## Absolute Value Inequalities

The forget the minus sign" definition of the absolute value is useless for our purposes. Instead, we will mostly use the geometric definition of the absolute value:

The absolute value of a number measures its distance to the origin on the real number line.

Since 5 is at 5 units distance from the origin 0, the absolute value of 5 is 5, |5|=5

Since -5 is also at a distance of 5 units from the origin, the absolute value of -5 is 5, |-5|=5:

We are ready for our first inequality. Find the set of solutions for

|x|<5.

Translate into English: we are looking for those real numbers x whose distance from the origin is less than 5 units.

Obviously we are talking about the interval (-5,5):

What about the solutions to ?

In English: which numbers, x, are at least 2 units away from the origin? On the left side, real numbers less than or equal to -2 qualify, on the right all real numbers greater than or equal to 2:

We can write this interval notation as

What is the geometric meaning of |x-y|?

|x-y| is the distance between x and y on the real number line.

Consider the example |(-4)-3|. The distance on the real number line between the points -4 and +3 is 7, thus

|(-4)-3|=7.

Let's find the solutions to the inequality:

In English: Which real numbers are not more than 1 unit apart from 2?

We're talking about the numbers in the interval [1,3].

Let's rewrite this as

which we can translate into the quest for those numbers x whose distance to -1 is at least 3.

The set of solutions is

With a little bit of tweaking, our method can also handle inequalities such as

|2x-5|<2.

We first divide both sides by 2. Note that absolute values interact nicely with multiplication and division:

as long as a is positive.

Thus we obtain

after simplification, we get the inequality

asking the question, which numbers are less than 1 unit apart from

So the original inequality has as its set of solutions the interval .

Consider the example

Let's divide by 3:

which is the same as

Which numbers have distance at least from ? The set of solutions is given by

Our method fails for more contrived examples.

Let us consider the inequality

|x-3|<2x-4

It's back to basic algebra with a twist.

The standard definition for the absolute value function is given by:

Thus we can get rid of the sign in our inequality if we know whether the expression inside, x-3, is positive or negative.

We will do exactly that!

Let's first consider only those values of x for which :

Case 1:

In this case we know that |x-3|=x-3, so our inequality becomes

x-3<2x-4

Solving the inequality, we obtain

x>1.

We have found some solutions to our inequality:

x is a solution if and x>1 at the same time! We're talking about numbers .

What if x-3<0?

Case 2: x<3

This time x-3<0, so |x-3|=-(x-3)=3-x, so our inequality reads as

3-x<2x-4.

Applying the standard techniques, this can be simplified to

Our inequality has some more solutions:

Under our case assumption x<3, solutions are those real numbers which satisfy .

We're talking about numbers in the interval

Combining the solutions we found for both cases, we conclude that the set of solutions for the inequality

|x-3|<2x-4

are the numbers in the interval

#### Exercise 1.

Find the solutions of the inequality

#### Exercise 2.

Find the solutions of the inequality

#### Exercise 3.

Find the solutions of the inequality

|x-3|>5.

#### Exercise 4.

Find the solutions of the inequality

#### Exercise 5.

Find the solutions of the inequality

|2x-5|>x+1.

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