 0$

If $\mid x \mid = 5$, we can have two values for $x$.

We can have $x = 5$ or $x = -5$

Based on this concept, take a look at this:

$\mid x + 8 \mid = 5$

Here, we can have two values of $x$.

This makes two equations.

Because as we know earlier, $x + 8 = 5$ or $x + 8 = -5$

First, let’s solve the left equation.

$x + 8 = 5$
$x + 8 - 8 = 5 - 8$
$x = -3$

Let’s go to the other equation.

$x + 8 = -5$
$x + 8 - 8 = -5 - 8$
$x = -13$

If we substitute these two values into $x$, we’ll know they are both true.

Let’s do it!

Using $x = -3$,

$\mid -3 + 8 \mid = \mid 5 \mid = 5$

And using $x = -13$

$\mid -13 + 8 \mid = \mid -5 \mid = 5$

Let’s take a look at another example.

$\mid x + 9 \mid + 3 = 12$

Before we get our two equations, we have to get the absolute value by itself.

Now, let’s get rid of $3$ by subtracting it on both sides.

$\mid x + 9 \mid + 3 - 3 = 12 - 3$
$\mid x + 9 \mid = 9$

From here, we have two equations.

$x + 9 = 9$ and $x + 9 = -9$

Let’s solve both of the equations individually.

$x + 9 = 9$
$x + 9 - 9 = 9 - 9$
$x = 0$

And

$x + 9 = -9$
$x + 9 - 9 = -9 - 9$
$x = -18$

Another keypoint is there are some equations that has no solution.

For example:

$\mid x + 6 \mid = -7$

Here, it’s impossible to get the value of $x$ to end up with a negative answer.

Because the absolute value of anything is always positive or zero.

Let’s have one more example.

$3 \mid x + 5 \mid + 4 = 25$

In order to solve this, we have to have $\mid x + 5 \mid$ by itself.

Let’s solve now by subtracting $4$ on both sides.

$3 \mid x + 5 \mid + 4 - 4 = 25 - 4$
$3 \mid x + 5 \mid = 21$

Then divide by $3$

$\dfrac{3}{3} \mid x + 5 \mid = \dfrac{21}{3}$
$\mid x + 5 \mid = 7$

Now, let’s get the two equations.

$x + 5 = 7$ and $x + 5 = -7$

Let’s solve $x$ on each equation

$x + 5 = 7$
$x + 5 - 5 = 7 - 5$
$x = 2$

And

$x + 5 - 5 = -7 - 5$
$x = -12$