Inequalities

An equation asks "what value makes these two things equal?" An inequality asks a richer, more realistic question: "what values keep this quantity below a limit, above a threshold, or between two bounds?" Almost every real decision — a budget you must not exceed, a temperature that must stay safe, a dose that must be large enough to work but small enough to be safe — is naturally an inequality, not an equation. Learning to solve them fluently is learning to describe ranges of acceptable answers rather than single points.

The one twist that trips up nearly every student is that inequalities are not just "equations with a different symbol." They carry a hidden rule about direction, and one operation — multiplying or dividing by a negative — flips that direction. Understand why that flip happens and the rest of the topic falls into place.

Learning Objectives

• Solve linear inequalities and state solutions in both inequality and interval notation.
• Apply the sign-flip rule correctly when multiplying or dividing by a negative number, and explain why it is required.
• Solve compound inequalities involving "and" (intersection) and "or" (union).
• Solve absolute-value inequalities by translating them into compound inequalities.
• Solve polynomial and rational inequalities using sign analysis on a number line.
• Graph solution sets accurately, using open and closed circles correctly.

Quick Answer

An inequality compares two expressions with $<$, $>$, $\le$, or $\ge$, and its solution is usually a range of values, not one number. You solve linear inequalities almost exactly like equations, with one critical exception: whenever you multiply or divide both sides by a negative number, you must reverse the inequality symbol. Solutions are written compactly in interval notation, where a square bracket includes an endpoint and a parenthesis excludes it. Compound inequalities combine conditions with "and" (both must hold) or "or" (either may hold). Absolute-value inequalities split into compound inequalities. Polynomial and rational inequalities are solved by finding the values where each factor is zero or undefined, then testing the sign in each resulting interval.

Where It Came From

For most of history, mathematicians worked with equalities. Comparison of magnitudes was expressed in words ("greater than," "less than") going back to Euclid, but there was no compact symbolism. The modern symbols $<$ and $>$ were introduced by the English mathematician Thomas Harriot, published posthumously in 1631 — the wide end of the wedge points at the larger quantity. The "or equal to" forms $\le$ and $\ge$ came later, popularized in the 18th and 19th centuries.

But symbols alone do not explain why inequalities became central. Two needs drove that. The first was the maturing idea of the ordered number line: the recognition that the real numbers are not just a set but a totally ordered set, where any two distinct numbers can be compared. Once numbers are laid out in order, "the set of numbers to the left of 7" is as natural an object as "the number 7 itself." The second, and more forceful, driver was optimization. As calculus, economics, and later linear programming developed (culminating in George Dantzig's simplex method in 1947 for allocating scarce wartime resources), the central questions became "maximize profit subject to these resource limits" or "minimize cost while meeting these demands." Every one of those constraints is an inequality. Inequalities are the mathematical language of limits, budgets, and feasibility — which is why they sit at the heart of applied mathematics.

Linear Inequalities and the Sign-Flip Rule

A linear inequality looks like $2x - 5 < 9$. You may add or subtract any quantity from both sides, and you may multiply or divide both sides by any positive number, all without changing the solution — exactly as with equations.

The one new rule: multiplying or dividing both sides by a negative number reverses the inequality symbol. Here is why, made concrete. Start with a true statement:

$3 < 5$

Multiply both sides by $-1$. If you kept the symbol, you would claim $-3 < -5$, which is false — $-3$ is actually to the right of $-5$ on the number line. To keep the statement true, the symbol must flip:

$-3 > -5 \quad \checkmark$

Negation reflects the number line through zero, reversing left-right order, so it reverses every comparison.

Worked example. Solve $7 - 2x \ge 13$.

$7 - 2x \ge 13$ $-2x \ge 6 \quad \text{(subtract 7)}$ $x \le -3 \quad \text{(divide by } -2 \text{, flip the symbol)}$

The solution is all $x \le -3$. In interval notation: $(-\infty, -3]$. Check with $x = -5$: $7 - 2(-5) = 17 \ge 13$. Correct. Check that $x = 0$ fails: $7 - 0 = 7 \ge 13$ is false. Correct.

Interval Notation and Graphing Solution Sets

Interval notation is a compact way to name a range. A square bracket $[$ or $]$ means the endpoint is included (goes with $\le$ or $\ge$); a parenthesis $($ or $)$ means the endpoint is excluded (goes with $<$ or $>$). Infinity always takes a parenthesis, because you never "reach" infinity.

Inequality	Interval	Graph endpoint
$x < 4$	$(-\infty, 4)$	open circle at 4
$x \ge -1$	$[-1, \infty)$	closed dot at $-1$
$-2 \le x < 5$	$[-2, 5)$	closed at $-2$, open at 5

Graphing on a number line: draw an open circle at an excluded endpoint (strict $<$ or $>$) and a filled dot at an included endpoint ($\le$ or $\ge$), then shade the region that satisfies the inequality. For $x \le -3$ you place a filled dot at $-3$ and shade everything to the left.

Compound Inequalities: "And" versus "Or"

A compound inequality joins two conditions.

• "And" (intersection): both must be true simultaneously. This produces an overlap. The form $-1 \le x < 4$ is shorthand for "$x \ge -1$ and $x < 4$," giving $[-1, 4)$.
• "Or" (union): at least one must be true. This can produce two separate pieces.

Worked "and" example. Solve $-4 \le 3x - 1 < 8$. Operate on all three parts at once. Add 1 throughout: $-3 \le 3x < 9$. Divide by 3 throughout: $-1 \le x < 3$. Solution: $[-1, 3)$.

Worked "or" example. Solve $2x + 1 \le -3$ or $x - 4 > 0$. The first gives $x \le -2$; the second gives $x > 4$. Since either condition suffices, the solution is the union $(-\infty, -2] \cup (4, \infty)$ — two disconnected rays with a gap between them.

Absolute-Value Inequalities

The absolute value $|x|$ is distance from zero, and that distance interpretation dictates everything.

• "Less than" means trapped near zero. $|x| < a$ (with $a > 0$) means $x$ is within distance $a$ of zero, so $-a < x < a$ — a single "and" interval.
• "Greater than" means far from zero. $|x| > a$ means $x$ is more than $a$ away, so $x < -a$ or $x > a$ — a two-piece "or" solution.

Worked example ("less than"). Solve $|2x - 3| \le 7$. Rewrite as the compound inequality $-7 \le 2x - 3 \le 7$. Add 3: $-4 \le 2x \le 10$. Divide by 2: $-2 \le x \le 5$. Solution: $[-2, 5]$.

Worked example ("greater than"). Solve $|x + 4| > 6$. This splits into $x + 4 > 6$ or $x + 4 < -6$, giving $x > 2$ or $x < -10$. Solution: $(-\infty, -10) \cup (2, \infty)$.

One trap to note: if you ever get $|x| < -3$, there is no solution (distance can never be negative), while $|x| > -3$ is satisfied by all real numbers. Always check the sign of the constant first.

Polynomial and Rational Inequalities: Sign Analysis

For inequalities like $x^2 - x - 6 > 0$ you cannot simply "divide across," because you don't know the sign of the variable factors. Instead use sign analysis.

1. Move everything to one side so the other side is 0.
2. Factor, and find the critical points — where the expression equals 0 (and, for rationals, where it is undefined).
3. Mark these points on a number line; they split it into intervals.
4. Test one value in each interval to determine the sign of the whole expression there.
5. Select the intervals matching your inequality.

Worked polynomial example. Solve $x^2 - x - 6 > 0$. Factor: $(x - 3)(x + 2) > 0$. Critical points: $x = -2$ and $x = 3$. Test each interval:

• $x = -3$: $(-6)(-1) = 6 > 0$ ✓
• $x = 0$: $(-3)(2) = -6 > 0$? No.
• $x = 4$: $(1)(6) = 6 > 0$ ✓

Solution: $(-\infty, -2) \cup (3, \infty)$. Endpoints are excluded because the inequality is strict.

Worked rational example. Solve $\dfrac{x - 1}{x + 2} \ge 0$. Critical points: numerator zero at $x = 1$ (included, since $\ge$), denominator zero at $x = -2$ (always excluded — division by zero is undefined). Test intervals:

• $x = -3$: $\frac{-4}{-1} = 4 \ge 0$ ✓
• $x = 0$: $\frac{-1}{2} = -0.5 \ge 0$? No.
• $x = 2$: $\frac{1}{4} \ge 0$ ✓

Solution: $(-\infty, -2) \cup [1, \infty)$. Note the parenthesis at $-2$ and the bracket at $1$.

Real-World Applications

• Budgeting and finance: "Total spending must stay at or below $2000" is $x \le 2000$; break-even analysis asks when revenue exceeds cost, $R(x) > C(x)$.
• Engineering tolerances: a machined part with target 50 mm and tolerance $\pm 0.2$ mm is exactly the absolute-value inequality $|x - 50| \le 0.2$.
• Medicine and safety: a drug concentration must stay in a therapeutic window — above the minimum effective level and below the toxic level — a compound inequality.
• Linear programming: maximizing profit subject to labor, material, and time constraints, each written as a linear inequality, is the foundation of operations research and supply-chain optimization.

Common Mistakes

Mistake 1: Forgetting to flip the symbol when dividing by a negative. A student solves $-2x > 8$ and writes $x > -4$. This is wrong because dividing by $-2$ reverses the order; the correct answer is $x < -4$. Quick self-check: pick a test value from your answer and plug it into the original. Here $x = 0$ gives $-2(0) = 0 > 8$, false — revealing the error.

Mistake 2: Using the wrong bracket type. Writing $[-2, 5]$ for $x < 5$ wrongly includes the endpoint. A strict inequality ($<$, $>$) always takes a parenthesis; only $\le$ and $\ge$ take a bracket. Infinity is never included, so it always gets a parenthesis.

Mistake 3: Including a value that makes a denominator zero. In $\frac{x-1}{x+2} \ge 0$, a student includes $x = -2$ because the region around it tests positive. But $x = -2$ makes the expression undefined, so it must always be excluded regardless of the inequality direction. Numerator zeros can be included (for $\ge$/$\le$); denominator zeros never can.

Comparison and Connections

Inequalities are the "ordered" cousins of equations. The mechanics overlap heavily, but the sign-flip rule and the range-valued solutions are what set them apart.

Feature	Equation	Inequality
Typical solution	one or few points	a range (interval)
Multiply by negative	no special rule	must reverse the symbol
Solution display	list of values	interval notation / shaded line
Absolute value	$\vert x\vert = a$ gives 2 values	$\vert x\vert < a$ or $\vert x\vert > a$ gives ranges

Absolute-value inequalities connect directly to the concept of distance and later to limits and error bounds in calculus (the $\varepsilon$-$\delta$ definition is built entirely from $|f(x) - L| < \varepsilon$). Systems of linear inequalities extend into two dimensions as shaded regions of the plane, the setting for linear programming.

Practice Questions

Recall

State the sign-flip rule and give one numeric example proving it is necessary. Answer: When you multiply or divide both sides by a negative number, reverse the inequality. Example: $2 < 4$ is true, but multiplying by $-1$ without flipping gives the false $-2 < -4$; flipping gives the true $-2 > -4$.

Understanding

Write $-3 \le x < 6$ in interval notation and describe its graph. Answer: $[-3, 6)$ — a filled dot at $-3$, an open circle at 6, shaded between them.

Application

Solve $|3x + 2| < 11$. Answer: $-11 < 3x + 2 < 11 \Rightarrow -13 < 3x < 9 \Rightarrow -\frac{13}{3} < x < 3$, i.e. $\left(-\frac{13}{3}, 3\right)$.

Analysis

Solve $\dfrac{x + 3}{x - 1} \le 0$ and explain each endpoint choice. Answer: Critical points: $x = -3$ (numerator zero, included since $\le$) and $x = 1$ (denominator zero, excluded). Testing intervals shows the expression is $\le 0$ on $[-3, 1)$. The bracket at $-3$ includes the zero of the fraction; the parenthesis at $1$ excludes the undefined point.

FAQ

Why does the inequality flip only for negatives and not positives? Multiplying by a positive number scales the number line without reflecting it, so the left-right order is preserved. Multiplying by a negative reflects the line through zero, reversing which number is larger — so the comparison must reverse too.

How do I know if a compound inequality is "and" or "or"? "Less than" absolute-value inequalities and problems phrased "between" are "and" (one interval). "Greater than" absolute-value inequalities and problems using the word "or" produce unions (possibly two pieces). Reading the logic of the problem, not just the symbols, is key.

Can I just multiply out a rational inequality by the denominator? Not safely — because the denominator's sign is unknown, and you would have to flip the inequality only when it is negative. Sign analysis with a number line avoids this trap entirely and is the reliable method.

Why is infinity always a parenthesis? Infinity is not a number you can reach or include; it describes unbounded direction. So it never earns a bracket.

What does "no solution" versus "all real numbers" look like? $|x| < -2$ has no solution (distance can't be negative), written $\varnothing$. $|x| > -2$ is true for every $x$, written $(-\infty, \infty)$. Whenever the constant on a $|x|$ inequality is negative, check these two cases before doing algebra.

Quick Revision

• Solve like an equation, but flip the symbol when multiplying/dividing by a negative.
• Interval notation: bracket $[\ ]$ includes; parenthesis $(\ )$ excludes; infinity always $(\ )$.
• $|x| < a \Rightarrow -a < x < a$ (one interval); $|x| > a \Rightarrow x < -a$ or $x > a$ (two pieces).
• Polynomial/rational: set to 0, find critical points (denominator zeros always excluded), test each interval.
• Graph: open circle for strict, filled dot for $\le/\ge$, then shade.
• "And" = intersection (overlap); "or" = union (may be two rays).

Related Topics

Prerequisites

• Algebra overview
• Solving linear equations
• The real number line and ordering

Related Topics

• Absolute value
• Systems of equations and inequalities
• Functions and their graphs

Next Topics

• Quadratic functions and their inequalities
• Linear programming and optimization
• Limits and the $\varepsilon$-$\delta$ definition (Calculus)