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ICSE • Class 8 • Mathematics • Ch 15
Estimated Time: 45 Mins
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Linear Equations

In ICSE Class 8 Mathematics, "Linear Equations" provides an authoritative, algebraically rigorous master study guide investigating the solutions, properties, and word problems of linear equations in one variable. This comprehensive chapter explores What is a Linear Equation? (An algebraic equation where the highest exponent of the unknown variable is strictly $1$: $ax + b = c$ with $a \ne 0$; Concept of Root / Solution; Checking solutions by LHS = RHS verification), Solving Equations with Linear Expressions on One Side and Numbers on the Other, Solving Equations with Variables on Both Sides ($ax + b = cx + d$), Solving Complex Fractional Equations (Clearing brackets using the distributive law; Clearing fractions using the Least Common Multiple [LCM] of denominators; Cross-multiplication method for rational expressions: $\frac{ax + b}{cx + d} = \frac{p}{q} \implies q(ax + b) = p(cx + d)$), and Comprehensive Word Problems on Linear Equations: 1. Number Problems (Sum/difference of two numbers, two-digit numbers involving reversed digits: $10t + u$), 2. Age Problems (Present age $x$, age $n$ years ago $[x - n]$, age $n$ years hence $[x + n]$), 3. Perimeter and Area Geometric Problems (Rectangles, isosceles triangles), 4. Currency Denomination and Coin Problems, 5. Speed, Distance, and Time Problems (Boats in upstream and downstream currents; Trains crossing poles and bridges) aligned with the 2026–27 CISCE ICSE curriculum.

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How Did an Ancient Egyptian Scribe 3,700 Years Ago Solve Linear Equations in the Sand Using the Method of False Position?

In 1650 BCE, a royal Egyptian scribe named Ahmes unrolled a $17\text{-foot}$ papyrus scroll along the banks of the Nile. Today known as the Rhind Mathematical Papyrus, it contained Problem 24: "A quantity and its seventh part added together make 19. What is the quantity?" Ahmes had no modern algebraic notation, no letters $x$ or $y$, and no equal signs ($=$ was invented in 1557 by Robert Recorde because "no two things can be more equal than parallel lines"). Yet Ahmes solved it using the brilliant Method of False Position! In modern algebra, that ancient Egyptian problem is simply: $x + \frac{x}{7} = 19 \implies \frac{8x}{7} = 19 \implies x = \frac{133}{8} = 16.625$! Linear equations are the fundamental balance scales of the universe: whatever you do to the left side, you MUST do to the right side! How do you crack two-digit number problems where reversing digits changes the value? How do upstream and downstream boat currents translate into simple equations? Let's master linear equations.

Why This Chapter Matters

Linear equations are the gateway to all applied mathematics: engineering load balances, financial loan amortizations, chemical stoichiometry, circuit Kirchhoff loop laws, and computer simulation physics. Mastering cross-multiplication and word problem modeling is essential for scoring 100% in ICSE Class 8 and 10 mathematics.

Before You Begin (Prerequisites)

  • Algebraic expressions, coefficients, and like terms from Chapter 12.
  • Solving simple one-step equations from Class 7.
  • Operations on fractions and negative integers.

What You Will Learn (Core Objectives)

  • Define a linear equation in one variable and verify candidate roots.
  • Solve equations with variables on both sides using systematic transposition.
  • Solve fractional linear equations using the Cross-Multiplication Method.
  • Formulate and solve two-digit number problems involving reversed digits.
  • Model and solve age-related word problems using past and future time frames.
  • Solve commercial currency denomination and upstream/downstream speed word problems.

Chapter Roadmap & Progression

1 1. Concept & The Balance Scale Axio...
2 2. Equations with Variables on Both...
3 3. Two-Digit Number Word Problems
4 4. Age, Geometry & Speed Word Probl...

Complete Concept Guide (100% Curriculum Coverage)

1. Concept & The Balance Scale Axioms

Understand
A. What is a Linear Equation?

A Linear Equation in One Variable is an algebraic equation in which the highest power of the variable is strictly one ($1$):

$$\mathbf{ax + b = c \quad (a \ne 0, \; a, b, c \in \mathbb{R})}$$
  • The value of the variable that makes $\text{LHS} = \text{RHS}$ is called the Root (or Solution) of the equation. A linear equation has exactly one unique root.
B. The Golden Balance Scale Axioms:
  1. The same number can be added to both sides without changing equality.
  2. The same number can be subtracted from both sides.
  3. Both sides can be multiplied by the same non-zero number.
  4. Both sides can be divided by the same non-zero number.

2. Equations with Variables on Both Sides & Cross-Multiplication

Techniques
A. Variables on Both Sides ($ax + b = cx + d$):

Transpose variable terms to LHS and numerical constant terms to RHS by changing signs:

$$ax - cx = d - b \implies x(a - c) = d - b \implies \mathbf{x = \frac{d - b}{a - c}}$$
B. The Cross-Multiplication Method:

For rational equations in the standard form:

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

Example: Solve $\frac{x + 1}{2x + 3} = \frac{3}{8}$:
Cross-multiply: $8(x + 1) = 3(2x + 3) \implies 8x + 8 = 6x + 9 \implies 2x = 1 \implies \mathbf{x = \frac{1}{2}}$.

3. Two-Digit Number Word Problems

Number Problems
A. General Representation:
  • Let the ten's digit be $t$ and unit's digit be $u$.
  • $$\mathbf{\text{Original Number} = 10t + u}$$
  • When digits are reversed (interchanged): $$\mathbf{\text{Reversed Number} = 10u + t}$$

Setup: If sum of digits is 9, let unit digit be $x$, ten digit be $9 - x$. Original number $= 10(9 - x) + x = 90 - 9x$.

4. Age, Geometry & Speed Word Problems

Applied Problems
A. Age Problems:

Let present age be $x$ years:

  • Age $n$ years ago: $x - n$ years.
  • Age $n$ years hence (in future): $x + n$ years.
B. Upstream vs Downstream Boat Problems:

Let speed of boat in still water be $x\text{ km/h}$ and speed of stream be $y\text{ km/h}$:

  • $$\mathbf{\text{Downstream Speed (with current)} = (x + y)\text{ km/h}}$$
  • $$\mathbf{\text{Upstream Speed (against current)} = (x - y)\text{ km/h}}$$
  • $$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$$

Key Formulas, Identities & Theorems

Cross-Multiplication Identity
$$\frac{ax + b}{cx + d} = \frac{p}{q} \iff q(ax + b) = p(cx + d)$$
Converts rational fractional equation into linear form.
Two-Digit Number Form
$$\text{Number} = 10 \times \text{Ten's Digit} + \text{Unit's Digit}$$
Essential template for digit reversal problems.

Algebra: Cross-Multiplication & Upstream/Downstream Speeds

Linear Equations: Cross-Multiplication & Word Problem Modeling CROSS-MULTIPLICATION METHOD ax + b cx + d = p q q (ax + b) = p (cx + d) • Two-Digit Number: Number = 10t + u • Reversing Digits: Reversed = 10u + t • Transpose: Variables to LHS, Constants to RHS BOAT SPEEDS: UPSTREAM & DOWNSTREAM • Downstream Speed (Aided by stream): Speed = (x + y) km/h Faster speed • Takes LESS time • Upstream Speed (Opposed by stream): Speed = (x - y) km/h Slower speed • Takes MORE time • Age Problems: n years ago = (x - n) • n years hence = (x + n) • Time = Distance / Speed BALANCE SCALE PRINCIPLE • CROSS-MULTIPLICATION: q(ax+b)=p(cx+d) • NUMBER = 10t + u

Chapter Summary & 10 Key Takeaways

Takeaway 1
A linear equation in one variable has the highest variable power of 1 (ax + b = c).
Takeaway 2
Solving an equation preserves the balance scale: perform identical operations on both sides.
Takeaway 3
Transpose variable terms to the left side and constant numbers to the right side, changing signs.
Takeaway 4
Cross-multiplication: (ax + b) / (cx + d) = p / q becomes q(ax + b) = p(cx + d).
Takeaway 5
In two-digit number problems, express the number as 10t + u and its reverse as 10u + t.
Takeaway 6
In age problems, age n years ago is (x - n) and age n years in the future is (x + n).
Takeaway 7
Downstream speed is (x + y) km/h; upstream speed is (x - y) km/h, where y is stream speed.
Takeaway 8
Always check the calculated root by substituting back into the original LHS and RHS.
Takeaway 9
In coin/currency problems, Total Value = Number of Coins * Individual Denomination Value.
Takeaway 10
A linear equation in one variable always has exactly one unique solution root.

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: $\frac{x - 5}{3} = \frac{x - 3}{5}$. Check your result.
Reveal Answer & Explanation
Answer:

Step 1: Cross-multiply the fractions:

$$5(x - 5) = 3(x - 3)$$


Step 2: Expand the brackets:

$$5x - 25 = 3x - 9$$


Step 3: Transpose $3x$ to LHS and $-25$ to RHS:

$$5x - 3x = -9 + 25$$


$$2x = 16$$


$$x = \frac{16}{2} = \mathbf{8}$$



• Check / Verification:
LHS $= \frac{8 - 5}{3} = \frac{3}{3} = 1$
RHS $= \frac{8 - 3}{5} = \frac{5}{5} = 1$
Since $\text{LHS} = \text{RHS} = 1$, the solution $x = 8$ is verified!


Cross-multiply: $5(x - 5) = 3(x - 3) \implies 5x - 25 = 3x - 9 \implies 2x = 16 \implies x = 8$.
2
Solve for $x$: $\frac{6x + 1}{3} + 1 = \frac{x - 3}{6}$.
Reveal Answer & Explanation
Answer: Step 1: Multiply all terms by the LCM of denominators ($3$ and $6$), which is $6$:
$$6 \left( \frac{6x + 1}{3} \right) + 6(1) = 6 \left( \frac{x - 3}{6} \right)$$
$$2(6x + 1) + 6 = 1(x - 3)$$
Step 2: Expand brackets:
$$12x + 2 + 6 = x - 3$$
$$12x + 8 = x - 3$$
Step 3: Transpose variables to LHS and numbers to RHS:
$$12x - x = -3 - 8$$
$$11x = -11$$
$$x = \frac{-11}{11} = \mathbf{-1}$$.
Multiply through by 6: $2(6x + 1) + 6 = x - 3 \implies 12x + 8 = x - 3 \implies 11x = -11 \implies x = -1$.
3
The difference between two positive integers is $66$. The ratio of the two integers is $2 : 5$. What are the integers?
Reveal Answer & Explanation
Answer: Step 1: Let the two integers be $2x$ and $5x$ based on the given ratio $2 : 5$.
Step 2: According to the problem, their difference is $66$:
$$5x - 2x = 66$$
$$3x = 66$$
$$x = \frac{66}{3} = 22$$
Step 3: Calculate the two integers:
• Smaller integer $= 2x = 2(22) = \mathbf{44}$
• Larger integer $= 5x = 5(22) = \mathbf{110}$
*(Check: $110 - 44 = 66$. Correct!)*.
$5x - 2x = 66 \implies 3x = 66 \implies x = 22$. The numbers are $2(22) = 44$ and $5(22) = 110$.
4
The sum of the digits of a two-digit number is $9$. When we interchange the digits, the resulting new number is greater than the original number by $27$. Find the original two-digit number.
Reveal Answer & Explanation
Answer:

Step 1: Let the unit's digit be $x$.
Since the sum of digits is $9$, the ten's digit is $(9 - x)$.

$$\text{Original Number} = 10(9 - x) + x = 90 - 10x + x = \mathbf{90 - 9x}$$


Step 2: When digits are interchanged, the unit's digit becomes $(9 - x)$ and the ten's digit becomes $x$:

$$\text{New Number} = 10(x) + (9 - x) = 10x + 9 - x = \mathbf{9x + 9}$$


Step 3: According to the question, $\text{New Number} = \text{Original Number} + 27$:

$$9x + 9 = (90 - 9x) + 27$$


$$9x + 9 = 117 - 9x$$


$$9x + 9x = 117 - 9$$


$$18x = 108$$


$$x = \frac{108}{18} = 6$$


Step 4: Calculate digits and original number:
• Unit's digit $x = 6$
• Ten's digit $= 9 - 6 = 3$
• Original Number:

$$\mathbf{36}$$


(Check: Reversed number is $63$. Difference: $63 - 36 = 27$. Sum: $3 + 6 = 9$. Perfect!).


Original $= 90 - 9x$, New $= 9x + 9$. Equation: $9x + 9 = 90 - 9x + 27 \implies 18x = 108 \implies x = 6$. Number is 36.
5
A father is $30\text{ years}$ older than his son. In $12\text{ years}$, the father's age will be twice the age of his son. Find their present ages.
Reveal Answer & Explanation
Answer:

Step 1: Let the present age of the son be $x\text{ years}$.
Then the present age of the father is $(x + 30)\text{ years}$.
Step 2: Determine their ages after $12\text{ years}$:
• Son's age in 12 years $= x + 12$
• Father's age in 12 years $= (x + 30) + 12 = x + 42$
Step 3: According to the question, in 12 years: $\text{Father's Age} = 2 \times \text{Son's Age}$:

$$x + 42 = 2(x + 12)$$


$$x + 42 = 2x + 24$$


$$42 - 24 = 2x - x$$


$$x = \mathbf{18}$$


• Son's Present Age: $18\text{ years}$.
• Father's Present Age: $18 + 30 = \mathbf{48\text{ years}}$
(Check: In 12 years, son is 30 and father is 60; $60 = 2 \times 30$. Correct!).


Let son be $x$, father $x + 30$. In 12 years: $x + 42 = 2(x + 12) \implies x = 18$. Father is 48.
6
Solve: $\frac{7y + 4}{y + 2} = -\frac{4}{3}$.
Reveal Answer & Explanation
Answer: Step 1: Apply Cross-Multiplication:
$$3(7y + 4) = -4(y + 2)$$
Step 2: Expand the brackets:
$$21y + 12 = -4y - 8$$
Step 3: Transpose terms:
$$21y + 4y = -8 - 12$$
$$25y = -20$$
$$y = \frac{-20}{25} = \mathbf{-\frac{4}{5}}$$.
Cross-multiply: $3(7y + 4) = -4(y + 2) \implies 21y + 12 = -4y - 8 \implies 25y = -20 \implies y = -4/5$.
7
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 the breadth of the pool.
Reveal Answer & Explanation
Answer:

Step 1: Let the breadth of the pool be $x\text{ meters}$.
Then its length is $2x + 2\text{ meters}$.
Step 2: Perimeter of rectangle $= 2(\text{length} + \text{breadth}) = 154\text{ m}$:

$$2[(2x + 2) + x] = 154$$


$$2[3x + 2] = 154$$


$$3x + 2 = \frac{154}{2} = 77$$


$$3x = 77 - 2 = 75$$


$$x = \frac{75}{3} = \mathbf{25\text{ meters}}$$


Step 3: Calculate dimensions:
• Breadth: $x = \mathbf{25\text{ meters}}$
• Length: $2(25) + 2 = 50 + 2 = \mathbf{52\text{ meters}}$.


Let breadth be $x$, length $2x + 2$. Perimeter: $2(3x + 2) = 154 \implies 3x = 75 \implies x = 25\text{ m}$. Length is $52\text{ m}$.
8
A boat goes downstream and covers a distance between two ports in $4\text{ hours}$, while it covers the same distance upstream in $5\text{ hours}$. If the speed of the stream is $2\text{ km/h}$, find the speed of the boat in still water.
Reveal Answer & Explanation
Answer:

Let the speed of the boat in still water be $x\text{ km/h}$.
Given: Speed of stream $= 2\text{ km/h}$.
• Downstream Speed $= (x + 2)\text{ km/h}$
• Upstream Speed $= (x - 2)\text{ km/h}$
Since $\text{Distance} = \text{Speed} \times \text{Time}$ and the distance between the two ports is identical:

$$\text{Distance Downstream} = 4(x + 2)$$


$$\text{Distance Upstream} = 5(x - 2)$$


Equating distances:

$$4(x + 2) = 5(x - 2)$$


$$4x + 8 = 5x - 10$$


$$8 + 10 = 5x - 4x$$


$$x = \mathbf{18\text{ km/h}}$$

.
The speed of the boat in still water is $18\text{ km/h}$.


Distance is equal: $4(x + 2) = 5(x - 2) \implies 4x + 8 = 5x - 10 \implies x = 18\text{ km/h}$.
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