Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables

Pair of Linear Equations in Two Variables is Chapter 3 of NCERT Class 10 Maths. It shows how two linear equations in x and y can be solved together, either by drawing their graphs or by algebra, and how to turn everyday word problems into a pair of equations. Watch the ten video lessons below, then use these notes, conditions and solved examples to revise before your tests.

What you will learn

• The general form of a pair of linear equations in two variables
• What the graph of each equation looks like and what the point of intersection means
• The consistency conditions that tell you, without solving, whether there is one solution, none, or infinitely many
• Solving by the substitution method and the elimination method

Video Lessons

Pair of Linear Equations in Two Variables - Part 1

Pair of Linear Equations in Two Variables - Part 2

Pair of Linear Equations in Two Variables - Part 3

Pair of Linear Equations in Two Variables - Part 4

Pair of Linear Equations in Two Variables - Part 5

Pair of Linear Equations in Two Variables - Part 6

Pair of Linear Equations in Two Variables - Part 7

Pair of Linear Equations in Two Variables - Part 8

Pair of Linear Equations in Two Variables - Part 9

Pair of Linear Equations in Two Variables - Part 10

Key Concepts and Formulas

General Form of the Pair

A linear equation in two variables, ax + by + c = 0 (a and b not both zero), gives a straight line when plotted, and every point on that line is a solution.

a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0

• A solution of the pair is a value of (x, y) that satisfies both equations together

Graphical Method

Plot at least two points for each equation, draw both lines on the same axes and see how they behave.

• Intersecting lines: exactly one common point, so a unique solution
• Coincident lines: the lines lie on top of each other, so infinitely many solutions
• Parallel lines: the lines never meet, so no solution

Consistency Conditions

Comparing the ratios of coefficients predicts the type of solution without drawing a graph, a favourite one-mark question in the CBSE board exam.

a₁/a₂ ≠ b₁/b₂         → intersecting, unique solution
a₁/a₂ = b₁/b₂ = c₁/c₂ → coincident, infinitely many
a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → parallel, no solution

• The first two cases are consistent (the second is also called dependent); the third is inconsistent

Substitution Method

This method works best when one variable has coefficient 1 or −1 in either equation, so it is easy to isolate.

• Step 1: From one equation, express one variable in terms of the other, for example x = 11 − 2y
• Step 2: Substitute this into the other equation to get an equation in one variable only
• Step 3: Solve it, then put the value back to find the second variable

Elimination Method

Here you make the coefficients of one variable equal (or opposite) and then add or subtract the equations to remove that variable.

• Multiply the equations by suitable numbers so that the coefficients of x (or y) become equal
• Subtract if the signs are the same, add if the signs are opposite
• Solve for the remaining variable and substitute back

Forming Equations from Word Problems

Most marks come from word problems: choose two clear variables and write two separate conditions.

Two-digit number (tens x, units y) = 10x + y
Number with digits reversed       = 10y + x

• Let the two unknowns be x and y, and state clearly what each one stands for
• For age problems, "n years ago" means subtract n from each present age, "after n years" means add n

Solved Examples

Example 1: Without solving, find whether 3x + 4y = 10 and 6x + 8y = 25 have a solution.

a₁/a₂ = 3/6 = 1/2
b₁/b₂ = 4/8 = 1/2
c₁/c₂ = 10/25 = 2/5
a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → parallel lines
∴ Inconsistent, no solution

Example 2: Solve x + 2y = 11 and 3x − y = 5 by substitution.

From x + 2y = 11: x = 11 − 2y
3(11 − 2y) − y = 5
33 − 7y = 5
7y = 28, so y = 4
x = 11 − 2 × 4 = 3
∴ x = 3, y = 4

Example 3: Solve 4x + 3y = 25 and 5x − 3y = 11 by elimination.

Adding: (4x + 3y) + (5x − 3y) = 25 + 11
9x = 36, so x = 4
4 × 4 + 3y = 25
3y = 9, so y = 3
∴ x = 4, y = 3

Example 4: Find k so that 2x + ky = 7 and 6x + 15y = 21 have infinitely many solutions.

Need a₁/a₂ = b₁/b₂ = c₁/c₂
2/6 = k/15 = 7/21
2/6 = 1/3 and 7/21 = 1/3
k/15 = 1/3
∴ k = 5

Example 5: 3 pens and 2 notebooks cost ₹110, while 2 pens and 5 notebooks cost ₹165. Find the cost of each.

Let pen = ₹p, notebook = ₹n
3p + 2n = 110 ... (1)
2p + 5n = 165 ... (2)
(1) × 2: 6p + 4n = 220
(2) × 3: 6p + 15n = 495
Subtracting: 11n = 275, so n = 25
3p = 110 − 50 = 60, so p = 20
∴ Pen = ₹20, Notebook = ₹25

Example 6: The sum of a father's and son's ages is 50 years. Five years ago the father was 7 times as old as the son. Find their present ages.

Let father = f years, son = s years
f + s = 50
f − 5 = 7(s − 5), so f = 7s − 30
(7s − 30) + s = 50
8s = 80, so s = 10
f = 50 − 10 = 40
∴ Son = 10 years, Father = 40 years

Example 7: The digits of a two-digit number add up to 9, and reversing the digits increases the number by 27. Find the number.

Let tens digit = x, units digit = y
x + y = 9 ... (1)
(10y + x) − (10x + y) = 27
9y − 9x = 27, so y − x = 3 ... (2)
Adding (1) and (2): 2y = 12, so y = 6
x = 9 − 6 = 3
∴ Number = 36

Common Mistakes to Avoid

• Writing the equations in different forms before comparing ratios; first bring both to the form ax + by + c = 0 (or both to ax + by = c) so the signs of c match.
• Forgetting to multiply every term, including the constant, when scaling an equation in the elimination method.
• Stopping after finding one variable; always find both and verify by substituting into the original equations.
• In word problems, not defining x and y in words; examiners give a step mark for clearly stating the variables.

Quick Revision

Pair: a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0
a₁/a₂ ≠ b₁/b₂         → unique solution
a₁/a₂ = b₁/b₂ = c₁/c₂ → infinitely many
a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → no solution

• Intersecting lines are consistent, coincident lines are dependent, parallel lines are inconsistent
• Algebraic methods: substitution and elimination

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