Inequalities: definitions and rules

An inequality is a statement that one real expression is less than, less than or equal to, greater than, or greater than or equal to another. To solve an inequality is to find every value of the unknown that makes the statement true.

Definitions

For real numbers $a, b$:

• $a < b$ means "$a$ is strictly less than $b$".
• $a \le b$ means "$a$ is less than or equal to $b$".
• $a > b$ means "$a$ is strictly greater than $b$".
• $a \ge b$ means "$a$ is greater than or equal to $b$".

A strict inequality uses $<$ or $>$ (no equality permitted). A non-strict inequality uses $\le$ or $\ge$.

A linear inequality is one in which the variable(s) appear only to power one and not inside non-linear functions. Examples: $3x - 2 \le 5$; $2x + 3y > 12$. Non-examples: $x^2 \le 4$, $|x| > 1$ (these can be reduced to linear ones).

The order axioms

For all real numbers $a, b, c$:

1. Trichotomy. Exactly one of $a < b$, $a = b$, $a > b$ holds.
2. Transitivity. $a < b$ and $b < c$ imply $a < c$.
3. Addition. $a < b$ implies $a + c < b + c$ for any $c$.
4. Multiplication by positive. $a < b$ and $c > 0$ imply $ac < bc$.
5. Multiplication by negative. $a < b$ and $c < 0$ imply $ac > bc$ (sign reverses!).

The same rules hold with $\le, \ge$ , and combine appropriately.

The four manipulation rules

When solving an inequality, you may:

1. Add or subtract the same quantity from both sides , direction preserved.
2. Multiply or divide both sides by a positive quantity , direction preserved.
3. Multiply or divide both sides by a negative quantity , direction reverses.
4. Take reciprocals of both sides if both sides have the same sign , direction reverses.

So $x < 3 \iff x - 5 < -2 \iff 2x < 6$, but $-x > -3$ (sign reversed!) and (for $x > 0$) $\dfrac{1}{x} > \dfrac{1}{3}$.

Writing solutions as intervals

A solution to an inequality in one variable is typically a set of real numbers , usually an interval. Notation:

• $\{x : x > 2\} = (2, \infty)$.
• $\{x : -3 \le x < 1\} = [-3, 1)$.
• $\{x : x \ne 5\} = (-\infty, 5) \cup (5, \infty)$.

Open brackets exclude endpoints; closed brackets include them.

Worked examples

Example 1. Solve $3x - 7 < 5$ for $x \in \mathbb{R}$.

Add $7$: $3x < 12$. Divide by $3$ (positive): $x < 4$. Solution: $(-\infty, 4)$.

Example 2. Solve $-2x + 5 \ge 1$.

Subtract $5$: $-2x \ge -4$. Divide by $-2$ , sign reverses: $x \le 2$. Solution: $(-\infty, 2]$.

Example 3. Solve $\dfrac{2x - 3}{4} \le \dfrac{x + 1}{2}$.

Multiply by $4$: $2x - 3 \le 2(x + 1) = 2x + 2$. Subtract $2x$: $-3 \le 2$ , always true. Solution: all real numbers, $\mathbb{R}$.

Example 4. Solve $-1 < 3x + 2 \le 5$ (compound inequality).

Subtract $2$ from all parts: $-3 < 3x \le 3$. Divide by $3$: $-1 < x \le 1$. Solution: $(-1, 1]$.

Example 5 (harder). Solve $\dfrac{x - 3}{x + 1} \ge 2$ for $x \in \mathbb{R}$.

Cannot just multiply both sides by $x + 1$ , sign of $x + 1$ is unknown!

Bring to one side: $\dfrac{x - 3}{x + 1} - 2 \ge 0 \iff \dfrac{x - 3 - 2(x + 1)}{x + 1} \ge 0 \iff \dfrac{-x - 5}{x + 1} \ge 0 \iff \dfrac{x + 5}{x + 1} \le 0$ (multiplied numerator and denominator by $-1$, flipping the inequality).

Sign analysis: numerator zero at $x = -5$; denominator zero at $x = -1$.

We want $\le 0$: $x \in [-5, -1)$. Note $x = -5$ included (numerator zero is fine); $x = -1$ excluded (denominator zero is undefined). Solution: $[-5, -1)$.

Try it yourself

1. Solve $5x + 2 < 17$.
2. Solve $-3x + 4 \ge 10$.
3. Solve $\dfrac{x}{3} - \dfrac{x}{4} \le \dfrac{1}{2}$.
4. Solve $-2 \le 3 - 5x < 7$.
5. Solve $2(x - 1) > 5x + 4$.
6. Solve $\dfrac{2x + 1}{3} - \dfrac{x - 2}{2} \ge 1$.
7. Solve $\dfrac{1}{x} > 2$ for $x > 0$.
8. Solve $\dfrac{x + 2}{x - 3} < 0$.
9. Solve $\dfrac{1}{x - 1} \le 2$.
10. Find all integer solutions of $-1 \le 2 - x \le 3$.
11. A student must score at least $40$ marks in the third test to maintain an average of $60$ over three tests. If she scored $58$ and $63$ in the first two, find the inequality.
12. Solve $\dfrac{x^2 - 4}{x - 1} \ge 0$ (mixed).

Pitfalls / Tricks

• The sign reversal when multiplying by a negative is the single biggest source of mistakes. Always note the sign of the multiplier.
• Never multiply both sides by an expression whose sign you do not know without splitting into cases or moving everything to one side.
• $\dfrac{1}{x} > a$ is not the same as $1 > ax$ unless you know the sign of $x$.
• Insight. When in doubt, move everything to one side and analyse signs on intervals. This sign-table method works for every rational inequality.