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

So we already proved that both answers are correct.

Absolute Value Equations

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