Summarize this article:
Last updated on September 20, 2025
In mathematics, equations use the 'equal to' symbol to show that both sides are balanced. Sometimes, a βnot equal toβ relationship exists, where one expression is either greater than or less than the other. These mathematical expressions are referred to as inequalities and fall under algebra.
A mathematical statement that compares two expressions and indicates which is larger, smaller, or possibly equal to the other is called an inequality. This term is used to evaluate the numerical value or magnitude of two expressions. In contrast to equations, which require precise equality, inequalities show a range of potential values.
They define constraints or conditions, such as workable solutions in optimization problems or relationships in decision-making scenarios, and are crucial in domains like engineering and economics. Value sets, which are frequently displayed as intervals, are revealed by solving them.
Inequalities play an important role in mathematics, as they compare quantities which are not necessarily equal. Since their main function is to compare quantities, different symbols are used to help compare values. For example, the greater than symbol (>) is used to indicate a value is greater than another, and the not equal to symbol (≠) is used to convey that two quantities are different.
Inequality Name |
Symbol |
Expression |
Description |
Greater than |
> |
x > a |
x is greater than a |
Less Than |
< |
x < a |
x is lesser than a |
Greater than equal to |
≥ |
x ≥ a |
x is greater than or equal to a |
Less than equal to |
≤ |
x ≤ a |
x is lesser than or equal to a |
Not Equal To |
≠ | x ≠ a |
x is not equal to a |
Inequalities can be solved using specific rules. The following is a discussion of some of these rules:
Rule 1: The transitive property states that for three numbers, a, b, and c, the following rules apply:
Rule 2: Switching the LHS and RHS of the expressions causes the inequality to reverse. We refer to it as a symmetric property.
Rule 3: Adding or subtracting the same constant, k, from both sides of an inequality doesn’t affect the inequality.
The same is true for other inequalities as well.
This is an organized method for solving inequalities and accurately identifying the solution set.
Step 1: First, simplify the inequality to isolate the variable.
Step 2: Solve the equation to find one or more values.
Step 3: Write each value on the number line.
Step 4: In addition, use open circles to symbolize all excluded values on the number line.
Step 5: Determine the intervals.
Step 6: Choose a random number from each interval and enter it into the inequality to see if it is satisfied.
Step 7: The intervals that are satisfied are the solutions.
However, we typically use algebraic operations like addition, subtraction, multiplication, and division to solve simple inequalities (linear).
The following considerations must be made when writing the interval notation solution to an inequality.
Inequality |
Interval |
x < 2 |
(-∞, 2) |
x > 2 |
(2, ∞) |
x ≤ 2 |
(-∞, 2] |
x ≥ 2 |
[2, ∞) |
2 < x ≤ 6 |
(2, 6] |
Plotting the ‘equals’ line and shading the relevant region are the first steps in graphing inequalities involving two variables. Three steps are involved:
Place y on the equation's left side and all other values on the right.
Plot the y= line, displaying a solid line for y ≤ or y ≥ and a dashed line for y< or y>.
For a less-than inequality (y < or y ≤), shade below the line. For a greater-than inequality (y > or y ≥), shade above the line.
How to Solve Polynomial Inequalities?
Finding the values of a variable that satisfy an inequality when the expression is a polynomial or a rational function is known as solving polynomial inequalities. Linear inequalities (including one-step and two-step inequalities), compound inequalities, quadratic inequalities, absolute value inequalities, and rational inequalities will all be covered in this response, along with concise instructions, examples, and answers provided in interval and set notation.
Linear Inequalities
Polynomials of degree 1 (e.g., ππ₯ + π < π) are involved in linear inequalities. Like linear equations, they are solved by isolating the variable while paying attention to inequality signs.
One-Step Inequalities
These can be solved using just one operation to isolate the variable.
Let’s consider the inequality π₯ + 3 > 7. To separate x from the equation, subtract 3 from both sides:
x > 4.
In set notation, the solution is {x ∈ R β£ x > 4}. In interval notation, it’s written as (4, ∞) which means all real numbers greater than 4.
On a number line, this is expressed with the help of an open circle at 4 and an arrow pointing right, as 4 is not included. This method ensures the inequality keeps pointing in the correct direction and isolates the variable easily.
Two-Step Linear Inequalities
Let’s look at this with the help of an example. Consider the equation 2x − 5 ≤ 3.
Adding 5 to both sides → 2x ≤ 8.
Dividing by 2 → x ≤ 4.
In set notation, the solution is {π₯ ∈ π
β£ π₯ ≤ 4}
In interval notation, the solution is (−∞, 4), meaning all numbers less than or equal to 4.
On a number line, this is shown with the help of a closed circle at 4 and shading towards the left as 4 is included.
2. Compound Inequalities
Compound inequalities combine two inequalities with 'and' (intersection, where both conditions must hold) or 'or' (union, where at least one condition holds). Solve each inequality independently, then combine the outcomes according to the connector.
Take the "and" compound inequality −2 < π₯ ≤ 3 as an example. Since x > −2 and x ≤ 3, the intersection of these sets—that is, all numbers between -2 and 3, including 3—is the answer. This is represented in interval notation as (−2, 3) and in set notation as {x ∈ R β£ − 2 < x ≤ 3}. Draw a closed circle at π₯=3 and an open circle at π₯=−2 on a number line, then shade the area in between. The solution for an "or" example such as π₯<−1 or π₯>2 is (−∞, −1) ∪ (2, ∞), graphed with open circles at -1 and 2, shading left of -1 and right of 2. This method captures combined constraints well.
3. Inequalities that are quadratic
Analysis is necessary to determine whether the quadratic expression is positive, negative, or zero in quadratic inequalities, which involve second-degree polynomials. To determine whether the inequality holds, rewrite the inequality with zero on one side, solve the corresponding equation to determine the roots, divide the number line into intervals using these roots, and test points in each interval.
To solve π₯² − π₯ − 6 > 0, for instance, factor the equation x² − x − 6 = 0 as (x − 3)(x + 2) = 0, yielding roots x = 3 and π₯ = −2. (−∞, −2), (−2, 3), and (3, ∞) are the intervals. Testing points: the expression is positive at x=−3, negative at x=0, and positive at x=4. The answer is therefore x < −2 or x > 3, or in interval notation (−∞, −2) ∪ (3, ∞), and in set notation {x ∈ R β£ x < −2 or x > 3}. On a number line with open circles at -2 and 3 and shading outward, the parabola π¦ = π₯² − π₯ − 6 is graphically located above the x-axis to the right of π₯ = 3 and to the left of π₯ = −2. This technique uses the behavior of the quadratic to identify areas of the solution.
4. Inequalities of Absolute Value
Expressions of the form β£ππ₯ + πβ£ < π or β£ππ₯ + πβ£ > π are examples of absolute value inequalities. These are solved by transforming them into compound inequalities according to the definition of the absolute value. For β£ππ₯ + π β£<π, rewrite as −π < ππ₯ + π < π; for β£ππ₯ + π β£> π, rewrite as ππ₯ + π <−π or ππ₯ + π > π.
Consider β£2 π₯ − 1β£ ≤ 5.
Rewrite as follows: −5 ≤ 2 π₯ − 1 ≤ 5, add 1 to get −4 ≤ 2 π₯ ≤ 6, and divide by 2 to get −2 ≤ π₯ ≤ 3. The answer is {π₯ ∈ π
β£− 2 ≤ π₯ ≤ 3} in set notation and [−2, 3] in interval notation. This is represented graphically as a number line with closed circles shaded between π₯ = −2 and π₯ = 3. For β£π₯ + 1 β£ > 2, rewrite as π₯ + 1 < −2 or π₯ + 1 > 2, resulting in π₯ < −3 or π₯ > 1, or (−∞, −3) ∪ (1, ∞), graphed with open circles at -3 and 1 shading outward. Absolute value problems are reduced to manageable linear inequalities using this method.
5. Rational Inequalities
Ratios of polynomials, like π(π₯) π(π₯) > 0, are the subject of rational inequalities, which call for determining whether the expression is positive, negative, or zero while avoiding points where the denominator is zero. Divide the number line into intervals, test points within each interval, move all terms to one side, and determine the denominator and numerator zeros.
For instance, figure out (x + 1)/(x − 2) ≥ 0. The critical points are π₯ = −1 (numerator zero) and π₯ = 2 (denominator zero). The intervals are (−∞, −1), (−1, 2), and (2, ∞). Test points: the expression is positive at π₯ = −2, negative at π₯ = 0, and positive at π₯ = 3. Since ≥, include π₯ = −1, but leave out π₯ = 2. The answer is {x ∈ Rβ£ x ≤ −1 or x > 2} in set notation and (−∞, −1] ∪ (2, ∞) in interval notation. A number line is represented graphically by a closed circle at π₯ = −1, an open circle at π₯ = 2, and shading to the left of -1 and to the right of 2. The function's graph has a vertical asymptote at π₯ = 2 and is non-negative in the solution regions. Discontinuities in rational expressions are handled with care in this method.
Real-world applications of inequalities include comparing values, establishing boundaries, and making decisions in domains such as daily planning, business, and budgeting.
To avoid common pitfalls, solving inequalities—whether linear, compound, quadratic, absolute value, or rational—requires close attention to detail.
A carβs speed S must be within 5 mph of 60 mph. So, solve |S - 60| < 5.
55 < S < 65.
Step 1: Write the inequality:
|S-60| < 5
-5 < S -60 < 5
Step 2: Add 60 to all parts and solve:
-5 + 60 < S < 5 + 60
55 < S < 65
Therefore, the speed S will be between 55 mph and 65 mph, so the answer will be 55 < S < 65.
Parts produced by a machine have to fall within 0.5 mm of the desired length of 10 mm. Solve |L-10|0.5.
9.5L10.5 mm
Step 1: Analyze the inequality in absolute values.
\(|L - 10| \leq 0.5\)
\(-0.5 \leq L - 10 \leq 0.5\)
Step 2: Find the solution to the compound inequalities by adding 10 to each component.
\(-0.5 + 10 \leq L \leq 0.5 + 10\)
\(9.5 \leq L \leq 10.5\)
Step 3: The required range for the length πΏ is 9.5 mm to 10.5 mm, inclusive. Therefore, the answer will be
\(9.5 \leq L \leq 10.5 \ \text{mm}\).
Food must be kept in a refrigerator between 35Β°F and 40Β°F, inclusive. Determine 35 β€ T β€ 40.
T35, 40
The given inequality
35 ≤ T ≤ 40
Step 1: This inequality is compound, which means T35 and T40.
Step 2: The temperature π needs to fall between [35, 40], which includes endpoints 35 and 40.
Therefore, the answer will be T35, 40 (in Fahrenheit degrees).
A manufacturing unit manufactures tables represented by y and chairs represented by x. The unit has 12 hours within which it must manufacture at least one table and two chairs. Note that tables take three hours, while chairs take two hours to get manufactured. Resolve the system: 2x+3y12, x2, y1
y ≤ (4 − 2x)/3
Write the inequalities according to the statement.
Time : 2x+3y12
Chairs : x2
Tables : y1
Step 2: Find the value of π¦ within the time limit:
\(2x + 3y \leq 12 \implies 3y \leq 12 - 2x \implies y \leq \frac{12 - 2x}{3} = 4 - \frac{2x}{3}\)
Step 3: Apply the constraints:
x2, y1
\(1 \leq y \leq 4 - \frac{2x}{3}\)
Step 4: Establish boundaries
When x=2, then
\(y \leq 4 - \frac{2(2)}{3} = 4 - \frac{4}{3} = \frac{8}{3} \approx 2.67\)
When y=1:
\(2x + 3(1) \leq 12 \implies 2x + 3 \leq 12 \implies 2x \leq 9 \implies x \leq 4.5\)
Step 5:
\(x \geq 2, \quad x \leq 4.5, \quad 1 \leq y \leq 4 - \frac{2x}{3}\)
The manufacturing unit can make one table and two chairs with x2 and y1, by \(y = 4 - \frac{2x}{3}\). So, the answer will be \(y = 4 - \frac{2x}{3}\).
Solve the inequality 3x β 5 < 7.
x < 4
Step 1: To isolate the term with π₯, add 5 to both sides:
3x -5 + 5 < 7 + 5
3x < 12
Step 2: Both sides should be divided by 3.
3x < 12 ⇒ x < 4
This indicates that all real numbers smaller than four are included in the solution set. Therefore, the answer will be x < 4.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.