Lesson Objectives

- Demonstrate an understanding of absolute value
- Demonstrate an understanding of how to solve a linear equation in one variable
- Learn how to solve a compound equation with "or"
- Learn how to solve an absolute value equation
- Learn how to identify an absolute value equation with no solution
- Learn how to identify an absolute value equation with one solution

## How to Solve Absolute Value Equations

Before we learn how to solve an absolute value equation, we must first review the concept of absolute value. The absolute value of a number is defined as the distance between the number and 0 on the number line. As an example, suppose we wanted to think about the absolute value of 5.

|5| = 5

The absolute value of 5 is 5 since 5 is 5 units away from zero on the number line. Additionally, we should remember that opposites (additive inverses) have the same absolute value. This is because opposites have the same distance from zero on the number line, they just lie on opposite sides. One will be positive and the other negative. We can see this by thinking about the absolute value of -5.

|-5| = 5

The absolute value of -5 is 5 since -5 is 5 units away from zero on the number line. Now suppose we saw an equation such as:

|x| = 5

What number or numbers can replace x and give us a true statement?

Since we are taking the absolute value of what is plugged in for x, x can be 5 or x can be -5.

|5| = 5

|-5| = 5

We would have two solutions for this equation:

x = 5 or x = -5

|ax + b| = k

Our solution will be found by solving the compound equation:

ax + b = k

or

ax + b = -k

Let's look at some examples.

Example 1: Solve each equation.

|-6 - 2x| = 14

Step 1) Isolate the absolute value operation on one side of the equation.

In this case, the absolute value operation is already isolated on the left side.

Step 2) Set up and solve a compound equation with "or"

We take what is inside of the absolute value bars and set this equal to our value on the other side:

-6 - 2x = 14

We then take what is inside of the absolute value bars and set this equal to the opposite of our value on the other side:

-6 - 2x = -14

We will end up solving the following compound equation:

-6 - 2x = 14

or

-6 - 2x = -14

The solution for a compound equation with "or" is the union of the two solution sets.

-6 - 2x = 14

-2x = 20

x = -10

-6 - 2x = -14

-2x = -8

x = 4

This tells us that x can be 4 or x can be -10. Both of these solutions will work in the original equation.

x = 4 or x = -10

Let's check these two solutions in the original equation:

|-6 - 2x| = 14

|-6 - 2(4)| = 14

|-6 - 8| = 14

|-14| = 14

14 = 14 ✔

|-6 - 2x| = 14

|-6 - 2(-10)| = 14

|-6 + 20| = 14

|14| = 14

14 = 14 ✔

The reason each solution works is due to the absolute value operation. We can set what's inside of the absolute value bars equal to 14 or -14. The absolute value operation will keep 14 as 14 and change -14 into 14. Let's look at another example.

Example 2: Solve each equation.

-4 + |4x + 6| = 30

Step 1) Isolate the absolute value operation on one side of the equation.

We can add 4 to each side of the equation:

-4 + 4 + |4x + 6| = 30 + 4

|4x + 6| = 34

Step 2) Set up and solve a compound equation with "or"

4x + 6 = 34

or

4x + 6 = -34

4x + 6 = 34

4x = 28

x = 7

4x + 6 = -34

4x = -40

x = -10

x = 7 or x = -10

Check:

-4 + |4x + 6| = 30

-4 + |4(7) + 6| = 30

-4 + |28 + 6| = 30

-4 + 34 = 30

30 = 30 ✔

-4 + |4x + 6| = 30

-4 + |4(-10) + 6| = 30

-4 + |-40 + 6| = 30

-4 + |-34| = 30

-4 + 34 = 30

30 = 30 ✔

Example 3: Solve each equation.

4 + 6|2x - 9| = 46

Step 1) Isolate the absolute value operation on one side of the equation.

We can subtract 4 away from each side of the equation:

6|2x - 9| = 42

We can divide each side of the equation by 6:

|2x - 9| = 7

Step 2) Set up and solve a compound equation with "or"

2x - 9 = 7

or

2x - 9 = -7

2x - 9 = 7

2x = 16

x = 8

2x - 9 = -7

2x = 2

x = 1

x = 8 or x = 1

Check:

4 + 6|2x - 9| = 46

4 + 6|2(8) - 9| = 46

4 + 6|16 - 9| = 46

4 + 6|7| = 46

4 + 6(7) = 46

4 + 42 = 46

46 = 46 ✔

4 + 6|2(1) - 9| = 46

4 + 6|2 - 9| = 46

4 + 6|-7| = 46

4 + 6(7) = 46

4 + 42 = 46

46 = 46 ✔

|ax + b| = |cx + d|

For this scenario, we follow the same logic. Two expressions (ax + b) and (cx + d) can only have the same absolute value if they are equal or they are opposites of each other. We can set up the following compound equation with "or":

ax + b = cx + d

or

ax + b = -(cx + d)

Let's look at an example.

Example 4: Solve each equation.

|3x + 1| = |2x + 4|

We set up a compound equation:

3x + 1 = 2x + 4

or

3x + 1 = -(2x + 4)

Let's start with our top equation:

3x + 1 = 2x + 4

3x - 2x = 4 - 1

x = 3

Our bottom equation:

3x + 1 = -(2x + 4)

3x + 1 = -2x - 4

3x + 2x = -4 - 1

5x = -5

x = -1

x = 3 or x = -1

Check: |3x + 1| = |2x + 4|

|3(3) + 1| = |2(3) + 4|

|9 + 1| = |6 + 4|

|10| = |10|

10 = 10 ✔

|3(-1) + 1| = |2(-1) + 4|

|-3 + 1| = |-2 + 4|

|-2| = |2|

2 = 2 ✔

|x| = -5

Since the absolute value of a number is always non-negative (0 or positive), the result here is no solution. We can't find a number that has an absolute value of -5.

When the absolute value operation is set equal to a negative value, we will not have a solution. Let's take a look at an example.

Example 5: Solve each equation.

-8|5 - 4x| + 9 = 33

Step 1) Isolate the absolute value operation.

-8|5 - 4x| = 24

|5 - 4x| = -3

We can see our absolute value operation is set equal to -3, this means we won't have a solution. When this occurs, we can state our answer as:

"no solution" or ∅

Example 6: Solve each equation.

|x + 6| = 0

For this equation, we simply solve one equation:

x + 6 = 0

x = -6

Check:

|-6 + 6| = 0

|0| = 0

0 = 0 ✔

|5| = 5

The absolute value of 5 is 5 since 5 is 5 units away from zero on the number line. Additionally, we should remember that opposites (additive inverses) have the same absolute value. This is because opposites have the same distance from zero on the number line, they just lie on opposite sides. One will be positive and the other negative. We can see this by thinking about the absolute value of -5.

|-5| = 5

The absolute value of -5 is 5 since -5 is 5 units away from zero on the number line. Now suppose we saw an equation such as:

|x| = 5

What number or numbers can replace x and give us a true statement?

Since we are taking the absolute value of what is plugged in for x, x can be 5 or x can be -5.

|5| = 5

|-5| = 5

We would have two solutions for this equation:

x = 5 or x = -5

### Solving an Absolute Value Equation

- Isolate the absolute value operation on one side of the equation
- Set up and solve a compound equation with "or"

|ax + b| = k

Our solution will be found by solving the compound equation:

ax + b = k

or

ax + b = -k

Let's look at some examples.

Example 1: Solve each equation.

|-6 - 2x| = 14

Step 1) Isolate the absolute value operation on one side of the equation.

In this case, the absolute value operation is already isolated on the left side.

Step 2) Set up and solve a compound equation with "or"

We take what is inside of the absolute value bars and set this equal to our value on the other side:

-6 - 2x = 14

We then take what is inside of the absolute value bars and set this equal to the opposite of our value on the other side:

-6 - 2x = -14

We will end up solving the following compound equation:

-6 - 2x = 14

or

-6 - 2x = -14

The solution for a compound equation with "or" is the union of the two solution sets.

-6 - 2x = 14

-2x = 20

x = -10

-6 - 2x = -14

-2x = -8

x = 4

This tells us that x can be 4 or x can be -10. Both of these solutions will work in the original equation.

x = 4 or x = -10

Let's check these two solutions in the original equation:

|-6 - 2x| = 14

|-6 - 2(4)| = 14

|-6 - 8| = 14

|-14| = 14

14 = 14 ✔

|-6 - 2x| = 14

|-6 - 2(-10)| = 14

|-6 + 20| = 14

|14| = 14

14 = 14 ✔

The reason each solution works is due to the absolute value operation. We can set what's inside of the absolute value bars equal to 14 or -14. The absolute value operation will keep 14 as 14 and change -14 into 14. Let's look at another example.

Example 2: Solve each equation.

-4 + |4x + 6| = 30

Step 1) Isolate the absolute value operation on one side of the equation.

We can add 4 to each side of the equation:

-4 + 4 + |4x + 6| = 30 + 4

|4x + 6| = 34

Step 2) Set up and solve a compound equation with "or"

4x + 6 = 34

or

4x + 6 = -34

4x + 6 = 34

4x = 28

x = 7

4x + 6 = -34

4x = -40

x = -10

x = 7 or x = -10

Check:

-4 + |4x + 6| = 30

-4 + |4(7) + 6| = 30

-4 + |28 + 6| = 30

-4 + 34 = 30

30 = 30 ✔

-4 + |4x + 6| = 30

-4 + |4(-10) + 6| = 30

-4 + |-40 + 6| = 30

-4 + |-34| = 30

-4 + 34 = 30

30 = 30 ✔

Example 3: Solve each equation.

4 + 6|2x - 9| = 46

Step 1) Isolate the absolute value operation on one side of the equation.

We can subtract 4 away from each side of the equation:

6|2x - 9| = 42

We can divide each side of the equation by 6:

|2x - 9| = 7

Step 2) Set up and solve a compound equation with "or"

2x - 9 = 7

or

2x - 9 = -7

2x - 9 = 7

2x = 16

x = 8

2x - 9 = -7

2x = 2

x = 1

x = 8 or x = 1

Check:

4 + 6|2x - 9| = 46

4 + 6|2(8) - 9| = 46

4 + 6|16 - 9| = 46

4 + 6|7| = 46

4 + 6(7) = 46

4 + 42 = 46

46 = 46 ✔

4 + 6|2(1) - 9| = 46

4 + 6|2 - 9| = 46

4 + 6|-7| = 46

4 + 6(7) = 46

4 + 42 = 46

46 = 46 ✔

### Solving Absolute Value Equations with Two Absolute Values

We may also see an absolute value equation where one absolute value operation is set equal to another.|ax + b| = |cx + d|

For this scenario, we follow the same logic. Two expressions (ax + b) and (cx + d) can only have the same absolute value if they are equal or they are opposites of each other. We can set up the following compound equation with "or":

ax + b = cx + d

or

ax + b = -(cx + d)

Let's look at an example.

Example 4: Solve each equation.

|3x + 1| = |2x + 4|

We set up a compound equation:

3x + 1 = 2x + 4

or

3x + 1 = -(2x + 4)

Let's start with our top equation:

3x + 1 = 2x + 4

3x - 2x = 4 - 1

x = 3

Our bottom equation:

3x + 1 = -(2x + 4)

3x + 1 = -2x - 4

3x + 2x = -4 - 1

5x = -5

x = -1

x = 3 or x = -1

Check: |3x + 1| = |2x + 4|

|3(3) + 1| = |2(3) + 4|

|9 + 1| = |6 + 4|

|10| = |10|

10 = 10 ✔

|3(-1) + 1| = |2(-1) + 4|

|-3 + 1| = |-2 + 4|

|-2| = |2|

2 = 2 ✔

### Solving Absolute Value Equations with No Solution

We will also encounter two special case scenarios. The first scenario is when we have an absolute value equation that has no solution. Think about the following equation:|x| = -5

Since the absolute value of a number is always non-negative (0 or positive), the result here is no solution. We can't find a number that has an absolute value of -5.

When the absolute value operation is set equal to a negative value, we will not have a solution. Let's take a look at an example.

Example 5: Solve each equation.

-8|5 - 4x| + 9 = 33

Step 1) Isolate the absolute value operation.

-8|5 - 4x| = 24

|5 - 4x| = -3

We can see our absolute value operation is set equal to -3, this means we won't have a solution. When this occurs, we can state our answer as:

"no solution" or ∅

### Solving Absolute Value Equations with One Solution

The other special case scenario occurs when our absolute value equation has only one solution. This occurs when the absolute value operation is set equal to 0. Zero is the only number that has an absolute value of 0. For this scenario, we will only have to solve one equation. Let's look at an example.Example 6: Solve each equation.

|x + 6| = 0

For this equation, we simply solve one equation:

x + 6 = 0

x = -6

Check:

|-6 + 6| = 0

|0| = 0

0 = 0 ✔

#### Skills Check:

Example #1

Solve each equation. $$-1 + |2 - x|=7$$

Please choose the best answer.

A

$$x=\frac{19}{4}, 7$$

B

$$x=-2, 5$$

C

$$x=3, 16$$

D

$$x=-6, 10$$

E

$$x=-\frac{5}{2}, 8$$

Example #2

Solve each equation. $$3|3 - 4x|=99$$

Please choose the best answer.

A

$$x=-8, 9$$

B

$$x=-\frac{15}{2}, 9$$

C

$$x=-5, 7$$

D

$$x=-5, -1$$

E

$$x=\frac{2}{3}, 11$$

Example #3

Solve each equation. $$-\frac{1}{2}\left|\frac{1}{2}x + \frac{11}{10}\right|=-\frac{67}{40}$$

Please choose the best answer.

A

$$x=-13, 7$$

B

$$x=-21, 2$$

C

$$x=-\frac{89}{10}, \frac{9}{2}$$

D

$$x=-7, 5$$

E

$$x=-\frac{2}{3}, 4$$

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