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ICSE • कक्षा 7 • Mathematics • अध्याय 19
अनुमानित समय: 45 Mins
प्रगति: अध्ययनरत

रैखिक असमिकाएँ (Linear Inequations)

In ICSE Class 7 Mathematics, &quot;Linear Inequations&quot; provides an authoritative, algebraic master study guide analyzing the solution and graphical representation of mathematical inequality relations. This comprehensive chapter explores Concept of an Inequation (Statements involving inequality symbols: $, \ge$; Linear inequation in one variable of the form $ax + b < c$), The Replacement Set / Domain ($\mathbb{N}$ [Natural numbers], $\mathbb{W}$ [Whole numbers], $\mathbb{Z}$ [Integers], $\mathbb{R}$ [Real numbers]), The Solution Set (The subset of the replacement set whose elements satisfy the inequation), Fundamental Rules of Inequality (Rule 1: Adding or subtracting the same number on both sides preserves the inequality sign; Rule 2: Multiplying or dividing by a positive number preserves the inequality sign; CRITICAL Rule 3: Multiplying or dividing both sides by a NEGATIVE number REVERSES the inequality sign: $a < b \implies -a > -b$), Solving Inequations by Transposition, and Graphing Solution Sets on a Real Number Line (Representing discrete elements using bold solid dots $\bullet$ for $\mathbb{N}, \mathbb{W}, \mathbb{Z}$; Open circle $\circ$ for strict inequalities on $\mathbb{R}$, closed circle $\bullet$ for inclusive inequalities, and shaded bold rays) aligned with the 2026–27 CISCE curriculum.

अध्याय CBT ऑनलाइन टेस्ट दें सभी अध्याय देखें
प्रस्तावना अध्याय अवधारणाएँ सूत्र एवं सर्वसमिकाएँ अध्याय सारांश स्व-मूल्यांकन अभ्यास टेस्ट दें

Why Does Multiplying Both Sides of an Inequality by Minus One Completely Flip the Direction of the World Like a Magic Mirror?

Consider two simple numbers: $3$ and $5$. Everyone knows that $3 < 5$. But now, multiply both sides by $-1$: does $-3$ remain less than $-5$? Look at the integer number line: $-3$ lies to the RIGHT of $-5$, which means $-3 > -5$! The inequality sign completely flipped around! This is the most dangerous, fatal trap in all of algebra: whenever you multiply or divide an inequation by a negative number, the inequality sign MUST REVERSE! While a linear equation ($2x = 6$) has only one single unique root ($x = 3$), a linear inequation ($2x < 6$) opens up a boundless universe of solutions! But your final answer depends entirely on your "Replacement Set": if you are hunting in the realm of Natural Numbers ($\mathbb{N}$), the solutions are $\{1, 2\}$; if you are in Integers ($\mathbb{Z}$), they are $\{\dots, -2, -1, 0, 1, 2\}$; and if you are in Real Numbers ($\mathbb{R}$), there are infinite numbers plotted with a shaded arrow on a number line! What is the difference between an open circle ($\circ$) and a solid dot ($\bullet$) on a number line? Let's master linear inequations.

यह अध्याय क्यों महत्वपूर्ण है

Inequalities are essential in computer programming (conditional IF statements, loop bounds), business operations research (maximizing profit under budget constraints), speed limits, credit limits, and engineering safety margins. Mastering inequation sign reversals and number-line plotting is a signature ICSE mathematics requirement.

अध्ययन से पूर्व (आवश्यक ज्ञान)

  • Operations on integers and sign rules from Chapter 1.
  • Solving linear equations by transposition from Chapter 10.
  • Plotting numbers on the number line.

इस अध्याय के लक्ष्य

  • Define an inequation and identify inequality symbols ($<, \le, >, \ge$).
  • Differentiate between the Replacement Set (Domain) and the Solution Set.
  • Apply the golden rule of inequality reversal when multiplying or dividing by negative numbers.
  • Solve linear inequations in one variable using systematic transposition.
  • Graph solution sets on the number line for $\mathbb{N}, \mathbb{W}, \mathbb{Z}$, and $\mathbb{R}$ using open/solid dots.
  • Solve multi-step fractional linear inequations with double inequalities ($a < x \le b$).

अध्याय रूपरेखा एवं प्रगति

1 1. Concept of an Inequation & The F...
2 2. Fundamental Axioms of Inequality...
3 3. Solving Inequations by Transposi...
4 4. Graphing Solution Sets on the Nu...

सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन

1. Concept of an Inequation & The Four Inequality Symbols

Understand
A. What is an Inequation?

An Inequation (or Inequality) is an algebraic statement containing one of the four inequality relation symbols:

  • $<$ (Strictly Less Than): $x < 5$ (5 is not included).
  • $\le$ (Less Than or Equal To): $x \le 5$ (5 is included).
  • $>$ (Strictly Greater Than): $x > -2$ (-2 is not included).
  • $\ge$ (Greater Than or Equal To): $x \ge -2$ (-2 is included).
B. Replacement Set vs Solution Set:
  • Replacement Set (Domain): The overarching set of numbers from which values of the variable $x$ may be chosen (e.g., $\mathbb{N}, \mathbb{W}, \mathbb{Z}, \mathbb{R}$).
  • Solution Set: The set of all values chosen from the replacement set that make the inequation a true mathematical statement.

2. Fundamental Axioms of Inequality (The Reversal Rule)

Inequality Rules

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

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

    Multiplying or dividing both sides of an inequation by a NEGATIVE NUMBER ($c < 0$) REVERSES the inequality sign:

    $$\mathbf{a < b \implies ac > bc} \quad \text{and} \quad \mathbf{\frac{a}{c} > \frac{b}{c}}$$

    Numerical Proof: We know $2 < 5$. Multiply by $-1$: $-2 > -5$ (since $-2$ lies to the right of $-5$). The sign flipped!

3. Solving Inequations by Transposition

Transposition

To solve a linear inequation:

  1. Clear fractions by multiplying by the positive LCM of denominators.
  2. Collect variable terms on one side (usually LHS) and constant terms on the other side (RHS) using transposition.
  3. Divide by the coefficient of $x$. If the coefficient is negative, flip the inequality sign immediately!
  4. Write the final Solution Set filtered through the given Replacement Set.

4. Graphing Solution Sets on the Number Line

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

Plot a bold solid dot ($\\bullet$) exclusively over the specific discrete integers that belong to the solution set. Do not draw continuous lines between dots!

B. For Continuous Real Numbers ($\mathbb{R}$):
  • Open Circle ($\\circ$): Drawn at an endpoint if that number is excluded ($<$ or $>$).
  • Closed Solid Circle ($\\bullet$): Drawn at an endpoint if that number is included ($\le$ or $\ge$).
  • A thick shaded line or arrow is drawn over the continuous line segment representing all real numbers.

महत्वपूर्ण सूत्र, सर्वसमिकाएँ एवं प्रमेय

Golden Inequation Reversal Law
$$a < b \iff -a > -b \quad (\text{Multiplying or dividing by } -1)$$
Sign of inequality reverses on negative multiplication/division.
Replacement-Solution Set Inclusion
$$\text{Solution Set} \subseteq \text{Replacement Set}$$
Solutions must be elements of the specified domain.

Linear Inequations: Reversal Rule & Number Line Graphs

Linear Inequations: The Reversal Law & Number Line Graphing THE GOLDEN REVERSAL LAW • Multiplying/Dividing by a NEGATIVE: a < b ⇔ -a > -b Example: 3 < 5 ⇒ -3 > -5 (Sign Flips!) • Adding or Subtracting: Sign UNCHANGED • Multiplying by POSITIVE: Sign UNCHANGED • Replacement Sets: ℕ={1,2,...} • 𝕎={0,1,...} • ℤ={...,-1,0,1,...} • ℝ • Solution Set ⊆ Replacement Set NUMBER LINE GRAPHING 1. Discrete Sets (ℤ, 𝕎, ℕ): Plot individual BOLD DOTS (•) only! 1 2 3 2. Real Numbers (ℝ): Shaded continuous line/arrow 1 (Solid • = ≥) → ∞ • Open circle (ο) for < or > (excluded) CRITICAL LAW: MULTIPLYING OR DIVIDING BY A NEGATIVE REVERSES THE INEQUALITY SIGN!

अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष

मुख्य बिंदु 1
An inequation contains inequality relation symbols: $<, \le, >, \ge$.
मुख्य बिंदु 2
Replacement Set (Domain) is the universe from which the variable may be chosen.
मुख्य बिंदु 3
Solution Set is the collection of elements from the replacement set that satisfy the inequation.
मुख्य बिंदु 4
Adding or subtracting the same number on both sides preserves the inequality sign.
मुख्य बिंदु 5
Multiplying or dividing by a positive number preserves the inequality sign.
मुख्य बिंदु 6
CRITICAL: Multiplying or dividing by a NEGATIVE number REVERSES the inequality sign ($a < b \implies -a > -b$).
मुख्य बिंदु 7
For discrete domains ($\mathbb{N}, \mathbb{W}, \mathbb{Z}$), plot isolated solid dots $\bullet$ on the number line.
मुख्य बिंदु 8
For continuous domains ($\mathbb{R}$), use an open circle $\circ$ for strict inequality ($<, >$) and closed dot $\bullet$ for inclusive ($\le, \ge$).
मुख्य बिंदु 9
A double inequation $a \le x < b$ denotes elements bounded simultaneously between $a$ and $b$.
मुख्य बिंदु 10
Always filter the final algebraic inequality through the specific Replacement Set given in the problem.

स्व-मूल्यांकन अभ्यास (Check Your Understanding)

मूल वैचारिक स्पष्टता की जांच के लिए नैदानिक प्रश्न। पहले स्वयं हल करें, फिर उत्तर देखें।

1
Solve the inequation: $3 - 2x \ge x - 12$, where the replacement set is $x \in \mathbb{W}$ (Whole numbers). Write the solution set.
उत्तर एवं व्याख्या देखें
उत्तर:

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 $\mathbb{W} = \{0, 1, 2, 3, \dots\}$:

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

.


Transpose to get $-3x \ge -15$. Dividing by $-3$ reverses the sign to $x \le 5$. Whole numbers are $\{0, 1, 2, 3, 4, 5\}$.
2
Solve: $\frac{x}{2} - 5 < \frac{x}{3} - 4$, given that $x \in \mathbb{Z}$ (Integers). What is the greatest integer value of $x$?
उत्तर एवं व्याख्या देखें
उत्तर:

Step 1: Multiply all terms by the LCM of denominators ($2, 3$), which is $6$:

$$6 \left( \frac{x}{2} \right) - 6(5) < 6 \left( \frac{x}{3} \right) - 6(4)$$


$$3x - 30 < 2x - 24$$


Step 2: Transpose $2x$ to LHS and $-30$ to RHS:

$$3x - 2x < -24 + 30$$


$$\mathbf{x < 6}$$


Step 3: Since $x \in \mathbb{Z}$, the solution set is $\{\dots, 2, 3, 4, 5\}$.
• The greatest integer value of $x$ is $5$.


Multiply by 6 to get $3x - 30 < 2x - 24 \implies x < 6$. The greatest integer less than 6 is 5.
3
Solve the double inequation: $-3 \le 2x - 1 < 5$, given $x \in \mathbb{R}$ (Real numbers). Represent the solution set on a number line.
उत्तर एवं व्याख्या देखें
उत्तर:

Step 1: Add $1$ to all three parts of the inequality:

$$-3 + 1 \le 2x - 1 + 1 < 5 + 1$$


$$-2 \le 2x < 6$$


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

$$\frac{-2}{2} \le \frac{2x}{2} < \frac{6}{2}$$


$$\mathbf{-1 \le x < 3}$$


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


Add 1 to all sides ($-2 \le 2x < 6$), divide by 2 ($-1 \le x < 3$). Plot solid dot at -1, open circle at 3.
4
If the replacement set is $\mathbb{N} = \{1, 2, 3, 4, \dots\}$, find the solution set of: $4(x - 1) \le 3(x + 1) + 2$.
उत्तर एवं व्याख्या देखें
उत्तर: Step 1: Expand brackets using the Distributive Law:
$$4x - 4 \le 3x + 3 + 2$$
$$4x - 4 \le 3x + 5$$
Step 2: Transpose $3x$ to LHS and $-4$ to RHS:
$$4x - 3x \le 5 + 4$$
$$\mathbf{x \le 9}$$
Step 3: Filter by Replacement Set $\mathbb{N}$ (Natural numbers start from $1$):
$$\text{Solution Set} = \mathbf{\{1, 2, 3, 4, 5, 6, 7, 8, 9\}}$$.
Expand to $4x - 4 \le 3x + 5 \implies x \le 9$. Natural numbers start at 1, so the solution is $\{1, 2, \dots, 9\}$.
5
Explain with a mathematical proof why multiplying an inequation by a negative number reverses the inequality symbol.
उत्तर एवं व्याख्या देखें
उत्तर:

• Let $a$ and $b$ be real numbers such that $a < b$.
• Subtract $b$ from both sides:

$$a - b < 0$$


• Subtract $a$ from both sides:

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


Notice that starting from $a < b$, we 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
Solve: $7 - 3x > -2$, given that $x \in \mathbb{Z}^+$ (Positive integers).
उत्तर एवं व्याख्या देखें
उत्तर:

Step 1: Transpose $7$ to RHS:

$$-3x > -2 - 7$$


$$-3x > -9$$


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

$$x < \frac{-9}{-3} \implies \mathbf{x < 3}$$


Step 3: Filter by Replacement Set $\mathbb{Z}^+ = \{1, 2, 3, \dots\}$:

$$\text{Solution Set} = \mathbf{\{1, 2\}}$$

.


$-3x > -9 \implies x < 3$. Positive integers strictly less than 3 are $\{1, 2\}$.
7
On a number line, what is the graphical difference between representing $x > 4$ and $x \ge 4$ on the set of real numbers $\mathbb{R}$?
उत्तर एवं व्याख्या देखें
उत्तर:

• For $x > 4$ (Strict Inequality): An open circle ($\\circ$) is drawn at $4$ to indicate that the number $4$ itself is excluded from the solution set, with a bold shaded arrow extending infinitely to the right.
• For $x \ge 4$ (Inclusive Inequality): A solid, filled-in dot ($\\bullet$) is drawn at $4$ to signify that $4$ is included in the solution set, with a bold shaded arrow extending infinitely to the right.


Open circle ($\circ$) excludes the endpoint ($>$); solid dot ($ullet$) includes the endpoint ($\ge$).
8
Find the smallest integer $x$ that satisfies the inequation: $5x - 4 > 2(x + 1)$.
उत्तर एवं व्याख्या देखें
उत्तर:

Step 1: Expand brackets:

$$5x - 4 > 2x + 2$$


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

$$5x - 2x > 2 + 4$$


$$3x > 6$$


$$x > \frac{6}{3} \implies \mathbf{x > 2}$$


Since $x \in \mathbb{Z}$ and $x$ must be strictly greater than $2$, the smallest integer is $3$.


$3x > 6 \implies x > 2$. The smallest integer strictly greater than 2 is 3.
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कक्षा 7 Mathematics के सभी अध्याय

अध्याय 1: पूर्णांक (Integers) अध्याय 2: परिमेय संख्याएँ (Rational Numbers) अध्याय 3: भिन्न (Fractions) अध्याय 4: घातांक (Exponents) अध्याय 5: समुच्चय (Sets) अध्याय 6: अनुपात और समानुपात (Ratio and Proportion) अध्याय 7: प्रतिशतता (Percentage) अध्याय 8: चाल, दूरी और समय (Speed, Distance and Time) अध्याय 9: बीजीय व्यंजक (Algebraic Expressions) अध्याय 10: एक चर वाले रैखिक समीकरण (Linear Equations) अध्याय 11: आकारों की समझ (Understanding Shapes) अध्याय 12: त्रिभुज (Triangles) अध्याय 13: सममिति (Symmetry) अध्याय 14: त्रिविमीय आकारों का द्विविमीय निरूपण (Representing 3D in 2D) अध्याय 15: सर्वांगसमता (Congruence) अध्याय 16: रचनाएँ (Constructions) अध्याय 17: क्षेत्रमिति (Mensuration) अध्याय 18: आँकड़ों का प्रबंधन (Data Handling) अध्याय 19: रैखिक असमिकाएँ (Linear Inequations)

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