# Absolute Value Inequalities

In the last section, we solved absolute value equations. In this section, we
turn our

attention to inequalities involving absolute value.

**Solving |x| < a**

The solutions of

|x| < a

again depend upon the value and sign of the number a. To solve |x| < a
graphically,

we must determine where the graph of the left-hand side lies below the graph of
the

right-hand side of the inequality |x| < a. There are three cases to consider.

• Case I: a < 0

In this case, the graph of y = a lies strictly below the x-axis. As you can see
in

** Figure 1**(a), the graph of y = |x| never lies below the graph of y = a.
Hence, the

inequality |x| < a has no solutions.

• Case II: a = 0

In this case, the graph of y = 0 coincides with the x-axis. As you can see in

**(b), the graph of y = |x| never lies strictly below the x-axis. Hence, the**

**Figure 1**inequality |x| < 0 has no solutions.

• Case III: a > 0

In this case, the graph of y = a lies strictly above the x-axis. In

**(c), the**

**Figure 1**graph of y = |x| and y = a intersect at x = −a and x = a. In

**(c), we**

**Figure 1**also see that the graph of y = |x| lies strictly below the graph of y = a when x is

in-between −a and a; that is, when −a < x < a.

In

**(c), we’ve dropped dashed vertical lines from the points of intersection**

**Figure 1**of the two graphs to the x-axis. On the x-axis, we’ve shaded the solution of

|x| < a, that is, −a < x < a.

**Figure 1**. The solution of |x| < a has three cases.

This discussion leads to the following key property.

Property 1. The solution of |x| < a
depends upon the value and sign of a.• Case I: a < 0 The inequality |x| < a has no solution. • Case II: a = 0 The inequality |x| < 0 has no solution. • Case III: a > 0 The inequality |x| < a has solution set {x : −a < x < a}. |

Let’s look at some examples.

** Example 2.** Solve the inequality |x| <
−5 for x.

The graph of the left-hand side of |x| < −5 is the “V” of **Figure 1**(a). The graph

of the right-hand side of |x| < −5 is a horizontal line located 5 units below
the x-axis.

This is the situation shown in **Figure 1**(a). The graph of y = |x| is therefore
never

below the graph of y = −5. Thus, the inequality |x| < −5 has no solution.

An alternate approach is to consider the fact that the absolute value of x is
always

nonnegative and can never be less than −5. Thus, the inequality |x| < −5 has no

solution.

** ** **Example 3.** Solve the
inequality |x| < 0 for x.

This is the case shown in **Figure 1**(b). The graph of y = |x| is never strictly
below

the x-axis. Thus, the inequality |x| < 0 has no solution.

** ** **Example 4.** Solve the
inequality |x| < 8 for x.

The graph of the left-hand side of |x| < 8 is the “V” of **Figure 1**(c). The graph
of

the right-hand side of |x| < 8 is a horizontal line located 8 units above the
x-axis. This

is the situation depicted in **Figure 1**(c). The graphs intersect at (−8, 8) and
(8, 8) and

the graph of y = |x| lies strictly below the graph of y = 8 for values of x
in-between

−8 and 8. Thus, the solution of |x| < 8 is −8 < x < 8.

It helps the intuition if you check the results of the last example. Note that
numbers

between −8 and 8, such as −7.75, −3 and 6.8 satisfy the inequality,

| − 7.75| < 8 and | − 3| < 8 and |6.8| < 8,

while values that do not lie between −8 and 8 do not
satisfy the inequality. For example,

none of the numbers −9.3, 8.2, and 11.7 lie between −8 and 8, and each of the
following

is a false statement.

| − 9.3| < 8 and |8.2| < 8 and |11.7| < 8 (all are false)

If you reflect upon these results, they will help cement the notion that the
solution of

|x| < 8 is all values of x satisfying −8 < x < 8.

** ** **Example 5**. Solve the
inequality |5 − 2x| < −3 for x.

If the inequality were |x| < −3, we would not hesitate. This is the situation
depicted

in **Figure 1**(a) and the inequality |x| < −3 has no solutions. The reasoning
applied

to |x| < −3 works equally well for the inequality |5 − 2x| < −3. The left-hand
side of

this inequality must be nonnegative, so its graph must lie on or above the
x-axis. The

right-hand side of |5 − 2x| < −3 is a horizontal line located 3 units below the
x-axis.

Therefore, the graph of y = |5 − 2x| can never lie below the graph of y = −3 and
the

inequality |5 − 2x| < −3 has no solution.

We can verify this result with the graphing calculator. Load the left- and
right-hand

sides of |5−2x| < −3 into Y1 and Y2, respectively, as shown in
**Figure 2**(a). From
the

ZOOM menu, select 6:ZStandard to produce the image shown in **Figure 2**(b).

As predicted, the graph of y = |5−2x| never lies below the graph of y = −3, so
the

inequality |5 − 2x| < −3 has no solution.

**Figure 2**. Using the graphing calculator to solve the
inequality

|5 − 2x| < −3.

** Example 6.
**Solve the inequality |5 − 2x| < 0 for x.

We know that the left-hand side of the inequality |5 − 2x| < 0 has the “V” shape

indicated in **Figure 1**(b). The graph “touches” the x-axis when |5 − 2x| = 0, or
when

However, the graph of y = |5 − 2x| never falls below the
x-axis, so the inequality

|5 − 2x| < 0 has no solution.

Intuitively, it should be clear that the inequality |5−2x|
< 0 has no solution. Indeed,

the left-hand side of this inequality is always nonnegative, and can never be
strictly

less than zero.

** Example 7.**
Solve the inequality |5 − 2x| < 3 for x.

In this example, the graph of the right-hand side of the inequality |5 − 2x| < 3
is a

horizontal line located 3 units above the x-axis. The graph of the left-hand
side of the

inequality has the “V” shape shown in **Figure 3**(b) and (c). You can use the
intersect

utility on the graphing calculator to find the points of intersection of the
graphs of

y = |5 − 2x| and y = 3, as we have done in **Figures 3**(b) and (c). Note that the

calculator indicates two points of intersection, one at x = 1 and a second at x
= 4.

**Figure 3**. Using the graphing calculator to solve the
inequality |5 − 2x| < 3.

The graph of y = |5 − 2x| falls below the graph of y = 3 for all values of x
between 1

and 4. Hence, the solution of the inequality |5 − 2x| < 3 is the set of all x
satisfying

1 < x < 4; i.e. {x : 1 < x < 4}.

Expectations. We need a way of summarizing this
graphing calculator approach on our homework paper. First, draw a reasonable facsimile of your calculator’s viewing window on your homework paper. Use a ruler to draw all lines. Complete the following checklist. • Label each axis, in this case with x and y. • Scale each axis. To do this, press the WINDOW button on your calculator, then report the values of xmin, xmax, ymin, and ymax on the appropriate axis. • Label each graph with its equation. • Drop dashed vertical lines from the points of intersection to the x-axis. Shade and label the solution set of the inequality on the x-axis. |

Following the guidelines in the above checklist, we obtain
the image in **Figure 4**.

**Figure 4**. Reporting a graphical solution of

|5 − 2x| < 3.

**Algebraic Approach.** Let’s now explore an algebraic solution of the inequality

|5 − 2x| < 3. Much as |x| < 3 implies that −3 < x < 3, the inequality

|5 − 2x| < 3

requires that

−3 < 5 − 2x < 3.

We can subtract 5 from all three members of this last inequality, then simplify.

−3 − 5 < 5 − 2x − 5 < 3 − 5

−8 < −2x < −2

Divide all three members of this last inequality by −2, reversing the inequality
symbols

as you go.

4 > x > 1

We prefer that our inequalities read from “small-to-large,” so we write

1 < x < 4.

This form matches the order of the shaded solution on the number line in **Figure
4**,

which we found using the graphing calculator.

The algebraic technique of this last example leads us to the following property.

Property 8. If a > 0, then the inequality
|x| < a is equivalent to the inequality−a < x < a. |

This property provides a simple method for solving
inequalities of the form |x| < a.

Let’s apply this algebraic technique in the next example.

** Example 9. **Solve the inequality |4x + 5| < 7 for x.

The first step is to use **Property 8** to write that

|4x + 5| < 7

is equivalent to the inequality

−7 < 4x + 5 < 7.

From here, we can solve for x by first subtracting 5 from all three members,
then

dividing through by 4.

We can sketch the solution on a number line.

And we can describe the solution in both interval and set-builder notation as follows.

Assuming that a > 0, the inequality |x| ≤ a requires that
we find where the absolute

value of x is either “less than” a or “equal to” a. We know that |x| < a when −a
< x < a

and we know that |x| = a when x = −a or x = a. Thus, the solution of |x| ≤ a is
the

“union” of these two solutions.

This argument leads to the following property.

**Example 11.** Solve the inequality 5 − 3|x − 4| ≥ −4 for x.

At first glance, the inequality

5 − 3|x − 4| ≥ −4

has a form quite dissimilar from what we’ve done thus far. However, let’s
subtract 5

from both sides of the inequality.

−3|x − 4| ≥ −9

Now, let’s divide both sides of this last inequality by
−3, reversing the inequality sign.

|x − 4| ≤ 3

Aha! Familiar ground. Using **Property 10**, this last inequality is equivalent to

−3 ≤ x − 4 ≤ 3,

and when we add 4 to all three members, we have the solution.

1 ≤ x ≤ 7

We can sketch the solution on a number line.

And we can describe the solution with interval and
set-builder notation.

[1, 7] = {x : 1 ≤ x ≤ 7}

**Solving |x| > a**

The solutions of |x| > a again depend upon the value and sign of a. To solve |x|
> a

graphically, we must determine where the graph of y = |x| lies above the graph
of

y = a. Again, we consider three cases.

• Case I: a < 0

In this case, the graph of y = a lies strictly below the x-axis. Therefore, the
graph

of y = |x| in **Figure 5**(a) always lies above the graph of y = a. Hence, all real

numbers are solutions of the inequality |x| > a.

• Case II: a = 0

In this case, the graph of y = 0 coincides with the x-axis. As shown in** Figure
5**(b),

the graph of y = |x| will lie strictly above the graph of y = 0 for all values
of x

with one exception, namely, x cannot equal zero. Hence, every real number except

x = 0 is a solution of |x| > 0. In **Figure 5**(b), we’ve shaded the solution of |x|
> 0,

namely the set of all real numbers except x = 0.

• Case III: a > 0

In this case, the graph of y = a lies strictly above the x-axis. In **Figure 5**(c),
the

graph of y = |x| intersects the graph of y = a at x = −a and x = a. In **Figure
5**(c),

we see that the graph of y = |x| lies strictly above the graph of y = a if x is
less

than −a or greater than a.

In **Figure 5**(c), we’ve dropped dashed vertical lines from the points of
intersection

to the x-axis. On the x-axis, we’ve shaded the solution of |x| > a, namely the

set of all real numbers x such that x < −a or x > a.

** Figure 5.** The solution of |x| > a has three cases.

This discussion leads to the following property.

Property 12. The solution of |x| > a
depends upon the value and sign of a.• Case I: a < 0 All real numbers are solutions of the inequality |x| > a. • Case II: a = 0 All real numbers, with the exception of x = 0, are solutions of |x| > 0. • Case III: a > 0 The inequality |x| > a has solution set {x : x < −a or x > a}. |