Greater than and lesser than inequalities are applied on absolute value numbers for solving for x, the variable normally. The geometric definition finds good use in solving for x in absolute value inequalities in the variable x.

**Example 1: ***| x | < 3*

**Solution: **Before solving it for x in an algebraic way, let us apply the geometric definition of the absolute value of a number as below: From | x| < 3 we need the number x whose distance from 0 is less than 3. Mathematically, it is x – 3 < 0, i.e., numbers less than 3 whose distance from 0 is less than 3, and –3 < x, i.e., x > – 3, i.e. numbers greater than –3 so that they are within a distance of 3 from 0.

**Geometrically, the solutions are: ***x < 3 and x > – 3. *Combinedly, the solution for x is *–3 < x < 3*

Algebraic solution for the absolute value inequality: *| x | < a*

**The algebraic definition of absolute value is**

*| x | = ± x*

Now, write *± x in | x |< a. *We get *± x < a, i.e., x < a or – x < a*

**Now, inequality reverses on inverting signs on the two sides:**

So, if – x < a, then x > – a {since 2 < 3, but -2 > – 3}

Combining x < a and –a < x, we can write: -a < x < a

**Example 2:**

| x | > a

**Solution:**

Algebraically, | x | = ± x

Apply this in | x | > a, ± x > a, i.e., x > a or – x < a

In – x < a, multiply both sides of the inequality with – 1, and change the sense of the inequality to > as below: —1(—x) > — (a), i.e., x < —a

Therefore, for the absolute value inequality | x | > a, the solutions are either x < —a Or x > a

**Example 3:**

Solve for x the absolute value inequality: | 2x — 1| < 5

**Solution:**

We know if |x | < a, then —a < x < a

Since | 2x — 1| < 5, so we get *—5 < 2x — 1 < 5* to solve further, isolate x in the middle.

Add 1 to every term* —5 + 1 < 2x — 1 + 1 < 5 + 1, —4 < 2x < 6*

Now, divide every term by 2 to leave x alone in the middle term *—4/2 < 2x/2 < 6/2 and —2 < x < 3.*

Therefore for the absolute value inequality |* 2x — 1| < 5*. The solution set interval is *—2 < x < 3.*

**Example 4:**

If y < x < —y, then which of the following two is greater: x2 or y2?

**Solution:**

We know that if *—a < x < a, then | x | < a*

Thus, for *y < x < —y*, we can write *| x | < —y*

Now squaring on both sides of the inequality *| x | < —y, we get | x |2 < (—y)2*

Remove bars in left and bracket with the negative sign in right, and write *x2 < y2*

Therefore, *y2 > x2*

**Example 5:**

Is | x | = —x if x < 0? Yes. Since x is negative, therefore —x will turn positive after substituting a negative number in x.

**Example 6:**

Find p if |p— 8|= 3p

**Solution:**

Since the absolute value equation has two solutions, i.e., | x | = + x Or | x | = − x, therefore, set up the following two equations: *|p — 8| = + (p – 8) or | p — 8| = − (p – 8). *From the first, we get *p – 8 = 3p, 2p = – 8, so p = -4*, and from the second, we get *− (p – 8) = 3p, so —p + 8 = 3p, i.e., 4p = 8, so p = 2. *The solution *p = —4*, on being substituted in |p— 8|= 3p renders absolute value in the left equal to a negative number in the right. But | x |≠ — ve. Therefore, —4 for p is to be discarded. The only valid solution for the absolute value equation *|p— 8|= 3p is p = 2.*