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

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