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.

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

—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.

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