Chapter 4: Linear Inequations (in One Variable)

A linear inequation is a statement of inequality involving linear algebraic expressions separated by $<$, $>$, $\le$, or $\ge$. Solving a linear inequation means finding all real or discrete values of the variable that satisfy the inequality. This master note covers all 6 transformation rules, replacement and solution sets, real vs discrete number line graphing conventions, compound double-sided inequations, set operations ($P \cap Q$, $P \cup Q$, $P - Q$), solved textbook master examples, and past ICSE board examination problems covering all 8 question types.

Official ICSE Class 10 Question Types Master Checklist

✔ Core Terminology: Inequation, Inequality signs ($<, >, \le, \ge$), Replacement Set, Solution Set, Discrete vs. Real Number Systems.
✔ The 6 Transformation Rules: Addition/subtraction of terms, multiplication/division by positive/negative numbers, sign reversals, and reciprocals.
✔ Number Line Conventions: Hollow Circle $\circ$ ($<$ or $>$) vs. Dark Solid Circle $\bullet$ ($\le$ or $\ge$) for Real numbers $\mathbb{R}$; Discrete Dots for $\mathbb{N}, \mathbb{W}, \mathbb{Z}$.
✔ Type 1 (Discrete Replacement Sets): Roster solution sets over Natural numbers $\mathbb{N}$, Whole numbers $\mathbb{W}$, and Integers $\mathbb{Z}$.
✔ Type 2 (Real Numbers $\mathbb{R}$ & Number Line): Set-builder notation $\{x : a \le x < b, x \in \mathbb{R}\}$ and real number line graphing.
✔ Type 3 (Double-Sided Compound Inequations): Solving $a \le f(x) < b$ by splitting into simultaneous inequations $a \le f(x)$ AND $f(x) < b$.
✔ Type 4 (Fractional Inequations & Sign Reversals): Solving complex rational linear expressions requiring LCM clearing and negative division sign flips.
✔ Type 5 (Smallest / Largest Integer Value): Finding extreme integer solutions satisfying given linear inequalities.
✔ Type 6 (Set Operations $P \cap Q, P \cup Q, P - Q, P \cap Q'$): Finding intersection, union, and complement regions of solution sets on number lines.
✔ Type 7 (Word Problems): Setting up and solving word problems involving consecutive numbers and inequality conditions.
✔ Type 8 (Quadratic Product Inequations): Solving product inequalities $(x-a)(x-b) < 0$ and $(x-a)(x-b) > 0$.

1. Fundamental Definitions & Number Systems

Core Definitions

Linear Inequation in One Variable: An algebraic inequality of the form $ax + b > c$, $ax + b < c$, $ax + b \ge c$, or $ax + b \le c$, where $a, b, c \in \mathbb{R}$ and $a \neq 0$.
Inequality Signs:
- $x > y$: $x$ is strictly greater than $y$.
- $x < y$: $x$ is strictly less than $y$.
- $x \ge y$: $x$ is greater than or equal to $y$.
- $x \le y$: $x$ is less than or equal to $y$.
Replacement Set (Domain): The set from which the values of the variable $x$ are chosen.
Solution Set: The subset of the replacement set whose elements satisfy the given inequation.

Standard Replacement Sets in ICSE Mathematics

Set Symbol	Set Name	Elements Included	Representation on Number Line
$\mathbb{N}$	Natural Numbers	$\{1, 2, 3, 4, 5, \dots\}$	Individual dark dots at discrete points $1, 2, 3 \dots$
$\mathbb{W}$	Whole Numbers	$\{0, 1, 2, 3, 4, \dots\}$	Individual dark dots at discrete points $0, 1, 2 \dots$
$\mathbb{Z}$ or $\mathbb{I}$	Integers	$\{\dots, -3, -2, -1, 0, 1, 2, 3, \dots\}$	Individual dark dots at integer tick marks
$\mathbb{R}$	Real Numbers	All rational & irrational numbers	Continuous thick line between boundary endpoints

2. The 6 Fundamental Rules of Transformation

Solving a linear inequation involves transforming it into an equivalent simpler inequation using the following 6 algebraic rules:

The 6 Rules of Transformation

Rule 1 (Addition of Term): Transferring a positive term to the other side changes its sign to negative without reversing the inequality. $$2x + 3 > 7 \implies 2x > 7 - 3 \implies 2x > 4$$
Rule 2 (Subtraction of Term): Transferring a negative term to the other side changes its sign to positive without reversing the inequality. $$2x - 3 > 7 \implies 2x > 7 + 3 \implies 2x > 10$$
Rule 3 (Mult/Div by POSITIVE Number): Multiplying or dividing both sides by a positive number PRESERVES the inequality sign. $$x < y \implies px < py \quad \text{and} \quad \frac{x}{p} < \frac{y}{p} \quad (\text{for } p > 0)$$
Rule 4 (Mult/Div by NEGATIVE Number — CRITICAL REVERSAL RULE!): Multiplying or dividing both sides by a negative number REVERSES the inequality sign! $$x < y \implies px > py \quad \text{and} \quad \frac{x}{p} > \frac{y}{p} \quad (\text{for } p < 0)$$ Example: $-4x \le 12 \implies x \ge \frac{12}{-4} \implies x \ge -3$.
Rule 5 (Sign Change of Both Sides): Changing the sign of both sides REVERSES the inequality sign. $$-x > 5 \iff x < -5 \qquad \text{and} \qquad -3y \le 15 \iff 3y \ge -15$$
Rule 6 (Reciprocals): If both sides have the same sign (both positive or both negative), taking reciprocals REVERSES the inequality sign. $$x > y > 0 \iff \frac{1}{x} < \frac{1}{y} \qquad \text{and} \qquad x \le y \iff \frac{1}{x} \ge \frac{1}{y}$$

⚠️ EXAM WARNING: WHEN DOES THE INEQUALITY SIGN REVERSE?

The inequality sign ($<, >, \le, \ge$) REVERSES in exactly 3 situations:

When multiplying or dividing both sides by a negative number.
When multiplying both sides by $-1$ (changing signs of both sides).
When taking reciprocals of both sides of the same sign.

It does NOT reverse when adding or subtracting numbers, or when dividing by positive numbers!

3. Number Line Representation Rules & Conventions

Graphing Conventions on the Real Number Line ($\mathbb{R}$)

Strict Inequalities ($<$ or $>$): Mark the boundary point with a Hollow Circle $\circ$ (indicates endpoint is EXCLUDED).
Inclusive Inequalities ($\le$ or $\ge$): Mark the boundary point with a Dark Solid Circle $\bullet$ (indicates endpoint is INCLUDED).
Continuous Line: Draw a thick continuous shaded line between the two boundary circles (or an arrow extending to $\infty$ or $-\infty$).

Examples of Set Notation vs. Number Line Graphing:

$\{x : x < 2, x \in \mathbb{R}\} \implies$ Hollow circle at 2, arrow extending left to $-\infty$

$\{x : x \ge 4, x \in \mathbb{R}\} \implies$ Solid circle at 4, arrow extending right to $+\infty$

$\{x : -2 < x \le 4, x \in \mathbb{R}\} \implies$ Hollow circle at -2, Solid circle at 4, thick line between -2 and 4

$\{x : -1 \le x \le 4, x \in \mathbb{N}\} \implies$ Dark dots ONLY at discrete points 1, 2, 3, 4 (NO continuous line!)

4. Solved Master Examples Covering All 8 Question Types

Type 1 — Discrete Replacement Sets ($\mathbb{N}$ & $\mathbb{W}$)

Problem: Find the solution set of: (i) $3x + 4 < 16$ if replacement set is $\mathbb{N}$, (ii) $8 - x \le 4x - 2$ if replacement set is $\mathbb{N}$.

Solution:

(i) $3x + 4 < 16 \implies 3x < 16 - 4 \implies 3x < 12 \implies x < 4$.
Since replacement set is $\mathbb{N} = \{1, 2, 3, 4, \dots\}$,
Solution Set = $\{1, 2, 3\}$.
(ii) $8 - x \le 4x - 2 \implies -x - 4x \le -2 - 8 \implies -5x \le -10$.
Dividing by $-5$ (reverses inequality sign): $x \ge \frac{-10}{-5} \implies x \ge 2$.
Since replacement set is $\mathbb{N}$,
Solution Set = $\{2, 3, 4, 5, 6, \dots\}$.

Type 2 — Real Numbers ($\mathbb{R}$) & Number Line Graphing

Problem: Solve the inequation $8 + 3x \ge 28 - 2x$, where $x \in \mathbb{R}$. Represent the solution set on the real number line.

Solution:

$8 + 3x \ge 28 - 2x \implies 3x + 2x \ge 28 - 8 \implies 5x \ge 20 \implies x \ge 4$.
Since $x \in \mathbb{R}$, solution set is written in set-builder form:
Solution Set = $\{x : x \ge 4, x \in \mathbb{R}\}$.
Number Line Representation: Solid circle at $x = 4$, thick line/arrow pointing to the right towards $+\infty$.

Type 3 — Double-Sided Compound Inequations (ICSE 2006)

Problem: Given that $x \in \mathbb{R}$, solve the following inequality and graph the solution set on the number line: $-1 \le 3 + 4x < 23$.

Solution:

Split the double-sided inequation into two parts:
$$\text{Part A: } -1 \le 3 + 4x \qquad \text{and} \qquad \text{Part B: } 3 + 4x < 23$$
Solve Part A: $-1 - 3 \le 4x \implies -4 \le 4x \implies -1 \le x \quad (\text{i.e. } x \ge -1)$.
Solve Part B: $4x < 23 - 3 \implies 4x < 20 \implies x < 5$.
Combining both parts: $-1 \le x < 5$.
Solution Set: $\{x : -1 \le x < 5, x \in \mathbb{R}\}$.
Number Line: Solid circle at $-1$, Hollow circle at $5$, thick line shaded between $-1$ and $5$.

Type 4 — Fractional Inequations with Negative Coefficients

Problem: Solve and graph the solution set on the real number line: $-\frac{1}{3} \le \frac{x}{2} - 1\frac{1}{3} < \frac{1}{6}, x \in \mathbb{R}$.

Solution:

Convert mixed fraction: $1\frac{1}{3} = \frac{4}{3}$. Inequation is $-\frac{1}{3} \le \frac{x}{2} - \frac{4}{3} < \frac{1}{6}$.
Add $\frac{4}{3}$ to all 3 parts:
$$-\frac{1}{3} + \frac{4}{3} \le \frac{x}{2} < \frac{1}{6} + \frac{4}{3}$$ $$\frac{3}{3} \le \frac{x}{2} < \frac{1 + 8}{6} \implies 1 \le \frac{x}{2} < \frac{9}{6} = \frac{3}{2}$$
Multiply all parts by 2:
$$1 \times 2 \le x < \frac{3}{2} \times 2 \implies 2 \le x < 3$$
Solution Set: $\{x : 2 \le x < 3, x \in \mathbb{R}\}$.
Number Line: Solid circle at $2$, Hollow circle at $3$, thick line between $2$ and $3$.

Type 5 — Smallest / Largest Integer Value Satisfying Inequation

Problem: Find the smallest value of $x$ for which $5 - 2x < 5\frac{1}{2} - \frac{5}{3}x$, where $x$ is an integer.

Solution:

$5 - 2x < \frac{11}{2} - \frac{5}{3}x \implies -\frac{5}{3}x + 2x > 5 - \frac{11}{2}$.
$\frac{-5x + 6x}{3} > \frac{10 - 11}{2} \implies \frac{x}{3} > -\frac{1}{2} \implies x > -\frac{3}{2} = -1.5$.
Since $x \in \mathbb{Z}$ (integers) and $x > -1.5$, the possible integers are $\{-1, 0, 1, 2, 3, \dots\}$.
Smallest Integer Value of x = $-1$.

Type 6 — Set Operations on Solutions ($P \cap Q, P \cup Q, P - Q$)

Problem: Given $P = \{x : -2 < x \le 6, x \in \mathbb{R}\}$ and $Q = \{x : 2 \le x < 8, x \in \mathbb{R}\}$, find: (i) $P \cap Q$, (ii) $P \cup Q$, (iii) $P - Q$, (iv) $P \cap Q'$.

Solution:

(i) $P \cap Q$ (Intersection / Common Region):
Overlapping region between $(-2, 6]$ and $[2, 8) \implies$ $\{x : 2 \le x \le 6, x \in \mathbb{R}\}$.
(ii) $P \cup Q$ (Union / Combined Region):
Entire region covered from $-2$ to $8 \implies$ $\{x : -2 < x < 8, x \in \mathbb{R}\}$.
(iii) $P - Q$ (In P but NOT in Q):
Region in $(-2, 6]$ excluding $[2, 8) \implies$ $\{x : -2 < x < 2, x \in \mathbb{R}\}$.
(iv) $P \cap Q'$ (Complement Intersection):
By definition, $P \cap Q' = P - Q \implies$ $\{x : -2 < x < 2, x \in \mathbb{R}\}$.

Type 7 — Word Problems Involving Linear Inequations

Problem: Find three smallest consecutive whole numbers such that the difference between one-fourth of the largest and one-fifth of the smallest is at least 3.

Solution:

Let the 3 consecutive whole numbers be $x, x+1, x+2$, where $x \in \mathbb{W}$.
Smallest = $x$, Largest = $x+2$.
As per statement: $\frac{x+2}{4} - \frac{x}{5} \ge 3$.
Take LCM = 20: $\frac{5(x+2) - 4x}{20} \ge 3 \implies \frac{5x + 10 - 4x}{20} \ge 3 \implies x + 10 \ge 60 \implies x \ge 50$.
Smallest whole number $x = 50$.
Three smallest consecutive whole numbers = 50, 51, and 52.

Type 8 — Quadratic Product Inequations ($(x-a)(x-b) < 0$)

Problem: Solve for $x \in \mathbb{R}$: (i) $(x-3)(x+5) < 0$, (ii) $(x-3)(x+5) > 0$.

Solution:

(i) Product of two terms is negative ($< 0$) when one factor is positive and the other is negative.
For $(x-3)(x+5) < 0$, $x$ must lie between the roots $-5$ and $3$.
Solution Set = $\{x : -5 < x < 3, x \in \mathbb{R}\}$.
(ii) Product of two terms is positive ($> 0$) when both factors have the same sign.
For $(x-3)(x+5) > 0$, $x$ must be less than smaller root or greater than larger root.
Solution Set = $\{x : x < -5 \text{ or } x > 3, x \in \mathbb{R}\}$.

5. Comprehensive ICSE Board Practice Problems

Textbook Exercises & Past Board Questions

Q1 (ICSE 2004). Solve the given inequation and graph the solution on the number line: $3 \ge \frac{x-4}{2} + \frac{x}{3} \ge 2$, where $x \in \mathbb{I}$.

Solution:
$\frac{3(x-4) + 2x}{6} = \frac{5x-12}{6}$. Inequation is $3 \ge \frac{5x-12}{6} \ge 2$.
Multiply by 6: $18 \ge 5x - 12 \ge 12 \implies 30 \ge 5x \ge 24 \implies 6 \ge x \ge 4.8$.
Since $x \in \mathbb{I}$ (integers), $4.8 \le x \le 6 \implies$ Solution Set = $\{5, 6\}$.

Q2 (ICSE 2005). Given $A = \{x : 11x - 5 > 7x + 3, x \in \mathbb{R}\}$ and $B = \{x : 18x - 9 \ge 15 + 12x, x \in \mathbb{R}\}$. Find the range of set $A \cap B$ and represent it on a number line.

Solution:
Set A: $11x - 7x > 3 + 5 \implies 4x > 8 \implies x > 2 \implies A = (2, \infty)$.
Set B: $18x - 12x \ge 15 + 9 \implies 6x \ge 24 \implies x \ge 4 \implies B = [4, \infty)$.
Intersection $A \cap B$: Region common to $x > 2$ and $x \ge 4 \implies$ $\{x : x \ge 4, x \in \mathbb{R}\}$.

Q3 (ICSE 2008). Solve the given inequation and graph the solution on the number line: $2y - 3 < y + 1 \le 4y + 7$, where $y \in \mathbb{R}$.

Solution:
Part A: $2y - 3 < y + 1 \implies y < 4$.
Part B: $y + 1 \le 4y + 7 \implies -6 \le 3y \implies y \ge -2$.
Combining: $-2 \le y < 4 \implies$ Solution Set = $\{y : -2 \le y < 4, y \in \mathbb{R}\}$.
Number Line: Solid circle at $-2$, Hollow circle at $4$, shaded line between $-2$ and $4$.

Q4 (ICSE 2010). Solve the following inequation and represent the solution set on the number line: $-3 < -\frac{1}{2} - \frac{2x}{3} \le \frac{5}{6}, x \in \mathbb{R}$.

Solution:
Add $\frac{1}{2}$ to all parts: $-3 + \frac{1}{2} < -\frac{2x}{3} \le \frac{5}{6} + \frac{1}{2} \implies -\frac{5}{2} < -\frac{2x}{3} \le \frac{8}{6} = \frac{4}{3}$.
Multiply by $-\frac{3}{2}$ (reverses inequality signs):
$$\left(-\frac{5}{2}\right)\left(-\frac{3}{2}\right) > x \ge \left(\frac{4}{3}\right)\left(-\frac{3}{2}\right) \implies \frac{15}{4} > x \ge -2 \implies -2 \le x < 3.75$$ Solution Set = $\{x : -2 \le x < 3.75, x \in \mathbb{R}\}$.

Q5 (ICSE 2012). Solve the following inequation and represent the solution set on the number line: $4x - 19 < \frac{3x}{5} - 2 \le -\frac{2}{5} + x, x \in \mathbb{R}$.

Solution:
Part A: $4x - \frac{3x}{5} < 19 - 2 \implies \frac{17x}{5} < 17 \implies x < 5$.
Part B: $\frac{3x}{5} - x \le -\frac{2}{5} + 2 \implies -\frac{2x}{5} \le \frac{8}{5} \implies -2x \le 8 \implies x \ge -4$.
Combining: $-4 \le x < 5 \implies$ Solution Set = $\{x : -4 \le x < 5, x \in \mathbb{R}\}$.

Q6 (ICSE 2013). Solve the following inequation, write the solution set and represent it on the number line: $-\frac{x}{3} \le \frac{x}{2} - 1\frac{1}{3} < \frac{1}{6}, x \in \mathbb{R}$.

Solution:
Part A: $-\frac{x}{3} \le \frac{x}{2} - \frac{4}{3} \implies \frac{4}{3} \le \frac{x}{2} + \frac{x}{3} = \frac{5x}{6} \implies 8 \le 5x \implies x \ge 1.6$.
Part B: $\frac{x}{2} - \frac{4}{3} < \frac{1}{6} \implies \frac{x}{2} < \frac{1}{6} + \frac{4}{3} = \frac{9}{6} \implies x < 3$.
Combining: $1.6 \le x < 3 \implies$ Solution Set = $\{x : 1.6 \le x < 3, x \in \mathbb{R}\}$.

Q7 (ICSE 2014). Find the values of $x$, which satisfy the inequation: $-2\frac{5}{6} < \frac{1}{2} - \frac{2x}{3} \le 2, x \in \mathbb{W}$. Graph the solution set on the number line.

Solution:
$-\frac{17}{6} < \frac{1}{2} - \frac{2x}{3} \le 2 \implies -\frac{17}{6} - \frac{1}{2} < -\frac{2x}{3} \le 2 - \frac{1}{2} \implies -\frac{20}{6} < -\frac{2x}{3} \le \frac{3}{2}$.
Multiply by $-\frac{3}{2}$ (reverses signs): $\left(-\frac{20}{6}\right)\left(-\frac{3}{2}\right) > x \ge \left(\frac{3}{2}\right)\left(-\frac{3}{2}\right) \implies 5 > x \ge -2.25$.
Since $x \in \mathbb{W} = \{0, 1, 2, 3, \dots\}$,
Solution Set = $\{0, 1, 2, 3, 4\}$.

Q8 (ICSE 2015). Solve the following inequation and write the solution set: $13x - 5 < 15x + 4 < 7x + 12, x \in \mathbb{R}$. Represent the solution on a real number line.

Solution:
Part A: $13x - 5 < 15x + 4 \implies -9 < 2x \implies x > -4.5$.
Part B: $15x + 4 < 7x + 12 \implies 8x < 8 \implies x < 1$.
Combining: $-4.5 < x < 1 \implies$ Solution Set = $\{x : -4.5 < x < 1, x \in \mathbb{R}\}$.
Number Line: Hollow circle at $-4.5$, Hollow circle at $1$, shaded line between $-4.5$ and $1$.