Absolute value of a number is the actual distance of the number from the point zero on the number line.

This means the distance of 8 from zero is 8 units to the right of zero and the distance of -8 from zero is also 8 units but to the left of zero. But both 8 and -8 have an absolute value of 8. This is because, the absolute value is distance and distance is a non-negative value.

Absolute vale of a non-negative number say 'a', is represented as |a| and its absolute value is a.

|-a| = a and |a| = a

The absolute value of any real number is defined as its overall measured distance from zero on a number line.

If a value ‘x’ represents a real number then the absolute value of ‘x’ is nothing but IxI.

As we see from the definition, every absolute value equation has two solutions. For instance, let us solve for x in the equation, |x| = 12.

Using the definition of absolute value, we know that x is at a distance of 12 units from 0 on the number line. Hence x can be 12 or -12 as both are at a distance of 12 units from 0.

Absolute Value Definition

The algebraic definition of absolute value, is

If |x| = a if x = 0 and if |x| = -x if x = 0.

Here -x is the opposite of a and not the negative of x.

While solving equations involving absolute values, we must replace the given relationship with two relationships

a) One with the quantity within the absolute value symbols as a positive quantity

b) The other with the quantity within the absolute value symbols as a negative quantity

Once this is done, we must solve the two relationships as any other algebraic equation. For example, the equation |y| = 6 must be replaced as two equations as y = 6 and y = -6. This means one equation, |y| = 6 is equivalent to two equations. Consider another equation involving absolute value |y - 2| = 6. This is equivalent to

y - 2 = 6 OR y - 2 = -6 {Now solve these two equations}

y = 8 OR y = -4

Now let us consider a little more complex situation.

Representing any set of numbers on the number line would give each number some important properties and they are (a) a direction from the position of zero and the distance from zero’s position.

The direction from zero would be represented by the symbol it carries in front of the number.

A number without any kind of symbol or sign is considered as a positive number.

The distance from the zero’s position is better known as the absolute value of the number.

The absolute value of a number is considered as negative. It is the overall distance the number is from zero regardless which direction it is from zero.

When we work with absolute values of sums and differences we apply the expression inside the absolute values and their symbols first and then the absolute value of the simplified version or expression.

Let us try a few examples:

Write each of the expression without absolute value symbols.

(a) I6I = 6 …………………..the number 6 is 6 units away from zero
(b) I-9I = 9 ………………….the number -9 is 9 units away from zero
(c) I$\frac{-3}{5}$I = $\frac{3}{5}$ ………….the fraction $\frac{- 3}{5}$ is $\frac{3}{5}$ units away from zero

Each negative number is considered as the opposite of some positive number and each of these positive numbers is the opposite of some negative number.
The opposite of a negative number is a positive number.
Symbol-wise it would be represented as follows:
If ‘x’ represents a positive number then – (- x) = x

The number $\sqrt{a^2 + b^2}$ is called as the absolute value of the complex number’ a =a + bi’ We denote the number as IaI and it is always non negative and in case ‘b’ = 0 then ‘a’ is real.

If and |a| = Ra and |a| = Ja ; so we have |a| = | a|

If a and ß are complex numbers then we have
|a| = $\sqrt{\alpha\overline{\alpha}}$
|aß| = |a| |ß|
|a + ß| = |a| + |ß|
||a| - |ß|| = |a| - |ß|

We note that aß + aß is real and consequently it is equal to its real part, and then we have

|a + ß|2 = (a + ß) (a + ß) = aa + ßß + R(aß + aß)
= |a|2 + |ß|2 + |aß | + |aß|
= |a|2 + |ß|2 + 2|a| |ß|
= ( |a| + |ß| )2

a and ß $\epsilon$ C
C is the set of numbers of the form a + bi, where ‘a’ ‘b’ are real and i2 = -1. C is a field.

Usually the absolute value changes the sign of a number to positive and so whenever we have a negative number we find the absolute value as positive and the positive number remain positive.

The variables ‘a’ and ‘b’ represent any real number, and then we have

I a I = 0 : The absolute value of a real number is positive or zero.

I a I = I -a I : The absolute value of a real number is equal to the absolute value of its opposite.

I a - b I = I b - a I : The expressions ‘a-b’ and ‘b-a’ are opposites of each other and hence their absolute values are equal.

As far zero is concerned it has no sign and so the value is always equal to its absolute value.

When the definition of absolute value is getting applied to any equation, the quantity within the absolute value symbol is considered to have two values.

This value could be either positive or negative before the absolute value is taken into consideration.
This results into having an absolute value equation actually containing two separate equations.

An equation such as IxI = 5 is quite easy to understand as it is asking for values of ‘x’ that would give us a value of 5 when we put the absolute value rule into place.

The two possible answers ‘5 and -5’ have an absolute value of 5.

Solving Absolute Value Equations


To solve an absolute value equation of the form Iax + bI = c, we need to change the absolute value equation to two equivalent linear equations and then solve it.
Iax + bI = c is equivalent to ‘ax + b = c or ax + b = -c.
The left side is the same in each equation but the value of C is positive in the first equation and then negative in the second because the expression inside the absolute value symbol could be positive or negative or the absolute value is making them both positive.

Given below are some examples based on absolute values.

Example 1:

|a - 5| = |a - 7|

Solution:

a - 5 must be equal to a - 7 or they must be opposites as their absolute values are equal.

Case 1: a - 5 = a - 7

-5 = -7

This is not true and hence there is no solution to the given equation when a - 5 = a - 7.

Case 2: a - 5 = -(a - 7)

2a = 12

a = 6

This is a solution to the given equation. Let us check.

|a - 5| = |a - 7|

|6 - 5| = |6 - 7 |

|1| = |-1|

1 = 1

This shows that a = 6 is a solution to the given equation.

Example 2:

|2x - 1| = 9

Solution:

The given equation says 2x - 1 is at a distance of 9 units from 0 on the number line. Hence, the quantity 2x - 1 is equal to +9 or -9 and so we have to solve this equation for both the possibilities.

2x - 1 = 9 OR 2x - 1 = -9

2x = 10 OR 2x = -8 [Add +1 on both sides of the equation]

x = 5 OR x = -4 [Multiply by ½ on both sides of the equation]

Therefore, the solution set for x is {5, -4}

Now let us check if the solution is right. Plug in x values in the given equation.

Put x = 5 in |2x - 1| = 9 Put x = -4 in |2x - 1| = 9

|2(5) - 1| = 9 |2(-4) - 1| = 9

|9| = 9 |-9| = 9

9 = 9 9 = 9

Hence, the solution is right!

Given below are some examples on absolute value inequalities

Simplifying Inequalities with an absolute value

An expression which contains ‘< ‘ or ‘>' sign is called an inequality.

1) x > 4 is an inequality
2) x < 5 is an inequality
3) x = 2 is an equality
4) x £ 3 is an inequality

Also, if you have an absolute value with inequality, we need to take two possibilities.

e.g. I x+7 I >5

then when we remove the absolute value sign, we break the expression as

x+7>5 or x+7 < - 5

Also, for the first expression keep the sign of inequality & the sign of the expression on the right.

For the second expression, change the sign of inequality as well as the sign of the right side expression.

So if it is

I x+3 I >5

Then the break up must be

x+3 >5 or x+3 < - 5

Solving Absolute Value Inequalities

Illustration 1:

Simplify I x+3 I >2

Here I x+3 I >2

So x+3 > 2 or x+3<-2

Subtracting from both sides we get

x+3 - 3 >2 - 3 or x + 3 - 3 < - 2 - 3

x > - 1 or x < - 5

Illustration 2:

Simplify I2x -1I >7

Now I 2x -1 I >7

2x - 1 >7 or 2x-1 < -7

Adding 1 on both sides

2x - 1 +1 > 7+1 or 2x -1 +1 < -7+1

2x > 8 or 2x < -6

Dividing by on both sides

$\frac{(2x)}{2}$ > $\frac{8}{2}$ or $\frac{(2x)}{2}$ < $\frac{(-6)}{2}$ $\rightarrow$ x > 4 or x < -3