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.