## Absolute value inequality

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 the algebraic way, let us apply the geometric definition of absolute value of a number as below:

From | x| < 3 we need 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.

—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

—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.