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अध्याय 11: एक चर वाले रैखिक समीकरण (Linear Equations)

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ICSE • कक्षा 6 • Mathematics • अध्याय 11

अनुमानित समय: 45 Mins

प्रगति: अध्ययनरत

एक चर वाले रैखिक समीकरण (Linear Equations)

Linear Equations in One Variable constitutes one of the most transformative conceptual bridges in the entire CISCE Class 6 Mathematics curriculum. While earlier arithmetic taught students to perform routine operations on concrete numbers, and basic algebra introduced letters as representations of generalised quantities, the study of equations elevates mathematics from mere computation to logical problem solving. An equation is fundamentally an open mathematical sentence of equality asserting that two distinct algebraic expressions have identical numerical values. In symbolic form, an equation consists of a Left Hand Side, abbreviated as LHS, connected to a Right Hand Side, abbreviated as RHS, by the universal sign of equality. Without an equality sign, a collection of terms such as four x plus seven is merely an algebraic expression; it possesses no truth value and cannot be solved. The moment an equality sign is introduced, as in four x plus seven equals twenty-three, the statement becomes an equation that can be investigated, evaluated, and solved for the unknown variable. To appreciate the exact scope of this chapter in ICSE Class 6, students must master two defining technical boundaries: the condition of being linear and the condition of containing one variable. The term linear signifies that the highest power or exponent of the variable appearing in the equation is strictly equal to one. Equations containing higher exponents such as x squared or x cubed, or variables trapped within square roots or denominators such as one divided by x, are non-linear and belong to advanced curricula. The phrase in one variable indicates that the mathematical statement contains only a single alphabetical letter representing an unknown quantity, such as x or y, rather than multiple unknown quantities like x and y simultaneously. Consequently, the standard canonical form of a linear equation in one variable can always be expressed in the general structure a times x plus b equals c, where a, b, and c are known constant numbers and the leading coefficient a is strictly non-zero. The central goal of studying an equation is finding its solution or root. The solution of a linear equation is that unique numerical value of the variable which, when substituted in place of the variable, renders the Left Hand Side strictly equal to the Right Hand Side. When this condition is met, the numerical value is said to satisfy the equation. Every linear equation in one variable possesses one and only one unique solution. In early pedagogical training, students explore the Trial and Error Method, also known as the inspection method. Under this intuitive technique, a student constructs a systematic table, substituting consecutive trial numbers such as one, two, three, and so forth into the variable, calculating the resulting value of the LHS for each trial until it matches the RHS. While the trial and error method offers profound conceptual appreciation of what satisfying an equation truly means, it is inherently inefficient, laborious, and practically impossible when the true solution is a negative integer, a large number, or a fraction. To achieve rapid, error-free mastery, algebra relies upon the Fundamental Axioms of Equality, which mirror the physical laws of a balanced pair of weighing scales. A balanced equation remains in perfect equilibrium provided that identical mathematical operations are applied to both sides simultaneously. Specifically, four universal balancing axioms govern all algebraic transformations: First, adding the same number to both sides preserves equality. Second, subtracting the same number from both sides preserves equality. Third, multiplying both sides by the same non-zero number preserves equality. Fourth, dividing both sides by the same non-zero number preserves equality. These balancing axioms establish the mathematical justification for the Method of Transposition, which serves as the universal workhorse of algebraic problem solving. Transposition is the systematic process of shifting a term from one side of an equation to the opposite side accompanied by an mandatory inversion of its mathematical sign. A positive term transposes as a negative term, a negative term transposes as a positive term, a multiplying factor transposes as a divisor, and a dividing quantity transposes as a multiplying factor. As equations grow in structural complexity, students learn to clear parentheses using the distributive property, collect like variable terms on the Left Hand Side, isolate constant numerals on the Right Hand Side, and simplify equations involving fractional coefficients. Finally, the ultimate real-world payoff of the chapter lies in solving practical Word Problems. By mastering a disciplined five-step protocol consisting of reading the problem, declaring the variable, formulating the algebraic equation, solving via transposition, and verifying the answer in the original word context, students gain the superpower of translating complex real-world situations into clear, soluble mathematical models.

अध्याय CBT ऑनलाइन टेस्ट दें सभी अध्याय देखें

प्रस्तावना अध्याय अवधारणाएँ सूत्र एवं सर्वसमिकाएँ हल उदाहरण सामान्य गलतियाँ अध्याय सारांश स्व-मूल्यांकन अभ्यास टेस्ट दें

क्या आपने कभी सोचा है?

Imagine you are holding an old-fashioned physical balance scale with identical brass weights on both pans. On the left pan sits a mysterious velvet pouch filled with gold coins along with seven loose silver coins. On the right pan rest exactly twenty-three identical silver coins, and the needle points dead centre in perfect equilibrium. How can you determine the exact number of coins hidden inside the velvet pouch without ever tearing it open? By removing seven loose coins from both pans simultaneously, the balance remains utterly undisturbed, revealing that four pouches equal sixteen coins, which means each pouch contains exactly four coins! This elegant, intuitive puzzle encapsulates the heart and soul of Linear Equations in One Variable. It is not merely an abstract mathematical topic; it is humanity's most powerful analytical technique for cracking hidden mysteries, decoding financial markets, balancing industrial chemical reactions, and programming the computer game physics of the digital world.

यह अध्याय क्यों महत्वपूर्ण है

Linear equations in one variable represent the true foundational cornerstone of analytical reasoning across all branches of quantitative human knowledge. In everyday adult life, you use linear equations intuitively whenever you calculate the remaining travel time on a highway road trip, estimate how many units of electricity you can consume within a monthly household budget, or compute the exact discounted price of an item during a seasonal shopping sale. In commercial banking and personal finance, calculating simple interest, determining loan repayment schedules, and calculating profit margins all rely directly on the principles of single-variable linear equations. In scientific laboratories, chemists balance chemical reaction equations, pharmacists calculate precise medication dosages proportional to patient body weight, and mechanical engineers calculate structural load distributions using these exact mathematical balances. Furthermore, modern computer science and software development are entirely built upon algorithmic equations; every video game engine tracking character collision physics, screen rendering coordinate transformations, and database search query algorithms uses linear equations as foundational code logic. For an ICSE Class 6 student, developing fluent comfort with forming and solving linear equations builds irreplaceable logical resilience, preparing you for simultaneous linear equations, quadratic equations, coordinate geometry, and higher physics in senior board classes.

अध्ययन से पूर्व (आवश्यक ज्ञान)

Mastery of fundamental operations on whole numbers, natural numbers, and integers (including addition, subtraction, multiplication, and division with positive and negative signs).
Clear understanding of algebraic notation, literal numbers, variables, constants, and the omission of multiplication signs in algebraic products.
Ability to distinguish between like terms and unlike terms, and proficiency in adding and subtracting like algebraic terms.
Familiarity with the distributive property of multiplication over addition and subtraction: a(b + c) = ab + ac.
Working knowledge of basic fractions, finding the Least Common Multiple (LCM) of numbers, and performing arithmetic with rational numbers.

इस अध्याय के लक्ष्य

Define an algebraic equation clearly and explain the fundamental distinction between an algebraic expression and an algebraic equation.
Identify and formulate linear equations in one variable, recognizing why the degree of the variable must strictly equal one.
Demonstrate the concept of a solution or root of an equation and verify whether a given numerical candidate satisfies LHS = RHS.
Solve introductory linear equations using the intuitive Trial and Error tabular method and evaluate its conceptual merits and practical limitations.
State and apply the four fundamental axioms of equality based on the weighing balance scale model to maintain mathematical equilibrium.
Execute the Method of Transposition with complete accuracy by correctly reversing operational signs when transferring terms across the equality symbol.
Solve multi-step linear equations containing brackets, parentheses, and variable terms distributed across both sides of the equation.
Formulate and solve real-life ICSE board word problems across number puzzles, age scenarios, geometric perimeter applications, and monetary transactions using a disciplined 5-step algebraic strategy.

अध्याय रूपरेखा एवं प्रगति

1 Module 1: Concept of an Equation &...

2 Module 2: Solution / Root of an Equ...

3 Module 3: Solving Equations by the...

4 Module 4: Fundamental Properties of...

5 Module 5: The Method of Transpositi...

6 Module 6: Solving Multi-Step Equati...

7 Module 7: Word Problems on Linear E...

सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन

Module 1: Concept of an Equation & Linear Equations in One Variable

The Statement of Equality: Understanding Equations

In algebra, we frequently combine numbers, operations, and variables. But when two algebraic expressions are joined together by an equality sign, we unlock an equation — a mathematical statement that can be evaluated and solved for an unknown quantity.

1. What is an Algebraic Equation?

An equation is a mathematical statement asserting that two algebraic expressions are equal in value. It contains two sides separated by the equality sign (=):

Left Hand Side (LHS)
$4x + 7$

Right Hand Side (RHS)
$23$

The equality sign acts as an unbreakable bridge stating that the numerical value of the LHS expression matches the RHS expression.

2. Fundamental Distinction: Algebraic Expression vs Algebraic Equation

One of the most persistent confusions among ICSE Class 6 students is treating an expression and an equation as the same thing. The table below outlines the essential differences:

Dimension	Algebraic Expression	Algebraic Equation
Definition	A mathematical phrase consisting of numbers, variables, and operational signs.	A mathematical sentence stating that two expressions are equal.
Equality Sign (=)	Does NOT contain an '=' sign.	MUST contain an '=' sign.
Sides	Has no LHS or RHS; it is a single standalone entity.	Has two clear sides: Left Hand Side (LHS) and Right Hand Side (RHS).
Solvability	Cannot be "solved"; can only be simplified or evaluated for given values.	Can be solved to determine the unique value of the variable.
Examples	$3x + 5$, $x^2 - 4$, $7y$	$3x + 5 = 14$, $x - 2 = 9$, $2y + 1 = y + 7$

3. Defining "Linear" and "In One Variable"

What does "Linear" mean?

An equation is called linear when the highest power (exponent or degree) of the variable involved is strictly one (1).

• $5x + 3 = 18$ is linear (power of $x$ is 1).
• $x^2 + 4 = 20$ is NOT linear (power is 2; it is quadratic).
• $y^3 - 1 = 26$ is NOT linear (power is 3; it is cubic).

What does "In One Variable" mean?

The equation contains only one single unknown letter (variable).

• $3x + 8 = 23$ is in one variable ($x$).
• $2y - 5 = 11$ is in one variable ($y$).
• $x + y = 10$ is NOT in one variable (it contains two variables, $x$ and $y$).

Standard Canonical Form:

Every linear equation in one variable can be written in the canonical form: \[ ax + b = c \quad ext{or} \quad ax + b = 0 \] where $a, b,$ and $c$ are known real constants, $x$ is the unknown variable, and the coefficient of the variable must never equal zero ($a eq 0$).

Module 2: Solution / Root of an Equation & Verification (LHS = RHS)

The Meaning of a Solution: Satisfying the Equation

Solving an equation means determining the specific number that makes the equation true. In this module, you will learn what it means for a number to be a solution or root, and how to verify answers using the ICSE two-column verification method.

1. What is a Solution or Root?

The solution (also called the root) of an equation is the specific numerical value of the variable which, when substituted in place of the variable, makes the Left Hand Side (LHS) strictly equal to the Right Hand Side (RHS).

Key Board Principle — Satisfying the Equation:

When a value of the variable makes $ ext{LHS} = ext{RHS}$, we say that this value satisfies the equation.
A linear equation in one variable has one and only one unique solution.

2. The Two-Column ICSE Verification Protocol

In ICSE examinations, questions often carry 1 mark specifically for verification. To secure full marks, never write LHS and RHS together in a single line; evaluate them in two distinct steps:

Worked Example: Verify whether $x = 5$ is a solution of $3x - 4 = 11$

Step 1: Evaluate LHS for $x = 5$

LHS = 3x - 4
= 3(5) - 4
= 15 - 4
LHS = 11

Step 2: State RHS and Compare

RHS = 11

LHS = RHS = 11

Conclusion: Since LHS = RHS, $x = 5$ satisfies the equation and is confirmed as the correct solution.

Testing a Non-Solution: Is $x = 3$ a root of $3x - 4 = 11$?

• $ ext{LHS} = 3(3) - 4 = 9 - 4 = 5$.
• $ ext{RHS} = 11$.
• Since $ ext{LHS} eq ext{RHS}$ ($5 eq 11$), $x = 3$ does NOT satisfy the equation and is NOT a solution.

Module 3: Solving Equations by the Trial and Error Method

The Intuitive Approach: Trial and Error Method

Before learning algebraic formulas, students first solve equations through inspection and experimentation. The Trial and Error Method (or Substitution Method) constructs a systematic numerical table to test consecutive values until the solution is discovered.

1. The 4-Column Tabular Algorithm

To solve an equation by trial and error in an ICSE Class 6 test, construct a 4-column table:

Trial Value ($x$)	LHS Value: $2x + 5$	RHS Value	Is LHS = RHS?
$x = 1$	$2(1) + 5 = 2 + 5 = 7$	13	No ($7 eq 13$)
$x = 2$	$2(2) + 5 = 4 + 5 = 9$	13	No ($9 eq 13$)
$x = 3$	$2(3) + 5 = 6 + 5 = 11$	13	No ($11 eq 13$)
$x = 4$	$2(4) + 5 = 8 + 5 = 13$	13	YES! ($13 = 13$)

Conclusion: The trial table proves that for $x = 4$, LHS becomes equal to RHS. Therefore, the unique solution is $x = 4$.

2. Critical Evaluation: Merits vs Limitations

Pedagogical Merits:

• Develops direct intuition for what an equation represents.
• Reinforces mental arithmetic and basic integer operations.
• Requires no formal algebraic rules or formula memorization.

Practical Limitations:

• Extremely time-consuming and tedious for large roots (e.g., $x = 78$).
• Fails completely when the solution is a fraction (e.g., $x = rac{3}{7}$).
• Difficult to guess when roots are negative integers (e.g., $x = -14$).

Why We Need Systematic Algebraic Methods:

Because trial and error is limited to small whole numbers, mathematicians created systematic techniques: the Fundamental Axioms of Equality (balancing scales) and the Method of Transposition, which allow us to solve any linear equation in seconds!

Module 4: Fundamental Properties of Equality (Axioms of Equation Balancing)

The Weighing Balance: Axioms of Equality

An equation behaves exactly like a mechanical weighing balance in perfect equilibrium. As long as identical mathematical operations are performed on both pans simultaneously, the balance remains true. These universal rules are called the Fundamental Axioms of Equality.

1. The Four Golden Axioms of Equality

Suppose two quantities are equal: $A = B$. The following four operations maintain perfect mathematical equilibrium:

Axiom 1: Addition Property

If the same number is added to both sides of an equation, equality is preserved: \[ A = B \implies A + c = B + c \] Application: To cancel $-8$, add $+8$ to both sides.

Axiom 2: Subtraction Property

If the same number is subtracted from both sides of an equation, equality is preserved: \[ A = B \implies A - c = B - c \] Application: To cancel $+11$, subtract $11$ from both sides.

Axiom 3: Multiplication Property

If both sides of an equation are multiplied by the same non-zero number, equality is preserved: \[ A = B \implies A imes c = B imes c \quad (c eq 0) \] Application: To cancel a divisor $4$, multiply both sides by $4$.

Axiom 4: Division Property

If both sides of an equation are divided by the same non-zero number, equality is preserved: \[ A = B \implies rac{A}{c} = rac{B}{c} \quad (c eq 0) \] Application: To cancel a multiplier $5$, divide both sides by $5$.

2. Solving One-Step Equations using Axioms

Four Representative ICSE Board Models:

Model A: $x - 7 = 15$
Add 7 to both sides (Axiom 1):
$x - 7 + 7 = 15 + 7$
$x = 22$

Model B: $y + 9 = 24$
Subtract 9 from both sides (Axiom 2):
$y + 9 - 9 = 24 - 9$
$y = 15$

Model C: $6p = 48$
Divide both sides by 6 (Axiom 4):
$rac{6p}{6} = rac{48}{6}$
$p = 8$

Model D: $rac{m}{5} = 9$
Multiply both sides by 5 (Axiom 3):
$rac{m}{5} imes 5 = 9 imes 5$
$m = 45$

Module 5: The Method of Transposition (Side-Changing Method)

The Algebraic Powerhouse: Method of Transposition

While balancing axioms are mathematically rigorous, writing addition or subtraction on both sides for multi-step equations is slow. Algebra provides an elegant shortcut: Transposition — shifting terms across the '=' sign with an inverse sign change.

1. What is Transposition?

Transposition is the systematic process of shifting a term or factor from one side of an equation to the other side. Whenever a term moves across the equality sign (=), its mathematical operation is strictly inverted.

The Four Universal Transposition Inversion Rules:

1. Addition becomes Subtraction:
$+ a$ on LHS $\implies - a$ on RHS
Example: $x + 7 = 19 \implies x = 19 - 7 = 12$

2. Subtraction becomes Addition:
$- b$ on LHS $\implies + b$ on RHS
Example: $x - 5 = 14 \implies x = 14 + 5 = 19$

3. Multiplication becomes Division:
$ imes c$ on LHS $\implies \div c$ on RHS
Example: $4x = 28 \implies x = rac{28}{4} = 7$

4. Division becomes Multiplication:
$\div d$ on LHS $\implies imes d$ on RHS
Example: $rac{x}{3} = 8 \implies x = 8 imes 3 = 24$

2. The 4-Step Transposition Algorithm

Step 1: Collect variable terms on the LHS: Transpose all terms containing the variable to the Left Hand Side, reversing their signs.
Step 2: Collect constant terms on the RHS: Transpose all constant numbers to the Right Hand Side, reversing their signs.
Step 3: Combine like terms: Add or subtract like terms on both LHS and RHS.
Step 4: Divide by the coefficient: Divide both sides by the numerical coefficient of the variable to isolate $x$.

            Solve: 5x - 7 = 2x + 8

            Step 1 & 2: 5x - 2x = 8 + 7  (2x transposed to LHS as -2x, -7 transposed to RHS as +7)

            Step 3:     3x = 15          (Combined like terms)

            Step 4:     x = 15 / 3 = 5   (Coefficient 3 transposed as divisor)

            Root: x = 5

Module 6: Solving Multi-Step Equations & Equations with Brackets

Expanding Parentheses & Handling Fractional Forms

In board examinations, equations rarely appear in simple one-step formats. They frequently feature brackets, negative multipliers, and fractional expressions. Mastering bracket expansion and cross-multiplication guarantees error-free performance.

1. Removing Brackets using the Distributive Property

When an equation contains brackets, expand them first using the distributive law: $a(b + c) = ab + ac$.

Examiner Warning — Negative Multiplier Outside Brackets:

A negative multiplier changes the sign of EVERY term inside the bracket:
• $-3(2x - 5) = -6x + 15$ (NOT $-6x - 15$!)
• $-(4x + 7) = -4x - 7$
Failing to change the inner sign is the single most common cause of lost marks in ICSE Class 6 algebra!

Worked Problem 1: Solving Equations with Multiple Brackets

Problem: Solve for $x$: $4(2x - 3) - 3(x - 5) = 22$

            Step 1 (Expand brackets):   8x - 12 - 3x + 15 = 22

            Step 2 (Combine LHS terms): (8x - 3x) + (-12 + 15) = 22

                                       5x + 3 = 22

            Step 3 (Transpose +3):      5x = 22 - 3

                                       5x = 19

            Step 4 (Transpose 5):       x = 19 / 5 = 3.8  (or 3 4/5)

2. Equations with Fractions & Cross-Multiplication

When an equation involves a single fraction on one side, such as $rac{ax + b}{c} = d$, we transpose the denominator $c$ by cross-multiplying: \[ rac{ax + b}{c} = d \implies ax + b = c imes d \]

Worked Problem 2: Clearing Fractional Denominators

Problem: Solve $rac{3x - 1}{4} = 5$

Step 1: Transpose denominator 4 to RHS as a multiplier:
$3x - 1 = 5 imes 4 \implies 3x - 1 = 20$

Step 2: Transpose $-1$ to RHS as $+1$:
$3x = 20 + 1 \implies 3x = 21$

Step 3: Transpose 3 as a divisor:
$x = rac{21}{3} = \mathbf{7}$.

Module 7: Word Problems on Linear Equations (Practical Real-Life Applications)

Translating Reality into Algebra: Word Problems

The ultimate test of algebraic thinking is translating an everyday English situation into a mathematical equation. By following a disciplined 5-step strategy, you can solve any ICSE word problem with absolute confidence.

1. The 5-Step ICSE Word Problem Strategy

Step 1: Read and Identify: Read the problem carefully twice. Clearly identify the unknown quantity to be found.

Step 2: Assign a Variable: Let the unknown quantity be $x$ (or any letter), specifying its units (e.g., "Let the required number be $x$", "Let son's age be $x$ years").

Step 3: Form the Equation: Translate the English conditions sentence-by-sentence into an algebraic equation connecting LHS and RHS.

Step 4: Solve by Transposition: Isolate the variable using standard transposition rules.

Step 5: Verify & State Units: Check that your numerical answer satisfies the original problem description and state the final answer clearly with units (e.g., cm, kg, ₹, years).

2. Four Classic ICSE Word Problem Models

Category 1: Consecutive Number Puzzles

Problem: The sum of two consecutive natural numbers is 43. Find the numbers.

• Let first number = $x$.
• Second consecutive number = $x + 1$.
• Equation: $x + (x + 1) = 43 \implies 2x + 1 = 43$
• $2x = 42 \implies x = 21$.
Numbers: 21 and 22.

Category 2: Age Scenarios

Problem: A father is 28 years older than his son. In 5 years, he will be three times as old as his son. Find their present ages.

• Let son's present age = $x$ yrs.
• Father's present age = $(x + 28)$ yrs.
• In 5 yrs: Father is $(x + 33)$, Son is $(x + 5)$.
• Equation: $x + 33 = 3(x + 5) \implies x + 33 = 3x + 15$
• $33 - 15 = 3x - x \implies 18 = 2x \implies x = 9$.
Son: 9 yrs, Father: 37 yrs.

Category 3: Geometric Perimeter

Problem: The length of a rectangle is 4 cm more than its breadth. If the perimeter is 48 cm, find its dimensions.

• Let breadth = $x$ cm. Length = $(x + 4)$ cm.
• Perimeter = $2( ext{Length} + ext{Breadth}) = 48$
• $2(x + 4 + x) = 48 \implies 2(2x + 4) = 48$
• $4x + 8 = 48 \implies 4x = 40 \implies x = 10$.
Breadth: 10 cm, Length: 14 cm.

Category 4: Monetary Sharing

Problem: ₹300 is divided between Rohan and Sohan such that Rohan gets ₹40 more than Sohan. Find each share.

• Let Sohan's share = $x$.
• Rohan's share = $x + 40$.
• Equation: $x + (x + 40) = 300$
• $2x + 40 = 300 \implies 2x = 260 \implies x = 130$.
Sohan: ₹130, Rohan: ₹170.

महत्वपूर्ण सूत्र, सर्वसमिकाएँ एवं प्रमेय

Canonical Structure of a Linear Equation in One Variable

$$ax + b = c \quad \text{or} \quad ax + b = 0 \quad (a, b, c \in \mathbb{R}, \; a \neq 0)$$

The Root / Solution Existence Identity

$$x = r \text{ is a root} \iff \text{LHS}(r) = \text{RHS}(r)$$

Fundamental Axioms of Equality (Weighing Scale Invariants)

$$A = B \implies A \pm c = B \pm c, \quad A \times c = B \times c, \quad \frac{A}{c} = \frac{B}{c} \quad (c \neq 0)$$

Universal Transposition Sign Inversion Law

$$(+) \xleftrightarrow{\text{crosses =}} (-), \qquad (\times) \xleftrightarrow{\text{crosses =}} (\div)$$

Parentheses Expansion and Distributive Sign Law

$$k(ax + b) = k \cdot ax + k \cdot b; \quad -k(ax - b) = -kax + kb$$

Single-Variable Isolation Algorithm

$$ax = d \implies x = \frac{d}{a} \quad (a \neq 0)$$

Cross-Multiplication Fractional Equation Rule

$$\frac{ax + b}{c} = d \implies ax + b = c \times d \quad (c \neq 0)$$

5-Step Word Problem Modeling Pipeline

$$\text{Read} \to \text{Assign Variable } x \to \text{Formulate } \text{LHS} = \text{RHS} \to \text{Solve via Transposition} \to \text{Verify with Units}$$

अवधारणात्मक हल उदाहरण एवं अनुप्रयोग (Solved Examples)

उदाहरण 1

Solve the equation $3x - 2 = 10$ using the trial and error method by testing values $x = 2, 3, 4$.

विस्तृत समाधान / उत्तर:

Step 1: Construct the systematic trial evaluation table
We evaluate the Left Hand Side ($\text{LHS} = 3x - 2$) for each consecutive trial value and compare it with the Right Hand Side ($\text{RHS} = 10$):

Trial Value	LHS Calculation: $3x - 2$	RHS	Is LHS = RHS?
$x = 2$	$3(2) - 2 = 6 - 2 = 4$	$10$	No ($4 \neq 10$)
$x = 3$	$3(3) - 2 = 9 - 2 = 7$	$10$	No ($7 \neq 10$)
$x = 4$	$3(4) - 2 = 12 - 2 = 10$	$10$	Yes ($10 = 10$)

[1 Mark for complete trial table]

Step 2: State the final solution
Since $x = 4$ makes $\text{LHS} = \text{RHS} = 10$, the root of the equation is: $$\mathbf{x = 4}$$ [1 Mark for concluding root]

Final Answer: $\mathbf{x = 4}$

उदाहरण 2

Solve the following equations by stating the specific axiom of equality applied in each step: (a) $x - 9 = 14$ (b) $\frac{y}{6} = 7$

विस्तृत समाधान / उत्तर:

Part (a): Solve $x - 9 = 14$ [1.5 Marks]
Given equation: $x - 9 = 14$
• To eliminate $-9$ from the LHS, add $9$ to both sides using Axiom 1 (Addition Property of Equality):
$$x - 9 + 9 = 14 + 9$$ $$x + 0 = 23$$ $$\mathbf{x = 23}$$ [1.5 Marks for stating Axiom 1 and finding $x = 23$]

Part (b): Solve $\frac{y}{6} = 7$ [1.5 Marks]
Given equation: $\frac{y}{6} = 7$
• To eliminate the divisor $6$ from the LHS, multiply both sides by $6$ using Axiom 3 (Multiplication Property of Equality):
$$\frac{y}{6} \times 6 = 7 \times 6$$ $$y = 42$$ $$\mathbf{y = 42}$$ [1.5 Marks for stating Axiom 3 and finding $y = 42$]

Final Answer: (a) $\mathbf{x = 23}$, (b) $\mathbf{y = 42}$

उदाहरण 3

Solve the linear equation $5x - 7 = 3x + 9$ using the method of transposition. Verify your answer by evaluating LHS and RHS.

विस्तृत समाधान / उत्तर:

Step 1: Solve using transposition [2 Marks]
Given: $5x - 7 = 3x + 9$
• Transpose variable term $3x$ from RHS to LHS as $-3x$:
$$5x - 3x - 7 = 9$$ • Transpose constant $-7$ from LHS to RHS as $+7$:
$$5x - 3x = 9 + 7$$ • Combine like terms on both sides:
$$2x = 16$$ • Transpose coefficient $2$ to RHS as a divisor:
$$x = \frac{16}{2} = \mathbf{8}$$ [2 Marks for correct transposition steps and solution]

Step 2: Verification Protocol [1 Mark]
Substitute $x = 8$ into the original equation:
• $\text{LHS} = 5x - 7 = 5(8) - 7 = 40 - 7 = \mathbf{33}$
• $\text{RHS} = 3x + 9 = 3(8) + 9 = 24 + 9 = \mathbf{33}$
Since $\text{LHS} = \text{RHS} = 33$, the root $\mathbf{x = 8}$ is verified to be strictly correct.
[1 Mark for complete verification showing LHS = RHS]

Final Answer: Root is $\mathbf{x = 8}$ (Verified)

उदाहरण 4

Solve the equation for $y$: $3(2y - 1) - 2(y - 5) = 19$. Check your solution.

विस्तृत समाधान / उत्तर:

Step 1: Expand brackets using the distributive property [1.5 Marks]
Given: $3(2y - 1) - 2(y - 5) = 19$
• Expand first bracket: $3 \times 2y - 3 \times 1 = 6y - 3$
• Expand second bracket with negative multiplier: $-2 \times y - 2 \times (-5) = -2y + 10$
Equation becomes: $$6y - 3 - 2y + 10 = 19$$ [1.5 Marks for correct expansion with $+10$ sign]

Step 2: Group like terms and transpose [1.5 Marks]
$$(6y - 2y) + (-3 + 10) = 19$$ $$4y + 7 = 19$$ • Transpose $+7$ to RHS as $-7$:
$$4y = 19 - 7$$ $$4y = 12$$ • Transpose coefficient $4$ as divisor:
$$y = \frac{12}{4} = \mathbf{3}$$ [1.5 Marks for combining like terms and isolating $y$]

Step 3: Verification Check [1 Mark]
Substitute $y = 3$ into original LHS:
$$\text{LHS} = 3[2(3) - 1] - 2[(3) - 5]$$ $$= 3(6 - 1) - 2(-2) = 3(5) - (-4) = 15 + 4 = \mathbf{19}$$ $$\text{RHS} = \mathbf{19}$$ Since $\text{LHS} = \text{RHS} = 19$, the solution is verified.
[1 Mark for verification]

Final Answer: $\mathbf{y = 3}$

उदाहरण 5

(a) The sum of three consecutive odd natural numbers is $57$. Find the three numbers. (b) The length of a rectangular park is twice its breadth. If the perimeter of the park is $72\text{ m}$, find its length and breadth.

विस्तृत समाधान / उत्तर:

Part (a): Consecutive Odd Numbers Puzzle [2 Marks]
Let the smallest odd natural number be $x$.
Since consecutive odd numbers differ by $2$, the three numbers are $x$, $(x + 2)$, and $(x + 4)$.
According to the problem: $$x + (x + 2) + (x + 4) = 57$$ $$3x + 6 = 57$$ Transpose $+6$ to RHS: $$3x = 57 - 6 = 51$$ $$x = \frac{51}{3} = \mathbf{17}$$ The three consecutive odd numbers are: • First number = $x = \mathbf{17}$
• Second number = $x + 2 = 17 + 2 = \mathbf{19}$
• Third number = $x + 4 = 17 + 4 = \mathbf{21}$
(Check: $17 + 19 + 21 = 57$, correct!)
[2 Marks for part (a)]

Part (b): Rectangular Park Dimensions [2 Marks]
Let the breadth of the park be $b\text{ metres}$.
Then the length is twice the breadth: $l = 2b\text{ metres}$.
Formula for perimeter of a rectangle: $$\text{Perimeter} = 2(l + b) = 72$$ $$2(2b + b) = 72$$ $$2(3b) = 72$$ $$6b = 72$$ Transpose $6$ as divisor: $$b = \frac{72}{6} = \mathbf{12\text{ m}}$$ Length $l = 2b = 2 \times 12 = \mathbf{24\text{ m}}$.
[2 Marks for part (b)]

Final Answer: (a) Three numbers are $\mathbf{17, 19, 21}$; (b) Breadth = $\mathbf{12\text{ m}}$, Length = $\mathbf{24\text{ m}}$

उदाहरण 6

(a) Solve the rational linear equation: $\frac{4x + 1}{3} - \frac{x - 2}{2} = 3$. (b) Ten years ago, a mother was four times as old as her daughter. If the sum of their present ages is $70\text{ years}$, find their present ages. Verify the solution with the original age conditions.

विस्तृत समाधान / उत्तर:

Part (a): Solve Fractional Equation [2.5 Marks]
Given: $\frac{4x + 1}{3} - \frac{x - 2}{2} = 3$
Step 1: Clear fractions by multiplying every term by the LCM of denominators $3$ and $2$, which is $6$:

$$6 \times \left(\frac{4x + 1}{3}\right) - 6 \times \left(\frac{x - 2}{2}\right) = 6 \times 3$$

$$2(4x + 1) - 3(x - 2) = 18$$

Step 2: Expand brackets carefully (notice $-3 \times (-2) = +6$):

$$8x + 2 - 3x + 6 = 18$$

$$(8x - 3x) + (2 + 6) = 18$$

$$5x + 8 = 18$$

Step 3: Transpose $+8$ and isolate $x$:

$$5x = 18 - 8 = 10$$

$$x = \frac{10}{5} = \mathbf{2}$$

[2.5 Marks for solving fractional equation]

Part (b): Age Scenario Problem [2.5 Marks]
Let the daughter's present age be $y\text{ years}$.
Since the sum of their present ages is $70$, the mother's present age is $(70 - y)\text{ years}$.

Ten years ago:
• Daughter's age was: $(y - 10)\text{ years}$
• Mother's age was: $(70 - y - 10) = (60 - y)\text{ years}$

According to the problem condition (Mother was $4$ times daughter's age):

$$60 - y = 4(y - 10)$$

$$60 - y = 4y - 40$$

Transpose variable terms and constant numbers:

$$60 + 40 = 4y + y$$

$$100 = 5y$$

$$y = \frac{100}{5} = \mathbf{20}$$

Therefore: • Daughter's present age: $y = \mathbf{20\text{ years}}$
• Mother's present age: $70 - y = 70 - 20 = \mathbf{50\text{ years}}$

Verification of conditions:

Sum of present ages: $20 + 50 = 70$ (Correct)
Ten years ago: Daughter was $20 - 10 = 10$, Mother was $50 - 10 = 40$. $40 = 4 \times 10$ (Correct!)
[2.5 Marks for age formulation, solution, and verification]

Final Answer: (a) $\mathbf{x = 2}$; (b) Daughter's age = $\mathbf{20\text{ years}}$, Mother's age = $\mathbf{50\text{ years}}$

सामान्य गलतियाँ एवं परीक्षक के जाल (Examiner Traps)

सामान्य भ्रम / गलत उत्तर

सही वैज्ञानिक तथ्य

Always apply the inversion rule whenever a term crosses '=': addition transposes as subtraction (x = 19 - 7 = 12), and subtraction transposes as addition.

सामान्य भ्रम / गलत उत्तर

सही वैज्ञानिक तथ्य

Remember that a negative multiplier flips every single internal sign: -2 * (-5) = +10. Thus, -2(x - 5) = -2x + 10.

सामान्य भ्रम / गलत उत्तर

सही वैज्ञानिक तथ्य

In 4x, the number 4 is a multiplier (4 * x). A multiplier transposes as a divisor: x = 20 / 4 = 5. Only added or subtracted terms transpose as opposites.

सामान्य भ्रम / गलत उत्तर

सही वैज्ञानिक तथ्य

'Subtracted from' places the original quantity first: '7 subtracted from x' means x - 7. Writing 7 - x inverts the entire equation and yields the wrong root.

सामान्य भ्रम / गलत उत्तर

सही वैज्ञानिक तथ्य

An equation is a weighing balance in equilibrium. Whatever operation is performed on the LHS MUST be executed identically on the RHS to preserve mathematical truth.

सामान्य भ्रम / गलत उत्तर

सही वैज्ञानिक तथ्य

Always write a brief 2-line verification (LHS = RHS) to secure the designated verification mark and ensure real-world units are appended to word problem answers.

चित्रात्मक व्याख्या एवं मॉडल

अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष

मुख्य बिंदु 1

An algebraic equation is an open mathematical statement asserting that two algebraic expressions are equal, joined by an equality sign (=).

मुख्य बिंदु 2

An equation is classified as linear in one variable when the highest exponent of the variable is strictly one and it contains only a single letter symbol.

मुख्य बिंदु 3

The standard canonical form of a linear equation in one variable is ax + b = c (or ax + b = 0), where a, b, and c are constants with a != 0.

मुख्य बिंदु 4

The solution or root of an equation is the unique numerical value of the variable which satisfies the equation by making LHS equal to RHS.

मुख्य बिंदु 5

Every linear equation in one variable has exactly one unique solution in the real number system.

मुख्य बिंदु 6

The trial and error method constructs a systematic table to test consecutive values of the variable until LHS matches RHS.

मुख्य बिंदु 7

The Fundamental Axioms of Equality state that adding, subtracting, multiplying, or dividing both sides by the same non-zero quantity preserves equality.

मुख्य बिंदु 8

Transposition is the algebraic shortcut of moving a term across the '=' sign with an inversion of its operational sign (+ to -, - to +, * to /, / to *).

मुख्य बिंदु 9

Multi-step equations are solved by expanding brackets with the distributive law, grouping variable terms on the LHS, constants on the RHS, and isolating x.

मुख्य बिंदु 10

Word problems are solved using a disciplined 5-step strategy: reading, declaring the variable, formulating the equation, solving via transposition, and verifying with units.

स्व-मूल्यांकन अभ्यास (Check Your Understanding)

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

Verify whether x = -4 is a solution of the linear equation 2x + 15 = 7.

उत्तर एवं व्याख्या देखें

उत्तर:

LHS = 2x + 15 = 2(-4) + 15 = -8 + 15 = 7.
RHS = 7.
Since LHS = RHS = 7, x = -4 satisfies the equation and is confirmed as the correct solution.

Substitute x = -4 into the LHS: calculate 2(-4) + 15 and compare the result with the RHS (7).

Solve the equation 4x - 9 = 19 using the trial and error method by testing values x = 5, 6, 7.

उत्तर एवं व्याख्या देखें

उत्तर:

• For x = 5: LHS = 4(5) - 9 = 20 - 9 = 11 != 19. • For x = 6: LHS = 4(6) - 9 = 24 - 9 = 15 != 19. • For x = 7: LHS = 4(7) - 9 = 28 - 9 = 19 = RHS. Therefore, the root is x = 7.

Create a table evaluating 4x - 9 for each trial value until LHS equals 19.

Solve the linear equation using transposition: 7x + 8 = 3x + 28.

उत्तर एवं व्याख्या देखें

उत्तर:

7x - 3x = 28 - 8.
4x = 20.
x = 20 / 4 = 5.

Transpose 3x to the LHS as -3x, and transpose +8 to the RHS as -8.

Solve the multi-step equation: 5(x - 2) - 2(x + 1) = 12.

उत्तर एवं व्याख्या देखें

उत्तर:

Expand brackets: 5x - 10 - 2x - 2 = 12.
Combine like terms: 3x - 12 = 12.
Transpose -12: 3x = 12 + 12 = 24.
x = 24 / 3 = 8.

Expand brackets carefully with the distributive law: 5(x - 2) = 5x - 10, and -2(x + 1) = -2x - 2.

Solve the fractional equation: (2x + 5) / 3 = 7.

उत्तर एवं व्याख्या देखें

उत्तर:

Multiply both sides by 3: 2x + 5 = 7 * 3 = 21.
Transpose +5: 2x = 21 - 5 = 16.
x = 16 / 2 = 8.

Transpose the denominator 3 across '=' as a multiplier first.

The perimeter of an isosceles triangle is 32 cm. The length of each equal side is 3 cm more than the base. Find the length of each side.

उत्तर एवं व्याख्या देखें

उत्तर:

Equation: x + (x + 3) + (x + 3) = 32 => 3x + 6 = 32 => 3x = 26 => wait, let's make it natural numbers: if perimeter is 33 cm, then 3x + 6 = 33 => 3x = 27 => x = 9. For perimeter = 32 cm: 3x = 26 => x = 26/3 = 8 2/3 cm. Base = 8 2/3 cm and equal sides = 11 2/3 cm.

Let the base be x cm. Then each equal side is (x + 3) cm. The perimeter is x + (x + 3) + (x + 3) = 32.

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एक चर वाले रैखिक समीकरण (Linear Equations) में कोई संदेह या प्रश्न है? हमारे AI अध्ययन मित्र से तुरंत समझें।

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Trial Value (\(x\))	LHS Value: \(2x + 5\)	RHS Value	Is LHS = RHS?
\(x = 1\)	\(2(1) + 5 = 2 + 5 = 7\)	13	No (\(7 eq 13\))
\(x = 2\)	\(2(2) + 5 = 4 + 5 = 9\)	13	No (\(9 eq 13\))
\(x = 3\)	\(2(3) + 5 = 6 + 5 = 11\)	13	No (\(11 eq 13\))
\(x = 4\)	\(2(4) + 5 = 8 + 5 = 13\)	13	YES! (\(13 = 13\))

एक चर वाले रैखिक समीकरण (Linear Equations)

क्या आपने कभी सोचा है?

यह अध्याय क्यों महत्वपूर्ण है

अध्ययन से पूर्व (आवश्यक ज्ञान)

इस अध्याय के लक्ष्य

अध्याय रूपरेखा एवं प्रगति

सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन

Module 1: Concept of an Equation & Linear Equations in One Variable

The Statement of Equality: Understanding Equations

1. What is an Algebraic Equation?

2. Fundamental Distinction: Algebraic Expression vs Algebraic Equation

3. Defining "Linear" and "In One Variable"

Module 2: Solution / Root of an Equation & Verification (LHS = RHS)

The Meaning of a Solution: Satisfying the Equation

1. What is a Solution or Root?

2. The Two-Column ICSE Verification Protocol

Worked Example: Verify whether \(x = 5\) is a solution of \(3x - 4 = 11\)

Module 3: Solving Equations by the Trial and Error Method

The Intuitive Approach: Trial and Error Method

1. The 4-Column Tabular Algorithm

2. Critical Evaluation: Merits vs Limitations

Module 4: Fundamental Properties of Equality (Axioms of Equation Balancing)

The Weighing Balance: Axioms of Equality

1. The Four Golden Axioms of Equality

2. Solving One-Step Equations using Axioms

Four Representative ICSE Board Models:

Module 5: The Method of Transposition (Side-Changing Method)

The Algebraic Powerhouse: Method of Transposition

1. What is Transposition?

The Four Universal Transposition Inversion Rules:

2. The 4-Step Transposition Algorithm

Module 6: Solving Multi-Step Equations & Equations with Brackets

Expanding Parentheses & Handling Fractional Forms

1. Removing Brackets using the Distributive Property

Worked Problem 1: Solving Equations with Multiple Brackets

2. Equations with Fractions & Cross-Multiplication

Module 7: Word Problems on Linear Equations (Practical Real-Life Applications)

Translating Reality into Algebra: Word Problems

1. The 5-Step ICSE Word Problem Strategy

2. Four Classic ICSE Word Problem Models

महत्वपूर्ण सूत्र, सर्वसमिकाएँ एवं प्रमेय

अवधारणात्मक हल उदाहरण एवं अनुप्रयोग (Solved Examples)

सामान्य गलतियाँ एवं परीक्षक के जाल (Examiner Traps)

सामान्य भ्रम / गलत उत्तर

सही वैज्ञानिक तथ्य

सामान्य भ्रम / गलत उत्तर

सही वैज्ञानिक तथ्य

सामान्य भ्रम / गलत उत्तर

सही वैज्ञानिक तथ्य

सामान्य भ्रम / गलत उत्तर

सही वैज्ञानिक तथ्य

सामान्य भ्रम / गलत उत्तर

सही वैज्ञानिक तथ्य

सामान्य भ्रम / गलत उत्तर

सही वैज्ञानिक तथ्य

चित्रात्मक व्याख्या एवं मॉडल

अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष

स्व-मूल्यांकन अभ्यास (Check Your Understanding)

ऑनलाइन CBT टेस्ट देकर तैयारी का मूल्यांकन करें

कक्षा 6 Mathematics के सभी अध्याय

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