**Absolute Value Equations and Inequalities**

**Absolute Value Equations**

Sometimes negative numbers appear in our bank accounts, elevation measurements, and temperature; however, most of the time we only deal with positive numbers. For instance, we would probably say, "My checking account is overdrawn by fifty dollars", rather than "There is negative fifty dollars in my checking account." When we measure the distance that we have travelled or the speed that we went, it only makes sense to cite a positive number. The math symbol that tells us to drop the negative sign is the absolute value symbol. For example the absolute value of -5, written |-5| is 5. The absolute value tells us to make the result positive, it does not turn subtraction into addition. For example, do not make the mistake of thinking that

| x - 3|

is the same as

x + 3

For example if x = 5, then

|x - 3| = |5 - 3|

= | 2 |

= 2

while

x + 3 = 5 + 3

= 8

If we know what the absolute value of a number is, then we do not necessarily know what the original number is. For example if the absolute value of a number is 3, then the original number can be either 3 or -3. Symbolically if|x| = 3

then

x = -3 Or x = 3

We will call the above equation an "Or equation". This is typical of equations that involve absolute values when the right hand side is positive. In this discussion we will consider only this type of equation.

Example 1

Solve

|2x - 3| = 5

Solution

We write this as an "or" equation

2x - 3 = -5 Or 2x - 3 = 5

Now just add 3 to both sides of each equation

2x = -2 Or 2x = 8

Finally, divide both sides of each equation by 2

x = -1 Or x = 4

Now it's your turn. If you want to see the answer, place your mouse over the yellow rectangle.

|3x + 5| = 4

Notice that in Example 1 and Exercise 1, the absolute value was isolated on the left hand side. This is a requirement for turning the absolute value equation into an Or equation. We must first isolate the absolute value before tuning the absolute value equation into an Or equation. Here is an example that illustrates this.

Solve

10|x| = 71

Since the absolute value is multiplied by 10, we must first divide both sides by 10. Recall that dividing by 10 is the same as moving the decimal place one digit to the left. This gives us

|x| = 7.1

Now that the absolute value is isolated on the left hand side, we can convert the absolute value equation into an Or equation.

x = -7.1 Or x = 7.1

Solve

4|x| = 20

Now for another example where we first must isolate the absolute value before turning it into an Or equation.

Solve

5|x + 4| + 2 = 12

Since the absolute value is multiplied by 5 and then 2 is added to it, we must first deal with the 5 and the 2 before we turn the absolute value equation into an Or equation. Start by subtracting 2 from both sides to get

5|x +4| = 10

Next divide both sides by 5 to get

|x + 4| = 2

Now that the absolute value is isolated on the left hand side, we can convert the absolute value equation into an Or equation.

x + 4 = -2 Or x + 4 = 2

Finally, subtract 4 from both sides of both equations to get

x = -6 Or x = -2

Solve

7 - 2|x + 5| = 1

Now that we understand how to solve absolute value equations, we will look at the process of solving absolute value inequalities. The process will be very similar to solving absolute value equations, but there will be some extra details that we will need to pay attention to. Here is an example.

**Example 4**

If x is an integer, what is the solution to

|x - 4| < 3

**Solution**

In order for |x - 4| to be less than 3, x - 4 cannot be greater than or equal to 3 and it cannot be less than or equal to -3. That is, x - 4 must be between -3 and 3. We can write this as

-3 < x - 4 < 3

We want x by itself, so we add 4 to all three sides

1 < x < 7

Since x is an integer, we take all the integers between and not including 1 and 7. Notice that we do not include 1 and 7 since the inequality is a "<" instead of a "<". The solution is

{2,3,4,5,6}

Now you try one.

**Exercise 4**

If x is an integer, what is the solution to

|x - 5| < 2

**Solution**

If the inequality is "<" or ">", then we include the endpoints. Here is an example where this occurs.

**Example 5**

If x is an integer, what is the solution to

|x - 2| + 1 < 5

**Solution**

First isolate the absolute value by subtracting 1 from both sides of the inequality.

|x - 2| < 4

In order for |x - 2| to be less or equal to 4, x - 2 must be between -4 and 4. We can write this as

-4 < x - 2 < 4

We want x by itself, so we add 2 to all three sides

-2 < x < 6

Since x is an integer, we take all the integers between and including -2 and 6. Notice that we include -2 and 6 since the inequality is a "<" instead of a "<". The solution is

{-2,-1,0,1,2,3,4,5,6}

Now it's your turn

**Exercise 5**

If x is an integer, what is the solution to

2|x + 3| < 2

**Solution**

The last two examples and exercises asked to find * integer*
solutions to the inequality. The answers were lists of numbers. If
instead we are asked to find

**Example 6**

Find all real number solutions to

|x + 1| < 4

and sketch the solution on a number line.

**Solution**

As with the prior two examples, we first change the absolute value inequality into a compound inequality.

-4 < x+1 < 4

Next, subtract 1 from all three sides to get

-5 < x < 3

We can now place this on a number line, using square brackets at the endpoints.

Now you try one.

**Exercise 6**

Which of the following displays all real number solutions to

|x + 3| < 2

A.

B.

C.

D.

**Solution**

So far, all of the examples that have come up involved the absolute value being less than a number. If the absolute value is greater than a number, then all numbers far to the left and all numbers far to the right will be a solution. Here is an example.

Find all real number solutions to

|x + 2| > 4

and sketch the solution on a number line.

**Solution**

In order for x to be a solution we must have either

x + 2 < -4 Or x + 2 > 4

Subtracting 2 from both sides of each inequality gives

x < -6 Or x > 2

On the number line, the solution will be two rays, one for each inequality. It is shown below

Now you try one.

**Exercise 7**

Which of the following displays all real number solutions to

|x + 1| > 2

A.

B.

C.

D.

**Solution**