Pair of Linear Equations in Two Variables Chapter 3 | Mathematics

Complete NCERT notes, graphical method, algebraic methods (substitution, elimination, cross-multiplication), consistency conditions, word problems, and key points for CBSE Class 10 Board Exam 2025-26.

📖 Introduction

An equation of the form ax + by + c = 0 where a, b, c are real numbers and a, b are not both zero, is called a linear equation in two variables (x and y).

A pair of linear equations in two variables is written as:

a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0

where a₁, b₁, c₁, a₂, b₂, c₂ are real numbers, and a₁² + b₁² ≠ 0, a₂² + b₂² ≠ 0.

The solution of a pair of linear equations is the ordered pair (x, y) that satisfies both equations.

📊 Graphical Method of Solution

In the graphical method, we plot both equations on the coordinate plane. The point(s) where the two lines intersect represent the solution(s).

🟢 Case 1: Intersecting Lines (Unique Solution)

If a₁/a₂ ≠ b₁/b₂, the lines intersect at a unique point → Consistent system

Example: x + y = 5 and x - y = 1 intersect at (3, 2)

🟡 Case 2: Coincident Lines (Infinite Solutions)

If a₁/a₂ = b₁/b₂ = c₁/c₂, the lines coincide → Dependent system (infinitely many solutions)

Example: 2x + 4y = 8 and x + 2y = 4 are the same line

🔴 Case 3: Parallel Lines (No Solution)

If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel → Inconsistent system (no solution)

Example: x + y = 5 and x + y = 7 are parallel lines

💡 Graphical Insight: The solution is the intersection point of the two lines.
🔄 Substitution Method

Steps:

📝 Example

Solve: x + y = 14, x - y = 4
From first: x = 14 - y
Substitute: (14 - y) - y = 4 → 14 - 2y = 4 → -2y = -10 → y = 5
Then x = 14 - 5 = 9
∴ Solution: (9, 5)
Elimination Method

Steps:

📝 Example

Solve: 3x + 4y = 10, 2x + 3y = 8
Multiply eq1 by 3: 9x + 12y = 30
Multiply eq2 by 4: 8x + 12y = 32
Subtract: (9x - 8x) + (12y - 12y) = 30 - 32 → x = -2
Substitute: 3(-2) + 4y = 10 → -6 + 4y = 10 → 4y = 16 → y = 4
∴ Solution: (-2, 4)
✖️ Cross-Multiplication Method

For equations: a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

📌 Cross-Multiplication Formula

x / (b₁c₂ - b₂c₁) = y / (c₁a₂ - c₂a₁) = 1 / (a₁b₂ - a₂b₁)

Therefore:

x = (b₁c₂ - b₂c₁) / (a₁b₂ - a₂b₁)
y = (c₁a₂ - c₂a₁) / (a₁b₂ - a₂b₁)

📝 Example

2x + 3y - 8 = 0, 3x + 2y - 7 = 0
a₁=2, b₁=3, c₁=-8; a₂=3, b₂=2, c₂=-7
x = (3×-7 - 2×-8)/(2×2 - 3×3) = (-21 + 16)/(4 - 9) = (-5)/(-5) = 1
y = (-8×3 - -7×2)/(4 - 9) = (-24 + 14)/(-5) = (-10)/(-5) = 2
∴ Solution: (1, 2)
🎯 Consistency Conditions
ConditionNature of LinesNumber of SolutionsConsistency
a₁/a₂ ≠ b₁/b₂IntersectingUnique solutionConsistent
a₁/a₂ = b₁/b₂ = c₁/c₂CoincidentInfinite solutionsDependent (Consistent)
a₁/a₂ = b₁/b₂ ≠ c₁/c₂ParallelNo solutionInconsistent
💡 Remember:
• Unique Solution: a₁/a₂ ≠ b₁/b₂
• No Solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂
• Infinite Solutions: a₁/a₂ = b₁/b₂ = c₁/c₂
📐 All Important Formulas

1️⃣ General Form

a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0

2️⃣ Cross-Multiplication

x = (b₁c₂ - b₂c₁)/(a₁b₂ - a₂b₁)
y = (c₁a₂ - c₂a₁)/(a₁b₂ - a₂b₁)

3️⃣ Unique Solution Condition

a₁/a₂ ≠ b₁/b₂

4️⃣ No Solution Condition

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

5️⃣ Infinite Solutions Condition

a₁/a₂ = b₁/b₂ = c₁/c₂

6️⃣ Slope of Line

m = -a/b (for ax + by + c = 0)
📝 NCERT Solved Examples (In-depth)

Example 1: Solve by Substitution

Solve: 2x + 3y = 11 and 2x - 4y = -24

From eq1: 2x = 11 - 3y → x = (11 - 3y)/2
Substitute in eq2: 2[(11 - 3y)/2] - 4y = -24 → 11 - 3y - 4y = -24 → 11 - 7y = -24 → -7y = -35 → y = 5
Then x = (11 - 15)/2 = -4/2 = -2
∴ Solution: (-2, 5)

Example 2: Solve by Elimination

Solve: 3x + 4y = 25 and 5x - 6y = -9

Multiply eq1 by 3: 9x + 12y = 75
Multiply eq2 by 2: 10x - 12y = -18
Add: 19x = 57 → x = 3
Substitute in eq1: 3(3) + 4y = 25 → 9 + 4y = 25 → 4y = 16 → y = 4
∴ Solution: (3, 4)

Example 3: Word Problem

The sum of two numbers is 35 and their difference is 13. Find the numbers.

Let numbers be x and y (x > y)
x + y = 35 ...(1)
x - y = 13 ...(2)
Adding (1) and (2): 2x = 48 → x = 24
Then y = 35 - 24 = 11
∴ Numbers are 24 and 11.

Example 4: Condition for Consistency

For what value of k does the system have a unique solution?
2x + 3y = 7 and (k + 1)x + (2k - 1)y = 4k + 1

For unique solution: a₁/a₂ ≠ b₁/b₂
2/(k+1) ≠ 3/(2k-1)
2(2k-1) ≠ 3(k+1)
4k - 2 ≠ 3k + 3 → 4k - 3k ≠ 3 + 2 → k ≠ 5
∴ k can be any real number except 5.
Key Points for Board Exam (Deep Research)

📌 Quick Reference Card

MethodWhen to UseKey Formula
SubstitutionOne variable isolated easilySolve for x or y, substitute
EliminationCoefficients are multiplesAdd/subtract to eliminate
Cross-MultiplicationAll methods workx = (b₁c₂ - b₂c₁)/Δ, y = (c₁a₂ - c₂a₁)/Δ

📌 Important Identities for Word Problems

Speed = Distance / Time
Simple Interest = (P × R × T)/100
Age problems: Present age + x = Future age
Number problems: Two-digit number = 10x + y
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