algebra 2 solve absolute value inequality and graph the solution

## |x – 3| < 5

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# 4 Answers

*Let's start with definition of absolute value*

*1. l a l = a , if a ≥ 0 (if "a" is positive or 0) , for example*

*l 5 l = 5*

*2. l a l = -a , if a < 0 (if "a" is negative) , for example*

*l -5 l = - (-5) = 5*

*So, absolute value is always positive or zero.*

**~~~~~~~~~~~**So, we need to consider two cases, because expression on the left side of inequality contain variable and can be positive, or negative.

*Let's assume that (x - 3) ≥ 0 , then*

**1.**x - 3 < 5 ---> x < 5 + 3 --->

**x < 8***If (x - 3) < 0 , then we have to put the "–" before (x - 3)*

**2.**– (x - 3) < 5 or

x - 3 > - 5 ---> x > - 5 + 3 --->

**x > - 2****x ∈ (-2, 8)**

<-------|----

**o**

**===|=======***-----|------->*

**o***- ∞ - 3 - 2*

*0*

*8 9 ∞*

*Now let's check our answer:*

*For example:*

if x = -3, then

| - 3 - 3 | < 5

| - 6| < 5

6 < 5 (false)

if x = 0, then

| 0 - 3 | < 5

| - 3 | < 5

3 < 5 (true)

if x = 9, then

| 9 - 3 | < 5

| 6 | < 5

6 < 5 (false)

The expression | x - 3 | < 5 means that the value of the expression x - 3 is between -5 and 5. While it is conventionally taught that this inequality is solved by solving two separate inequalities, I teach my students to solve both inequalities using
a method I call Stop, Drop, and Roll.

Consider the inequality | x + 4 | < 7.

First, stop the solution from escaping by using the opposite of 7. Think of a good cop / bad cop duo where the bad cop covers the back door to make sure the suspect does not escape. I call this step the STOP step, because you are stopping the solution
from escaping out the back door.

Now you have the following:

-7 < | x + 4 | < 7

Now drop the absolute value symbols. You don't need them any more. I call this the DROP step because you are dropping the absolute value symbols.

Now you have the following:

-7 < x + 4 < 7

Now roll to the solution like you would any other inequality, treating all three sides of the inequality the same. This is called the ROLL step because you roll to the solution.

-7 - 4 < x + 4 - 4 < 7 - 4

-11 < x < 3

Apply the same process to | x - 3 | < 5.

You should also note that this same process works for inequalities like | x - 2 | > 6, with one minor adjustment.

-6 > | x - 2 | > 6 (STOP)

-6 > x - 2 > 6 (DROP)

-6 + 2 > x - 2 + 2 > 6 + 2 (ROLL)

-4 > x > 8

The adjustment comes when you notice that -4 > x > 8 makes no sense because -4 is not > 8. When this happens, you split the inequality: -4 > x OR x > 8, which is the correct solution.

For legal purposes, I should state that if your clothes are on fire, you should not start solving absolute value inequalities.

|x-3|<5;

This is equivalent to the following system of inequalities:

1) x<3

3-x<5

2) x≥3

x-3<5

The first one transforms into x<3; x>-2

The second one transforms into x≥3; x<8

So, the solution for the first part can be written in the form of double inequality like this:

-2<x<3

For the second part, it can be written as follows:

3≤x<8.

The solution to the original problem is a union of both sets, that is:

**-2<x<8**

### |x – 3| < 5

### for absolute value inequalities, the first step is to be sure that no values are outside the "curtain", such as |x-3| + 7 < 5. In this situation, the 7 would move to the other side by subtracting 7.

### In your case, 2 cases will need to be solved:

## (x-3) < 5 and -(x-3) < 5

## x-3 < 5 and -x + 3 < 5

## x < 8 -x < 2

## x > -2

## Therefore -2 < x < 8

## Be sure to circle the 2 and 8 in the graph. Then draw a line from 2 to 8 in a number line.

**I hope this helps. If so, please vote.**