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ICSE • Class 7 • Mathematics • Ch 10
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Linear Equations in One Variable

In ICSE Class 7 Mathematics, "Linear Equations in One Variable" provides an authoritative, algebraic master study guide analyzing the solution, balance, and problem-solving applications of degree-one equations. This comprehensive chapter explores Concept of an Equation (An open mathematical statement of equality containing an equal sign "$=$", an algebraic expression on the Left-Hand Side [LHS], and an expression on the Right-Hand Side [RHS]; Definition of a Linear Equation in one variable: highest power of the variable is strictly 1: $ax + b = 0$ with $a \neq 0$), The Concept of a Solution / Root (A numerical value of the variable which satisfies the equation by making $\text{LHS} = \text{RHS}$), Rules of Solving Equations (The Balance Scale model; Four fundamental properties of equality; The Method of Transposition: shifting a term from one side to the other with reversed sign $[+ \leftrightarrow -, \times \leftrightarrow \div]$; Clearing denominators using LCM; Cross-multiplication method for fractional linear equations $\frac{ax+b}{cx+d} = \frac{p}{q}$), and Word Problems on Linear Equations (Translating real-world situations into algebraic equations: Number and digit problems, Age problems [present age, age $n$ years ago, age $n$ years hence], Perimeter and geometry problems, Commercial currency/coin denomination problems) aligned with the 2026–27 CISCE curriculum.

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How Did an Ancient Greek Riddle on a Tombstone Reveal the Exact Lifespan of the "Father of Algebra" Using a Single Equation?

In the 3rd century CE in Alexandria, Egypt, lived a brilliant mathematician named Diophantus. When he died, his sorrowful admirers carved no dates on his marble tombstone. Instead, they engraved a fascinating mathematical riddle: "Here lies Diophantus. God granted him to be a boy for the sixth part of his life; after a twelfth more, his cheeks grew a beard; after a seventh part, he wed, and five years later, a son was born. Alas! The child lived only half his father's life, and four years after his son's death, Diophantus met his end." How old was Diophantus when he died? By translating this ancient poetic riddle into a single Linear Equation in One Variable—$\frac{x}{6} + \frac{x}{12} + \frac{x}{7} + 5 + \frac{x}{2} + 4 = x$—mathematicians solved it in minutes: Diophantus lived to exactly 84 years old! An equation is like a perfectly balanced weighing scale: whatever operation you perform on one side, you must perform on the other to preserve the balance. What is the Method of Transposition? How do you solve fractional equations using Cross-Multiplication? Let's master linear equations.

Why This Chapter Matters

Linear equations are the ultimate problem-solving engine in mathematics, physics (calculating forces, circuit currents via Ohm's Law $V = IR$), chemistry (balancing chemical reactions), and personal finance (budgeting, loan payments). Mastering algebraic transposition and word-problem translation is essential for ICSE exams and higher analytical thinking.

Before You Begin (Prerequisites)

  • Operations on integers and rational numbers.
  • Algebraic expressions, like terms, and distributive law from Chapter 9.
  • Basic arithmetic with fractions and LCM.

What You Will Learn (Core Objectives)

  • Define a linear equation in one variable and distinguish it from algebraic expressions.
  • Verify whether a given numerical value is a valid root / solution of an equation.
  • Solve linear equations using the Method of Transposition and clearing brackets.
  • Solve fractional linear equations using Cross-Multiplication: $\frac{ax+b}{cx+d} = \frac{p}{q}$.
  • Formulate and solve real-world word problems based on numbers, ages, geometry, and money.
  • Check and verify algebraic solutions by substituting back into LHS and RHS.

Chapter Roadmap & Progression

1 1. Concept of a Linear Equation & T...
2 2. The Method of Transposition & Ax...
3 3. Solving Fractional Linear Equati...
4 4. Word Problems: Translation to Eq...

Complete Concept Guide (100% Curriculum Coverage)

1. Concept of a Linear Equation & The Balance Model

Understand
A. Defining an Equation:
  • An Equation is a mathematical statement of equality containing an equal sign ($=$) connecting two expressions: $$\text{Left-Hand Side (LHS)} = \text{Right-Hand Side (RHS)}$$
  • Linear Equation in One Variable: An equation involving only one unknown variable whose highest exponent (degree) is strictly $1$. Its standard form is: $$ax + b = 0 \quad (a \neq 0)$$ (e.g., $3x - 5 = 16$ is linear; $x^2 + 4 = 0$ is NOT linear because the degree is 2).
B. Solution / Root of an Equation:

A number $k$ is called the solution or root of an equation if substituting $x = k$ makes the $\text{LHS} = \text{RHS}$.

Example: For $2x + 3 = 11$, check $x = 4$: $\text{LHS} = 2(4) + 3 = 8 + 3 = 11 = \text{RHS}$. Hence, $x = 4$ is the unique root.

2. The Method of Transposition & Axioms of Equality

Transposition Rules
A. The Balance Model:

An equation is like a balanced scale. Equality is preserved if you:

  • Add or subtract the same number on both sides.
  • Multiply or divide both sides by the same non-zero number.
B. Method of Transposition:

Instead of performing operations on both sides, we transpose (shift) terms across the "$=$" sign by reversing their operation:

Operation on One SideTransposed Operation on Other SideExample
Addition ($+$)Subtraction ($-$)$x + 7 = 15 \implies x = 15 - 7 = 8$
Subtraction ($-$)Addition ($+$)$x - 4 = 9 \implies x = 9 + 4 = 13$
Multiplication ($\times$)Division ($\div$)$5x = 35 \implies x = \frac{35}{5} = 7$
Division ($\div$)Multiplication ($\times$)$\frac{x}{3} = 8 \implies x = 8 \times 3 = 24$

3. Solving Fractional Linear Equations: Cross-Multiplication

Cross-Multiplication
A. The Cross-Multiplication Technique:

When an equation is in the form of a fraction equal to another fraction:

$$\frac{ax + b}{cx + d} = \frac{p}{q}$$

Multiply the numerator of the LHS by the denominator of the RHS, and equate it to the product of the denominator of the LHS and numerator of the RHS:

$$\mathbf{q(ax + b) = p(cx + d)}$$

Expand using the Distributive Law, collect variable terms on the LHS and constants on the RHS, and solve for $x$.

4. Word Problems: Translation to Equations

Word Problems Strategy

To solve applied word problems successfully:

  1. Identify the Unknown: Assign a variable (e.g., "Let the present age be $x$ years").
  2. Translate Words into Algebraic Expressions:
    • "Age $5$ years ago" $\to (x - 5)$
    • "Age $5$ years hence / after 5 years" $\to (x + 5)$
    • "Consecutive integers" $\to x, x+1, x+2$
    • "Consecutive even/odd integers" $\to x, x+2, x+4$
  3. Formulate the Equation: Connect the expressions using the given condition in the problem.
  4. Solve and Verify: Solve for $x$ and check if the result makes practical real-world sense.

Key Formulas, Identities & Theorems

Standard Linear Equation Form
$$ax + b = 0 \implies x = -\frac{b}{a} \quad (a \neq 0)$$
Has a unique single solution.
Cross-Multiplication Formulation
$$\frac{ax + b}{cx + d} = \frac{p}{q} \iff q(ax + b) = p(cx + d)$$
Valid provided cx + d != 0 and q != 0.

Linear Equations: The Balance Scale & Transposition Rules

Linear Equations: The Balance Scale & Transposition Rules LHS: 3x + 5 = RHS: 26 TRANSPOSITION RULES • (+) becomes (-) across = • (-) becomes (+) across = • (×) becomes (÷) across = • (÷) becomes (×) across = CROSS-MULTIPLICATION (ax+b) / (cx+d) = p / q ⇓ q(ax + b) = p(cx + d) Expand brackets & collect x on LHS WORD PROBLEMS • 1. Assign variable x • 2. Age n yrs ago = (x - n) • 3. Age n yrs hence = (x + n) • 4. Verify root: LHS = RHS AN EQUATION REMAINS BALANCED IF IDENTICAL OPERATIONS ARE EXECUTED ON BOTH SIDES

Chapter Summary & 10 Key Takeaways

Takeaway 1
An equation is a statement of mathematical equality: LHS = RHS.
Takeaway 2
A linear equation in one variable has a degree of strictly 1: $ax + b = 0$ ($a \neq 0$).
Takeaway 3
The solution (root) is the numerical value of the variable that makes LHS = RHS.
Takeaway 4
Transposition shifts terms across the equals sign by reversing operations ($+ \leftrightarrow -, \times \leftrightarrow \div$).
Takeaway 5
To solve fractional linear equations $\frac{ax+b}{cx+d} = \frac{p}{q}$, cross-multiply: $q(ax+b) = p(cx+d)$.
Takeaway 6
When brackets are preceded by a minus sign, change the sign of every term inside when expanding.
Takeaway 7
Always check the solution by substituting the obtained value back into the original equation.
Takeaway 8
In age problems, present age is $x$; age $n$ years ago is $x - n$; age $n$ years hence is $x + n$.
Takeaway 9
In consecutive integer problems, let the integers be $x, x+1, x+2$ (or $x, x+2, x+4$ for even/odd).
Takeaway 10
In geometric problems, use formulas like Perimeter of a Rectangle $= 2(l + b)$.

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 linear equation: $\frac{2x - 3}{4} - \frac{x - 1}{2} = \frac{1}{4} - \frac{x + 2}{3}$.
Reveal Answer & Explanation
Answer: Step 1: Find the LCM of all denominators ($4, 2, 3$). $\text{LCM} = 12$.
Multiply every term on both sides by $12$ to clear all fractions:
$$12 \left( \frac{2x - 3}{4} \right) - 12 \left( \frac{x - 1}{2} \right) = 12 \left( \frac{1}{4} \right) - 12 \left( \frac{x + 2}{3} \right)$$
$$3(2x - 3) - 6(x - 1) = 3(1) - 4(x + 2)$$
Step 2: Expand all brackets using the Distributive Law:
$$6x - 9 - 6x + 6 = 3 - 4x - 8$$
Step 3: Combine like terms on both sides:
• LHS: $(6x - 6x) + (-9 + 6) = 0 - 3 = -3$
• RHS: $-4x + (3 - 8) = -4x - 5$
Step 4: Solve for $x$ by transposition:
$$-3 = -4x - 5 \implies 4x = -5 + 3 \implies 4x = -2$$
$$x = \frac{-2}{4} = \mathbf{-\frac{1}{2}}$$.
Multiply every term by the LCM (12) to clear denominators, expand brackets, and solve for $x$.
2
Solve using cross-multiplication: $\frac{5x - 7}{3x + 1} = \frac{2}{3}$.
Reveal Answer & Explanation
Answer: Step 1: Cross-multiply the fractions:
$$3(5x - 7) = 2(3x + 1)$$
Step 2: Expand brackets:
$$15x - 21 = 6x + 2$$
Step 3: Transpose $6x$ to LHS and $-21$ to RHS:
$$15x - 6x = 2 + 21$$
$$9x = 23 \implies \mathbf{x = \frac{23}{9} = 2\frac{5}{9}}$$.
Check: $\text{LHS} = \frac{5(23/9) - 7}{3(23/9) + 1} = \frac{115/9 - 63/9}{69/9 + 9/9} = \frac{52/9}{78/9} = \frac{52}{78} = \frac{2}{3} = \text{RHS}$. Verified!
Cross-multiply: $3(5x - 7) = 2(3x + 1)$. Expand to $15x - 21 = 6x + 2$. Transpose and solve.
3
The sum of three consecutive multiples of $7$ is $357$. Find the three multiples.
Reveal Answer & Explanation
Answer: Let the three consecutive multiples of $7$ be $7x$, $7(x+1)$, and $7(x+2)$ (or $y, y+7, y+14$ where $y = 7x$).
According to the problem:
$$y + (y + 7) + (y + 14) = 357$$
$$3y + 21 = 357$$
Transpose $21$ to the RHS:
$$3y = 357 - 21 = 336$$
$$y = \frac{336}{3} = 112$$
The three consecutive multiples are:
• First multiple: $y = \mathbf{112}$
• Second multiple: $y + 7 = 112 + 7 = \mathbf{119}$
• Third multiple: $y + 14 = 112 + 14 = \mathbf{126}$
Check: $112 + 119 + 126 = 357$. All are multiples of 7!
Let the multiples be $y, y+7, y+14$. Add them: $3y + 21 = 357 \implies 3y = 336 \implies y = 112$.
4
A father is 30 years older than his son. In 12 years, the father will be twice as old as his son will be then. Find their present ages.
Reveal Answer & Explanation
Answer: Let the present age of the son be $x\text{ years}$.
Then, the present age of the father $= (x + 30)\text{ years}$.
Step 1: Ages after 12 years ($12\text{ years hence}$):
Son's age $= x + 12\text{ years}$
Father's age $= (x + 30) + 12 = x + 42\text{ years}$
Step 2: According to the given condition:
$$\text{Father's age} = 2 \times (\text{Son's age})$$
$$x + 42 = 2(x + 12)$$
$$x + 42 = 2x + 24$$
Step 3: Transpose $x$ to RHS and $24$ to LHS:
$$42 - 24 = 2x - x \implies \mathbf{x = 18}$$
• Present age of the son $= \mathbf{18\text{ years}}$
• Present age of the father $= 18 + 30 = \mathbf{48\text{ years}}$.
Check: In 12 years, son is 30, father is 60 ($60 = 2 \times 30$). Correct!
Let son be $x$, father be $x+30$. After 12 years, $x+42 = 2(x+12)$. Solve for $x=18$.
5
The perimeter of a rectangular swimming pool is $154\text{ meters}$. Its length is $2\text{ meters}$ more than twice its breadth. Find the length and breadth of the pool.
Reveal Answer & Explanation
Answer: Let the breadth of the pool be $b\text{ meters}$.
Then, length $l = (2b + 2)\text{ meters}$.
Perimeter formula for a rectangle: $P = 2(l + b)$
$$2(2b + 2 + b) = 154$$
$$2(3b + 2) = 154$$
Divide both sides by $2$:
$$3b + 2 = 77$$
$$3b = 77 - 2 = 75 \implies b = \frac{75}{3} = \mathbf{25\text{ meters}}$$
$$\text{Length } l = 2(25) + 2 = 50 + 2 = \mathbf{52\text{ meters}}$$.
Check: $2(52 + 25) = 2(77) = 154\text{ m}$. Correct!
Let breadth be $b$ and length be $2b+2$. Perimeter $2(2b+2+b) = 154 \implies 3b+2 = 77 \implies b=25$.
6
A sum of $\text{Rs. } 800$ is in the form of $\text{Rs. } 10$ and $\text{Rs. } 20$ notes. If the total number of notes is $50$, find the number of notes of each denomination.
Reveal Answer & Explanation
Answer: Let the number of $\text{Rs. } 10$ notes be $x$.
Since the total number of notes is $50$, the number of $\text{Rs. } 20$ notes $= (50 - x)$.
Step 1: Express total monetary value in terms of $x$:
$$\text{Value of Rs. 10 notes} = 10x$$
$$\text{Value of Rs. 20 notes} = 20(50 - x) = 1,000 - 20x$$
Step 2: Total value is $\text{Rs. } 800$:
$$10x + 1,000 - 20x = 800$$
$$-10x + 1,000 = 800$$
Transpose $1,000$ to RHS:
$$-10x = 800 - 1,000 = -200$$
$$x = \frac{-200}{-10} = \mathbf{20}$$
• Number of $\text{Rs. } 10$ notes $= \mathbf{20}$
• Number of $\text{Rs. } 20$ notes $= 50 - 20 = \mathbf{30}$.
Check: $(20 \times 10) + (30 \times 20) = 200 + 600 = \text{Rs. } 800$.
Let 10-rupee notes be $x$, 20-rupee notes be $50-x$. Equation: $10x + 20(50-x) = 800$.
7
Solve: $0.25(4x - 3) = 0.05(10x - 9)$.
Reveal Answer & Explanation
Answer: Step 1: Multiply both sides by $100$ to eliminate decimals:
$$100 \times 0.25(4x - 3) = 100 \times 0.05(10x - 9)$$
$$25(4x - 3) = 5(10x - 9)$$
Divide both sides by $5$:
$$5(4x - 3) = 1(10x - 9)$$
Step 2: Expand brackets:
$$20x - 15 = 10x - 9$$
Step 3: Transpose $10x$ to LHS and $-15$ to RHS:
$$20x - 10x = -9 + 15$$
$$10x = 6 \implies x = \frac{6}{10} = \mathbf{\frac{3}{5} = 0.6}$$.
Multiply the entire equation by 100 to clear decimals, giving $25(4x - 3) = 5(10x - 9)$.
8
Explain the difference between an "Algebraic Expression" and an "Algebraic Equation".
Reveal Answer & Explanation
Answer:

• Algebraic Expression: A combination of constants, variables, and mathematical operators ($+, -, \times, \div$) without an equality sign (e.g., $3x + 5$). It does not state a relation of equality and cannot be "solved" for a unique value of $x$; it can only be simplified or evaluated for given values of $x$.
• Algebraic Equation: A complete mathematical statement stating that two expressions are equal, containing an equal sign "$=$" (e.g., $3x + 5 = 20$). It restricts the variable to specific root(s) that satisfy the equality.


An expression has no equal sign and cannot be solved ($3x+5$); an equation has an equal sign ($3x+5=20$) and has a unique root.
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