Solving Inequalities | Definition, Examples, Rules, Graphing & Writing

What is an inequality?

In mathematics, inequality is defined as the relationship between any two numbers or algebraic expressions that are not equal. Inequalities can be viewed as either a mathematical question that can be manipulated to find the value of some variables or a statement of fact in the form of theorems.

We use different mathematical terms and symbols to compare in order to show inequality. The table below shows the four mathematical symbols we use in comparing and showing inequality of numbers or mathematical expressions.

What are the rules of inequality?

There are certain rules that we need to follow when solving inequalities. In this section, we will try to understand all the rules of inequality.

Law of Trichotomy

The law of trichotomy for real lines states that for any real numbers a and b, only one of

𝑎 < 𝑏, 𝑎 = 𝑏, and 𝑎 > 𝑏 is true.

Say, we have the statements, 7 < 9, 7 = 9, 7 > 9, only one of it is true.

Since we know that 9 is larger than 7, we can say that the only true statement is 7 < 9.

Converse Property

Converse property of inequality states that < and >, ≤ and ≥ are each other’s converse, respectively.

This means, for any real numbers a and b, 𝑎 < 𝑏 and 𝑏 > 𝑎 and 𝑎 ≤ 𝑏 and 𝑏 ≥ 𝑎 are equivalent, or we can simply say that:

𝑎 < 𝑏 ↔ 𝑏 > 𝑎

𝑎 ≤ 𝑏 ↔ 𝑏 ≥ 𝑎

For example, we have 6 < 7.

By converse property of inequality, 6 < 7 is the same as 7 > 6.

Transitive Property

For any real numbers a, b, and c,

If 𝑎 < 𝑏 and 𝑏 < 𝑐, then 𝑎 < 𝑐.
If 𝑎 > 𝑏 and 𝑏 > 𝑐, then 𝑎 > 𝑐.

Suppose 14 > 12 and 12 > 6.

By transitive property of inequality, 14 > 6.

Addition Property

In addition property of inequality, if a common constant term c is added to both sides of inequality then, for any real number a, b, and c:

if 𝑎 < 𝑏, then 𝑎 + 𝑐 < 𝑏 + 𝑐.
if 𝑎 ≤ 𝑏, then 𝑎 + 𝑐 ≤ 𝑏 + 𝑐.

For example,

If we add 15 in the inequality 10 > 5, using the addition property of inequality, it follows that:

10 > 5

10 + 15 > 5 + 15

25 > 20

The rule still holds for > and ≥.

Subtraction Property

Subtraction property of inequality states that if a common constant term c is subtracted to both sides of an inequality, then, for any real number a, b, and c:

if 𝑎 < 𝑏, then 𝑎 − 𝑐 < 𝑏 − 𝑐
if 𝑎 ≤ 𝑏, then 𝑎 − 𝑐 ≤ 𝑏 – 𝑐

Suppose we have the inequality 20 < 14, and we are tasked to subtract 7.

By subtraction property of inequality,

20 < 14

20 – 7 < 14 – 7

13 < 7

The rule still holds for > and ≥.

Multiplication Property

For any real numbers a, b, and c≠0,

if 𝑎 < 𝑏 and 𝑐 > 0, then 𝑎𝑐 < 𝑏𝑐
if 𝑎 < 𝑏 and 𝑐 < 0, then 𝑎𝑐 > 𝑏𝑐
if 𝑎 ≤ 𝑏 and 𝑐 > 0, then 𝑎𝑐 ≤ 𝑏𝑐
If 𝑎 ≤ 𝑏 and 𝑐 < 0, then 𝑎𝑐 ≥ 𝑏𝑐.

For example, we are asked to multiply 𝑐 = 8 to the inequality 2 < 4

By multiplication property of inequality,

2 < 4

2 x 8 < 4 x 8

16 < 32

Another example is when 5 < 7 and 𝑐 = −2. If we multiply -2 to the inequality 5 < 7, it follows that:

5 < 7

5 x -2 < 7 x -2

-10 > -14

Always remember that the inequality symbol needs to be reversed every time we multiply an inequality with a negative number.

Division Property

For any real numbers a, b, and c≠0,

if 𝑎 < 𝑏 and 𝑐 > 0, then $\frac{a}{c}$<$\frac{b}{c}$
if 𝑎 < 𝑏 and 𝑐 < 0, then $\frac{a}{c}$>$\frac{b}{c}$

For example,

If we divide c = 2 in the equality of 20 < 30. Then, by division property of inequality

$\frac{20}{2}$ < $\frac{30}{2}$

10 < 15

Additive Inverse Property

The additive inverse property of inequality states that for any real numbers a and b,

if 𝑎 < 𝑏, then −𝑎 > −𝑏.

if 𝑎 ≤ 𝑏, then −𝑎 ≥ −𝑏.

Multiplicative Inverse Property

The multiplicative inverse property of inequality states that if for any real numbers a and b that are both positive or both negative,

if a<b, then $\frac{1}{a}$>$\frac{1}{b}$.
if a>b, then $\frac{1}{a}$<$\frac{1}{b}$.

How to write inequalities in an interval notation?

The following considerations must be kept in mind while writing the solutions of an inequality in interval notation.

If the intervals of the solution of the inequality use < or >, use open parenthesis ‘(‘ or ‘)’.
If the intervals of the solution of the inequality use ≤ or ≥, use closed brackets ‘[‘ or ‘]’.
Always use an open parenthesis to represent or -.

Let’s take a look at some of the examples below.

INEQUALITY	INTERVAL NOTATION
x < 5	(-∞, 5)
x > 5	(5, ∞)
x ≤ 5	(-∞, 5]
x ≥ 5	[5, ∞)
1 < x ≤ 5	(1, 5]

How to graph inequalities in a number line?

When graphing inequalities in a number line, we need to look at the inequality symbol used as that will give us a hint as to where the number line is leading.

INEQUALITY SYMBOL	HOW TO GRAPH	ILLUSTRATION
>	To graph an inequality having greater than symbol, use an open circle to mark the starting value and point the arrow towards the positive infinity.
<	To graph an inequality having less than symbol, use an open circle to mark the starting value and point the arrow towards the negative infinity.
≥	To graph an inequality with greater than or equal to, use a closed circle to mark the starting value and point the arrow towards the positive infinity.
≤	To graph less than or equal to inequality, use a close circle to mark the starting value and point the arrow towards the negative infinity.

How to solve inequalities?

To find the solutions of any inequality, you may follow the following steps below:

Solve for the value of the variable/s using the rules of inequality.
Represent all the values on a number line.
Represent included and excluded values by using closed and open circles, respectively.
Identify the intervals.
Double-check the interval by picking a random number from the interval/s and substituting it to the inequality.

Example #1

Determine the solutions of the inequality x + 5 < 13.

Solution

Step-by-step process	Explanation
x + 5 < 13	Write the inequality and observe the possible rules that we need to do in order to find the value of x.
x + 5 – 5 < 13 – 5	In order to get the value of x, we need to remove 5 from the left-hand side. Hence, subtracting 5 to both sides of the inequality.
x < 8	Simplify.
Therefore, the solution of the inequality x + 5 < 13 is x < 8.

Example #2

Find the solution of the inequality 6x – 7 > 3x + 2.

Solution

Step-by-step process	Explanation
6x – 7 > 3x + 2	Write the inequality and observe the possible rules that we need to do in order to find the value of x.
6x – 3x – 7 > 3x – 3x + 2	Put all the x on the left-hand side of the inequality by subtracting 3x to both sides of the inequality.
3x – 7 > 2	Simplify.
3x – 7 + 7 > 2 + 7	Remove all the constants on the left-hand side using addition property of inequality.
3x > 9	Simplify.
$\frac{3x}{3}$ > $\frac{9}{3}$	Find the value of x by dividing 3 to both sides of the inequality.
x > 3	Simplify.
Therefore, the solution of the inequality 6x – 7 > 3x + 2 is x > 3.

How to solve linear inequalities?

A linear inequality is like a linear equation, except that the inequality sign replaces the equality sign. Hence, solving linear inequalities is almost the same as solving linear equations.

Solving one-step linear inequalities

Suppose we have an inequality of 6x > 24. For us to find the solution to this inequality using only one-step. Hence, we need to divide both side of the inequality by 6. Thus, we will have x > 6. Therefore, the solution of the inequality is x > 6 or by interval notation the solution is represented by (6, ).

Example

Using interval notation, what is the solution of the inequality 5x ≤ 70?

Solution

Step-by-step process	Explanation
5x ≤ 70	Write the inequality and observe the possible rules that we need to do in order to find the value of x.
$\frac{1}{5}$ ∙ 5x ≤ 70 ∙ $\frac{1}{5}$	Multiply both sides of the inequality by $\frac{1}{5}$.
x ≤ 14	Simplify.
Using interval notation, the solution of the inequality can be represent by (-∞, 14].

Solving two-step linear inequalities

Consider the inequality 8x – 11 < 5. To be able to get the solution of this inequality, we need to work it out using only two steps. The first step that we need to do is add 11 to both sides of the inequality. Hence, we will have 8x < 16. Then, let’s divide 8 to both sides of the inequality – which will result to x < 2. Using interval notation, we can write the solution of 8x – 11 < 5 as (-, 2).

Example

What is the solution of the inequality 4x – 17 ≥ 23?

Solution

Step-by-step process	Explanation
4x – 17 ≥ 23	Write the inequality and observe the possible rules that we need to do in order to find the value of x.
4x – 17 + 17 ≥ 23 + 17	Add 17 to the left and right hand side.
4x ≥ 40	Simplify.
$\frac{1}{4}$ ∙ 4x ≥ 40 ∙ $\frac{1}{4}$	Multiply both sides by $\frac{1}{4}$.
x ≥ 10	Simplify.
Using interval notation, the solution of the inequality can be represent by [10, ∞).

Solving compound inequalities

Compound inequality is a group of two or more inequalities with either “and” or “or”. The solution of a compound inequality, the solution to both parts must be true. Hence, to solve inequalities having this case, all we need to do is work on it independently and then find the final solution according to the these rules:

If the inequality have “and” between them, then the final solution is the intersection of the solutions of the two inequalities.
If the inequality have “or” between them, then the final solution is the union of the solutions of the two inequalities.

Example #1

Solve the solution of the compound inequality 3x + 7 > 28 and 5x ≤ 80.

Solution

Step-by-step process	Explanation
3x + 7 > 28	Work on the first given inequality.
3x + 7 – 7 > 28 – 7	Subtract 7 from both sides of the inequality.
3x > 21	Simplify.
$\frac{1}{3}$ ∙ 3x > 21 ∙ $\frac{1}{3}$	Multiply $\frac{1}{3}$ to both sides of the inequality.
x > 7	Simplify.
5x ≤ 80	Work on the second inequality.
$\frac{1}{5}$ ∙ 5x ≤ 80 ∙ $\frac{1}{5}$	Multiply $\frac{1}{5}$ to both sides of the inequality.
x ≤16	Simplify.
x > 7 : (7, ∞) x ≤ 16 : (-∞, 16]	Represent the solutions of the two inequalities using interval notation.
7 < x ≤ 16 : (7, 16]	Get the intersection of the solutions of the compound inequality.
Therefore, the solution of the given compound inequality is 7 < x ≤ 16 or (7, 16] using interval notation.

Example #2

What is the solution of the compound inequality 4x + 9 ≤ 29 or 9x + 27 > 108?

Solution

Step-by-step process	Explanation
4x + 9 ≤ 29	Work on the first given inequality.
4x + 9 – 9 ≤ 29 – 9	Subtract 9 from both sides of the inequality.
4x ≤ 20	Simplify.
$\frac{1}{4}$ ∙ 4x ≤ 20 ∙ $\frac{1}{4}$	Multiply both sides by $\frac{1}{4}$.
x ≤ 5	Simplify.
9x + 27 > 108	Work on the second inequality.
9x + 27 – 27 > 108 – 27	Subtract 27 from both sides.
9x > 81	Simplify.
$\frac{1}{9}$ ∙ 9x > 81 ∙ $\frac{1}{9}$	Multiply both sides by $\frac{1}{9}$.
x > 9	Simplify.
x ≤ 5 : (-∞, 5]x > 9 : (9, ∞)	Represent the solutions of the two inequalities using interval notation.
x ≤ 5 or x > 9 : (-∞, 5] ∪ (9, ∞)	Get the union of the solutions of the compound inequality.
Therefore, the solution of the given compound inequality is x ≤ 5 or x > 9 or (-∞, 5] ∪ (9, ∞) using interval notation.

How to solve quadratic inequalities?

Quadratic inequality is a mathematical statement involving quadratic expression and inequality symbols. To solve quadratic inequalities, these are some of the things that we should do:

Re-write the inequality as an equation.
Determine the values of x.
Represent the solutions using interval.
Check whether the inequality is true for the interval by taking any number from each interval.
If the solution of the inequality in each interval is true, then that is the solution of the quadratic inequality.

Example

What is the solution of the quadratic inequality x² – x – 12 ≤ 0?

Solution

Step 1: Re-write the inequality as a quadratic equation. Hence, x² – x – 12 = 0.

Step 2: Using the rules of finding the solutions of a quadratic equation, we can make the quadratic equation as (x + 3)(x – 4) = 0.

Step 3: The values of x are x = -3 and x = 4.

Step 4: Make a table that will represent the solutions of x using interval notation and check whether a random number will make it true.

INTERVAL NOTATION	RANDOM NUMBER	SUBSTITUTE X TO THE INEQUALITY x² – x – 12 ≤ 0
(-∞, -3]	x = -4	(-4)² – (-4) – 12 ≤ 0 16 + 4 – 12≤ 0 20 – 12 ≤ 0 8 ≤ 0; false
[-3, 4]	x = 0	0² – 0 – 12 ≤ 0 -12 ≤ 0; true
[4, ∞)	x = 5	5² – 5 – 12 ≤ 0 25 – 5 – 12 ≤ 0 20 – 12 ≤ 0 8 ≤ 0; false

Therefore, the solution of the quadratic inequality x² – x – 12≤ 0 is [-3, 4].

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