Linear Inequalities • Topic 1 of 3

Linear Inequalities in One Variable

An inequality is a statement that two quantities are not necessarily equal — one is less than, greater than, or comparable to the other. A linear inequality in one variable is any statement of the form $ax + b < 0$, $ax + b \le 0$, $ax + b > 0$ or $ax + b \ge 0$, where $a \ne 0$. The strict symbols $<$ and $>$ exclude equality; the weak symbols $\le$ and $\ge$ allow it.

Solving an inequality means finding every value of the variable that makes the statement true. Unlike an equation, which usually pins the variable to one or two numbers, an inequality almost always has a whole range of solutions — an interval on the number line.

Most of the rules you already use for equations carry over unchanged. You may add or subtract the same number from both sides, and you may multiply or divide both sides by the same positive number. The one rule that catches students out is the sign-flip:

$$\text{Multiply or divide both sides by a negative number} \Rightarrow \text{reverse the inequality}$$

For example, $-3x \ge 6 \Rightarrow x \le -2$ — dividing by $-3$ turns $\ge$ into $\le$. The table below summarises which operations preserve the direction of the inequality and which reverse it:

Operation on both sidesEffect on the symbolExample
Add or subtract any numberunchanged$x - 4 < 1 \Rightarrow x < 5$
Multiply / divide by a positive numberunchanged$2x \le 10 \Rightarrow x \le 5$
Multiply / divide by a negative numberreversed$-2x \le 10 \Rightarrow x \ge -5$

The solution is then written either in set-builder form, such as $\{x : x \le -2,\ x \in \mathbb{R}\}$, or as an interval, such as $(-\infty, -2]$. On the number line we use an open circle ($\circ$) at the endpoint for a strict inequality and a closed circle ($\bullet$) for a weak one, then shade the ray of all values that satisfy it.

When the variable is restricted to the integers ($x \in \mathbb{Z}$) or naturals ($x \in \mathbb{N}$) rather than all reals, the same algebra applies but the final answer is a discrete list, for instance $\{-2, -1, 0, 1\}$, not a continuous interval.

Deeper Insight — why the sign-flip is unavoidable: The reversal rule is not an arbitrary convention you must memorise; it is forced by what "less than" actually means on the number line. Consider the plain truth $2 < 5$. Now multiply both sides by $-1$: the numbers $-2$ and $-5$ sit on the negative side, where the order is mirrored — $-5$ lies further left than $-2$, so $-2 > -5$. Multiplying by any negative number reflects every point across zero, and reflection turns "to the left of" into "to the right of", which is exactly what flipping the symbol records. This is also why you should never multiply an inequality by an expression like $x$ whose sign you do not know — you would not know whether to flip. Keep the operations transparent: isolate the variable using additions and positive multipliers wherever possible, and reach for a negative multiplier only when you consciously remember to reverse the symbol.

Number line solution for x less than or equal to minus two Solution of −3x ≥ 6, i.e. x ≤ −2 −4 −2 0 2 closed dot (x = −2 included) Open circle for a strict inequality x greater than one Strict inequality: x > 1 −1 0 1 2 open dot (x = 1 excluded)
1
Worked Example
Solve $3x - 5 < 7$ for $x \in \mathbb{R}$ and show the solution on a number line.
Solution
  1. Add $5$ to both sides: $3x < 12$.
  2. Divide both sides by $3$ (positive, so the symbol stays): $x < 4$.
  3. On the number line, place an open circle at $4$ and shade everything to the left.

Answer: $x < 4$, i.e. $x \in (-\infty, 4)$.

2
Worked Example
Solve $-4x + 2 \ge 14$ for $x \in \mathbb{R}$.
Solution
  1. Subtract $2$ from both sides: $-4x \ge 12$.
  2. Divide both sides by $-4$. Because the divisor is negative, reverse the inequality: $x \le -3$.
  3. The endpoint is included, so use a closed circle at $-3$ and shade to the left.

Answer: $x \le -3$, i.e. $x \in (-\infty, -3]$.

3
Worked Example
Solve $\dfrac{x}{2} - 1 \le \dfrac{x}{3} + 1$ for $x \in \mathbb{R}$.
Solution
  1. Multiply every term by $6$ (positive — symbol unchanged): $3x - 6 \le 2x + 6$.
  2. Subtract $2x$ from both sides: $x - 6 \le 6$.
  3. Add $6$ to both sides: $x \le 12$.

Answer: $x \le 12$, i.e. $x \in (-\infty, 12]$.

4
Worked Example
Solve $5 - 2x < 11$ and list the solutions when $x \in \mathbb{Z}$ with $x > -5$.
Solution
  1. Subtract $5$ from both sides: $-2x < 6$.
  2. Divide by $-2$ and reverse the symbol: $x > -3$.
  3. Combine with the restriction $x > -5$ and $x \in \mathbb{Z}$: the binding condition is $x > -3$.
  4. Integers greater than $-3$: $-2, -1, 0, 1, 2, \dots$

Answer: $x > -3$; integer solutions are $\{-2, -1, 0, 1, 2, \dots\}$.

5
Worked Example
Solve $\dfrac{2x - 1}{3} \ge \dfrac{3x - 2}{4} - \dfrac{2 - x}{5}$ for $x \in \mathbb{R}$.
Solution
  1. The LCM of $3, 4, 5$ is $60$. Multiply every term by $60$: $20(2x-1) \ge 15(3x-2) - 12(2-x)$.
  2. Expand: $40x - 20 \ge 45x - 30 - 24 + 12x$.
  3. Simplify the right side: $40x - 20 \ge 57x - 54$.
  4. Bring terms together: $-20 + 54 \ge 57x - 40x \Rightarrow 34 \ge 17x$.
  5. Divide by $17$ (positive): $2 \ge x$, i.e. $x \le 2$.

Answer: $x \le 2$, i.e. $x \in (-\infty, 2]$.

6
Worked Example
A company makes a profit only when revenue exceeds cost. If revenue is $\text{₹}\,40x$ and cost is $\text{₹}(2500 + 15x)$ for $x$ units, find the smallest whole number of units that yields a profit.
Solution
  1. Profit needs revenue $>$ cost: $40x > 2500 + 15x$.
  2. Subtract $15x$ from both sides: $25x > 2500$.
  3. Divide by $25$ (positive): $x > 100$.
  4. The smallest whole number strictly greater than $100$ is $101$.

Answer: The company must sell at least $101$ units to make a profit.

Key Points

  • A linear inequality in one variable has the form $ax + b$ compared to $0$ using $<, \le, >$ or $\ge$, with $a \ne 0$.
  • Adding, subtracting, or multiplying / dividing by a positive number leaves the symbol unchanged.
  • Multiplying or dividing both sides by a negative number reverses the inequality.
  • Solutions form a range: write them in set-builder or interval form, e.g. $x \le -2$ as $(-\infty, -2]$.
  • On the number line use an open circle for $<,>$ and a closed circle for $\le, \ge$, then shade the solution ray.
Tap an option to check your answer0 / 4
Q1.The solution of $2x>6$ is:
Explanation: Divide both sides by $2$.
Q2.The solution of $-x<4$ is:
Explanation: Multiplying/dividing by $-1$ reverses the inequality.
Q3.The solution of $x+5\le9$ is:
Explanation: Subtract $5$.
Q4.The solution of $3x-1\ge5$ is:
Explanation: $3x\ge6\Rightarrow x\ge2$.