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ICSE • Class 8 • Mathematics • Ch 14
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Linear Inequations

In ICSE Class 8 Mathematics, &quot;Linear Inequations&quot; provides an authoritative, algebraically rigorous master study guide investigating the solution, algebraic properties, and number line representation of linear inequalities in one variable. This comprehensive chapter explores What is an Inequation? (Mathematical statement connecting algebraic expressions with inequality relation symbols: $, \ge, \ne$; Linear inequation in one variable: $ax + b < c$), The Replacement Set / Domain vs The Solution Set (Replacement set $\mathbb{N}$ [Natural numbers: $\{1, 2, 3, \dots\}$], $\mathbb{W}$ [Whole numbers: $\{0, 1, 2, \dots\}$], $\mathbb{Z}$ [Integers: $\{\dots, -1, 0, 1, \dots\}$], $\mathbb{R}$ [Real numbers: continuous continuum]; Solution set is the valid subset of the replacement set), Fundamental Axioms of Inequations (Rule 1: Adding or subtracting the same real quantity preserves the inequality sign; Rule 2: Multiplying or dividing by a positive quantity preserves the inequality sign; CRITICAL Rule 3 / The Golden Law: Multiplying or dividing by a NEGATIVE quantity REVERSES the inequality sign: $a < b \iff -a > -b$ and $\frac{a}{-k} > \frac{b}{-k}$), Systematic Solution by Transposition (Fractional inequations with cross-multiplication, clearing brackets), Double Inequations ($a \le f(x) < b$: solving both parts simultaneously), and Graphing on the Real Number Line (Discrete dots for $\mathbb{N}, \mathbb{W}, \mathbb{Z}$; Hollow open circle $\circ$ for strict exclusion [$$] vs solid dark dot $\bullet$ for inclusion [$\le, \ge$] with shaded thick segment/arrow for $\mathbb{R}$) aligned with the 2026–27 CISCE ICSE curriculum.

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Why Does Multiplying Both Sides of an Inequality by Minus One Completely Flip the Direction of the Universe?

Look at two numbers: $3$ and $5$. It is undeniable that $3 < 5$. Now multiply both sides by $-1$: does $-3$ stay less than $-5$? Look at the real number line: $-3$ sits to the RIGHT of $-5$, which means $-3 > -5$! The inequality sign completely flipped backwards! This is the most dangerous mathematical trap in all of algebra: whenever you multiply or divide an inequation by a negative number, the inequality sign MUST REVERSE! Unlike a linear equation ($2x = 6$) that has only one single unique root ($x = 3$), a linear inequation opens up an entire universe of infinite solutions! But your final answer depends completely on your Replacement Set: if you are hunting in Natural Numbers ($\mathbb{N}$), $x < 3$ gives $\{1, 2\}$; in Whole Numbers ($\mathbb{W}$), it gives $\{0, 1, 2\}$; and in Real Numbers ($\mathbb{R}$), it is an infinite continuous shaded ray! What is the difference between an open circle ($\circ$) and a solid dot ($\bullet$)? Let's master linear inequations.

Why This Chapter Matters

Inequalities are fundamental to computer programming (conditional if-else branching, array loop bounds), business operations research (profit maximization under budget constraints), speed limits, credit ratings, and structural load safety margins. Mastering sign reversal and number line graphing is a mandatory ICSE algebra topic.

Before You Begin (Prerequisites)

  • Operations on integers and the real number line from Class 7.
  • Solving simple linear equations by transposition.
  • Sets notation (roster form and domain subsets).

What You Will Learn (Core Objectives)

  • Differentiate between an inequation and an equation.
  • Distinguish between the Replacement Set (Domain) and the Solution Set.
  • Apply the Golden Rule of Inequations: reversing the inequality sign when multiplying or dividing by negative numbers.
  • Solve multi-step linear inequations with fractions and brackets using transposition.
  • Solve compound double inequations ($a \le ax + b < c$).
  • Graph solution sets accurately on the number line using solid dots for $\mathbb{Z}/\mathbb{W}/\mathbb{N}$ and open/closed circles with shaded rays for $\mathbb{R}$.

Chapter Roadmap & Progression

1 1. Inequations, Replacement Sets &...
2 2. Axiomatic Rules of Inequations &...
3 3. Solving Inequations & Double Ine...
4 4. Graphing Solution Sets on the Re...

Complete Concept Guide (100% Curriculum Coverage)

1. Inequations, Replacement Sets & Solution Sets

Understand
A. What is an Inequation?

An Inequation is a mathematical statement stating that one algebraic expression is greater than, less than, or not equal to another, using symbols: $\mathbf{<, \le, >, \ge, \ne}$.

B. Replacement Set vs Solution Set:
  • Replacement Set (Domain): The given universal set of numbers from which the variable $x$ is allowed to take values (e.g., $\mathbb{N}, \mathbb{W}, \mathbb{Z}, \mathbb{R}$).
  • Solution Set: The specific subset of values from the replacement set that make the inequation mathematically TRUE: $$\mathbf{\text{Solution Set} \subseteq \text{Replacement Set}}$$
  • Example: If $x < 4$:
    • If $x \in \mathbb{N}$: $\text{Solution Set} = \mathbf{\{1, 2, 3\}}$.
    • If $x \in \mathbb{W}$: $\text{Solution Set} = \mathbf{\{0, 1, 2, 3\}}$.
    • If $x \in \mathbb{Z}$: $\text{Solution Set} = \mathbf{\{\dots, -2, -1, 0, 1, 2, 3\}}$.
    • If $x \in \mathbb{R}$: $\text{Solution Set} = \mathbf{\{x : x \in \mathbb{R}, \, x < 4\}}$ (Continuous interval!).

2. Axiomatic Rules of Inequations & Sign Reversal

Inequality Rules

Let $a, b, c \in \mathbb{R}$:

  1. Addition / Subtraction Rule: Adding or subtracting the same number on both sides preserves the inequality sign: $$a < b \implies a + c < b + c \quad \land \quad a - c < b - c$$
  2. Positive Multiplication / Division: Multiplying or dividing both sides by a POSITIVE number ($c > 0$) preserves the inequality sign: $$a < b \implies ac < bc \quad \land \quad \frac{a}{c} < \frac{b}{c}$$
  3. THE GOLDEN RULE (Negative Multiplication/Division):

    Multiplying or dividing both sides by a NEGATIVE NUMBER ($c < 0$) REVERSES THE INEQUALITY SIGN:

    $$\mathbf{a < b \iff -a > -b \quad \land \quad \frac{a}{-c} > \frac{b}{-c}}$$

    Algebraic Proof: $a < b \implies a - b < 0 \implies -b < -a \implies -a > -b$!

3. Solving Inequations & Double Inequations

Transposition
A. Transposition Algorithm:
  1. Multiply all terms by the positive LCM of denominators to clear fractions.
  2. Collect variable terms on the LHS and constants on the RHS.
  3. Divide by the coefficient of $x$. If the coefficient is negative, flip the inequality sign immediately!
  4. Filter the algebraic result through the given Replacement Set.
B. Double Inequations:

To solve $a < f(x) \le b$, isolate $x$ in the middle by performing the same operation on all three parts:

$$-3 \le 2x - 1 < 5 \implies -3 + 1 \le 2x < 5 + 1 \implies -2 \le 2x < 6 \implies \mathbf{-1 \le x < 3}$$

4. Graphing Solution Sets on the Real Number Line

Number Line Graphing
A. Discrete Domains ($\mathbb{N}, \mathbb{W}, \mathbb{Z}$):

Plot a bold, solid dot ($\\bullet$) exclusively over the specific discrete integer numbers in the solution set. Do NOT draw lines connecting the dots!

B. Continuous Real Domain ($\mathbb{R}$):
  • Hollow Open Circle ($\\circ$): Drawn at an endpoint if that number is EXCLUDED ($<$ or $>$).
  • Solid Dark Dot ($\\bullet$): Drawn at an endpoint if that number is INCLUDED ($\le$ or $\ge$).
  • Draw a thick, bold shaded line segment or ray arrow representing all continuous real numbers.

Key Formulas, Identities & Theorems

Golden Inequation Reversal Law
$$a < b \iff -a > -b \quad \land \quad a \le b \iff -a \ge -b$$
Multiplication or division by a negative flips the inequality sign.
Double Inequation Sandwich
$$a \le x < b \implies x \in [a, b)$$
Simultaneous lower and upper bounds.

Inequalities: The Reversal Rule & Number Line Graphing

Linear Inequations: The Golden Reversal Rule & Graphing THE GOLDEN REVERSAL RULE a < b ⇔ -a > -b Multiply/Divide by Negative ⇒ FLIP INEQUALITY SIGN! • Example: -3x ≥ 12 x ≤ 12 / (-3) ⇒ x ≤ -4 (Sign Inverted!) • Replacement Set Filters: Natural ℕ={1,2...} • Whole 𝕎={0,1...} • Integers ℤ • Real ℝ • Solution Set is always a subset of Replacement Set NUMBER LINE GRAPHING CONVENTIONS 1. Discrete Domains (ℤ, 𝕎, ℕ): Bold solid dots (•) only over specific numbers 2. Real Numbers Domain (ℝ): Continuous Ray -1 (≤ Solid) 3 (< Open) • Solid • includes boundary (≤ or ≥) • Hollow ο excludes boundary (< or >) MULTIPLYING BY NEGATIVE REVERSES SIGN • SOLID DOT = INCLUDED • HOLLOW CIRCLE = EXCLUDED

Chapter Summary & 10 Key Takeaways

Takeaway 1
An inequation is an algebraic statement involving <, <=, >, >=, or !=.
Takeaway 2
The replacement set (domain) provides the numbers from which solutions may be selected.
Takeaway 3
The solution set is the subset of the replacement set that makes the inequation true.
Takeaway 4
Adding or subtracting the same number on both sides preserves the inequality sign.
Takeaway 5
Multiplying or dividing by a positive number preserves the inequality sign.
Takeaway 6
CRITICAL RULE: Multiplying or dividing by a negative number REVERSES the inequality sign.
Takeaway 7
For discrete domains (N, W, Z), plot isolated bold solid dots on the number line.
Takeaway 8
For real numbers (R), use a solid dot for <= and >=, and a hollow open circle for < and >.
Takeaway 9
Double inequations are solved by performing identical operations across all three parts.
Takeaway 10
Always check the final solution set against the given domain before writing the answer.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
Solve the inequation: $3 - 2x \ge x - 12$, given that the replacement set is $x \in \mathbb{W}$ (Whole numbers). Graph the solution on a number line.
Reveal Answer & Explanation
Answer:

Step 1: Transpose $x$ to LHS and $3$ to RHS:

$$3 - 2x \ge x - 12$$


$$-2x - x \ge -12 - 3$$


$$-3x \ge -15$$


Step 2: Divide both sides by $-3$. REVERSE the inequality sign from $\ge$ to $\le$:

$$x \le \frac{-15}{-3} \implies \mathbf{x \le 5}$$


Step 3: Filter by Replacement Set $\mathbb{W} = \{0, 1, 2, 3, 4, 5, \dots\}$:

$$\text{Solution Set} = \mathbf{\{0, 1, 2, 3, 4, 5\}}$$


• Number Line Graph: Plot distinct bold solid dots ($\\bullet$) exclusively over the numbers $0, 1, 2, 3, 4,$ and $5$ on the number line.


Transpose to $-3x \ge -15$. Dividing by $-3$ reverses the sign to $x \le 5$. Whole numbers are $\{0, 1, 2, 3, 4, 5\}$.
2
Solve the double inequation: $-1 \le 3 + 4x < 23$, given that $x \in \mathbb{R}$ (Real numbers). Represent the solution set on a number line.
Reveal Answer & Explanation
Answer:

Step 1: Subtract $3$ from all three parts:

$$-1 - 3 \le 4x < 23 - 3$$


$$-4 \le 4x < 20$$


Step 2: Divide all parts by $4$ (positive, sign unchanged):

$$\frac{-4}{4} \le \frac{4x}{4} < \frac{20}{4}$$


$$\mathbf{-1 \le x < 5}$$


Step 3: Number Line Representation:
• Draw a solid closed circle ($\\bullet$) at $-1$ (included).
• Draw a hollow open circle ($\\circ$) at $5$ (excluded).
• Draw a thick, bold shaded line segment connecting $-1$ and $5$.


Subtract 3 from all parts: $-4 \le 4x < 20$. Divide by 4: $-1 \le x < 5$. Solid dot at -1, open circle at 5.
3
Solve: $\frac{2x - 1}{3} \ge \frac{3x - 2}{4} - \frac{2 - x}{5}$, where $x \in \mathbb{Z}$ (Integers). What is the greatest value of $x$?
Reveal Answer & Explanation
Answer:

Step 1: Multiply all terms by the LCM of denominators ($3, 4, 5$), which is $60$:

$$60 \left( \frac{2x - 1}{3} \right) \ge 60 \left( \frac{3x - 2}{4} \right) - 60 \left( \frac{2 - x}{5} \right)$$


$$20(2x - 1) \ge 15(3x - 2) - 12(2 - x)$$


Step 2: Expand brackets:

$$40x - 20 \ge 45x - 30 - 24 + 12x$$


$$40x - 20 \ge 57x - 54$$


Step 3: Transpose $57x$ to LHS and $-20$ to RHS:

$$40x - 57x \ge -54 + 20$$


$$-17x \ge -34$$


Step 4: Divide by $-17$ and reverse the inequality sign:

$$x \le \frac{-34}{-17} \implies \mathbf{x \le 2}$$


• Since $x \in \mathbb{Z}$, the solution set is $\{\dots, -1, 0, 1, 2\}$.
• The greatest integer value of $x$ is $2$.


Multiply by 60: $40x - 20 \ge 57x - 54 \implies -17x \ge -34 \implies x \le 2$. Greatest integer is 2.
4
If the replacement set is $\mathbb{Z}^-$ (Negative integers), find the solution set of: $2(x - 3) < 4x - 2$.
Reveal Answer & Explanation
Answer:

Step 1: Expand brackets:

$$2x - 6 < 4x - 2$$


Step 2: Transpose $4x$ to LHS and $-6$ to RHS:

$$2x - 4x < -2 + 6$$


$$-2x < 4$$


Step 3: Divide by $-2$ and reverse the sign:

$$x > \frac{4}{-2} \implies \mathbf{x > -2}$$


Step 4: Filter by Replacement Set $\mathbb{Z}^- = \{-1, -2, -3, \dots\}$:
Negative integers strictly greater than $-2$ is only $-1$.

$$\text{Solution Set} = \mathbf{\{-1\}}$$

.


$-2x < 4 \implies x > -2$. Negative integers greater than -2 is only $\{-1\}$.
5
Explain algebraically why multiplying an inequation by a negative number reverses the inequality symbol.
Reveal Answer & Explanation
Answer:

• Let $a$ and $b$ be any two real numbers such that $a < b$.
• Subtract $(a + b)$ from both sides of the inequality:

$$a - (a + b) < b - (a + b)$$


$$a - a - b < b - a - b$$


$$-b < -a$$


• Which is mathematically equivalent to:

$$\mathbf{-a > -b}$$


• Notice that we started with $a < b$ and arrived at $-a > -b$.
• Therefore, multiplying both sides by $-1$ reverses the inequality symbol from $<$ to $>$.


Subtract $(a + b)$ from both sides: $a - (a+b) < b - (a+b) \implies -b < -a \implies -a > -b$.
6
On a number line, what is the graphical difference between $x \ge 2$ and $x > 2$ on the set of real numbers $\mathbb{R}$?
Reveal Answer & Explanation
Answer:

• For $x \ge 2$ (Inclusive Inequality): A solid, filled-in dark dot ($\\bullet$) is plotted at $2$ to indicate that the number $2$ itself is included in the solution set, with a thick shaded arrow extending to the right toward positive infinity.
• For $x > 2$ (Strict Inequality): A hollow open circle ($\\circ$) is drawn at $2$ to signify that the endpoint $2$ is strictly excluded from the solution set, with a thick shaded arrow extending to the right.


Solid dot ($ullet$) includes 2 ($\ge$); hollow circle ($\circ$) excludes 2 ($>$).
7
Find the smallest integer that satisfies the inequation: $4x - 3 \ge 2(x + 5)$.
Reveal Answer & Explanation
Answer:

Step 1: Expand brackets:

$$4x - 3 \ge 2x + 10$$


Step 2: Transpose $2x$ to LHS and $-3$ to RHS:

$$4x - 2x \ge 10 + 3$$


$$2x \ge 13$$


$$x \ge \frac{13}{2} \implies \mathbf{x \ge 6.5}$$


Since $x \in \mathbb{Z}$, the smallest integer greater than or equal to $6.5$ is $7$.


$2x \ge 13 \implies x \ge 6.5$. The smallest integer is 7.
8
Given that $x \in \mathbb{W}$, find the solution set of: $-2 \le x < 4$.
Reveal Answer & Explanation
Answer: • The given inequation is $-2 \le x < 4$.
• The replacement set is $\mathbb{W} = \{0, 1, 2, 3, 4, \dots\}$.
• Filter values: Whole numbers that are greater than or equal to $-2$ and strictly less than $4$ are:
$$\text{Solution Set} = \mathbf{\{0, 1, 2, 3\}}$$.
*(Negative numbers $-2$ and $-1$ are rejected because they are not whole numbers)*.
Whole numbers start at 0, so solutions are $\{0, 1, 2, 3\}$.
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