NCERT Solutions Class 10 Maths Chapter 3 – Pair of Linear Equations [2026-27]
Looking for complete NCERT Solutions for Class 10 Maths Chapter 3? You are in the right place. This guide covers every question from Exercise 3.1, Exercise 3.2, and Exercise 3.3 of the chapter "Pair of Linear Equations in Two Variables" with clear, step-by-step solutions and visual diagrams that make each concept easy to understand.
Chapter 3 is one of the most scoring chapters in CBSE Class 10 board exams. It teaches you how to solve two equations with two unknowns using graphical and algebraic methods. Whether you are solving age problems, speed-distance problems, or number puzzles, the techniques in this chapter form the foundation. Master them here, and you will approach your exam with confidence.
📑 Table of Contents
- Chapter Overview & Key Concepts
- Important Formulas & Conditions
- Exercise 3.1 Solutions (3 Questions — Graphical Representation)
- Exercise 3.2 Solutions (7 Questions — Graphical Method)
- Exercise 3.3 Solutions (2 Questions — Algebraic Methods)
- Exam Tips & Common Mistakes
- Frequently Asked Questions
Chapter Overview & Key Concepts
A pair of linear equations in two variables takes the general form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
where a₁, a₂, b₁, b₂, c₁, c₂ are real numbers and a₁² + b₁² ≠ 0 and a₂² + b₂² ≠ 0. Each equation, when plotted on a graph, represents a straight line. The solution to the pair is the point (or points) where these two lines meet.
🔑 Three Possible Cases When Two Lines Are in a Plane
Case 1 — Intersecting Lines: The pair has exactly one unique solution. The equations are called a consistent pair. This happens when a₁/a₂ ≠ b₁/b₂.
Case 2 — Coincident Lines: The pair has infinitely many solutions. The equations are called a dependent (consistent) pair. This happens when a₁/a₂ = b₁/b₂ = c₁/c₂.
Case 3 — Parallel Lines: The pair has no solution. The equations are called an inconsistent pair. This happens when a₁/a₂ = b₁/b₂ ≠ c₁/c₂.
Fig 3.1: The three possible cases for a pair of linear equations in two variables
Methods for Solving a Pair of Linear Equations
The NCERT textbook (2025-26 edition) teaches the following methods in this chapter:
| Method | When to Use | Exercise |
|---|---|---|
| Graphical Method | When you need to visualize solutions or when the question asks for a graph | Ex 3.1, 3.2 |
| Substitution Method | When one variable has coefficient 1 or is easily isolatable | Ex 3.3 |
| Elimination Method | When coefficients can be made equal easily by multiplication | Ex 3.3 |
Important Formulas & Conditions
For the pair of equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, the nature of the solution depends on the ratio of coefficients:
Condition Table
| Condition | Type of Lines | Number of Solutions | Consistency |
|---|---|---|---|
| a₁/a₂ ≠ b₁/b₂ | Intersecting | Exactly one (unique) | Consistent |
| a₁/a₂ = b₁/b₂ = c₁/c₂ | Coincident | Infinitely many | Dependent (Consistent) |
| a₁/a₂ = b₁/b₂ ≠ c₁/c₂ | Parallel | None | Inconsistent |
🎯 Exam Tip: Memorize this condition table. Almost every year, CBSE asks a question where you need to determine consistency or find the value of k for which a pair is consistent, inconsistent, or has infinitely many solutions.
Exercise 3.1 Solutions – Representing Situations as Linear Equations
Exercise 3.1 tests your ability to translate word problems into pairs of linear equations and represent them graphically. There are 3 questions in this exercise. Each question gives a real-life situation that you must express algebraically and then plot on a graph.
1 Exercise 3.1 — Question 1
Aftab tells his daughter, "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be." Represent this situation algebraically and graphically.
✅ Solution
Step 1: Define variables
Let the present age of Aftab's daughter = x years
Let the present age of Aftab = y years
Step 2: Form equations from the given conditions
Condition 1: Seven years ago, Aftab was seven times as old as his daughter.
Seven years ago: Aftab's age = (y − 7), Daughter's age = (x − 7)
y − 7 = 7(x − 7)
y − 7 = 7x − 49
7x − y = 42 ... (i)
y − 7 = 7(x − 7)
y − 7 = 7x − 49
7x − y = 42 ... (i)
Condition 2: Three years from now, Aftab will be three times as old as his daughter.
Three years later: Aftab's age = (y + 3), Daughter's age = (x + 3)
y + 3 = 3(x + 3)
y + 3 = 3x + 9
3x − y = −6 ... (ii)
y + 3 = 3(x + 3)
y + 3 = 3x + 9
3x − y = −6 ... (ii)
Step 3: Create tables of values for graphing
For Equation (i): y = 7x − 42
| x | 5 | 6 | 7 |
|---|---|---|---|
| y | −7 | 0 | 7 |
For Equation (ii): y = 3x + 6
| x | −2 | 0 | 2 |
|---|---|---|---|
| y | 0 | 6 | 12 |
Fig: Graphical representation of Aftab's age problem — the two lines intersect at the point (12, 42)
Answer: The algebraic representation is: 7x − y = 42 and 3x − y = −6. The two lines intersect at the point (12, 42), meaning the daughter's present age is 12 years and Aftab's present age is 42 years.
2 Exercise 3.1 — Question 2
The coach of a cricket team buys 3 bats and 6 balls for ₹3900. Later, she buys another bat and 3 more balls of the same kind for ₹1300. Represent this situation algebraically and graphically.
✅ Solution
Step 1: Define variables
Let the cost of one bat = ₹x
Let the cost of one ball = ₹y
Step 2: Form equations
Condition 1: 3x + 6y = 3900 → x + 2y = 1300 ... (i) (dividing by 3)
Condition 2: x + 3y = 1300 ... (ii)
Step 3: Tables for graphing
For Equation (i): x = 1300 − 2y
| y | 0 | 100 | 200 |
|---|---|---|---|
| x | 1300 | 1100 | 900 |
For Equation (ii): x = 1300 − 3y
| y | 0 | 100 | 200 |
|---|---|---|---|
| x | 1300 | 1000 | 700 |
Answer: The algebraic representation is: x + 2y = 1300 and x + 3y = 1300. From the graph (or solving algebraically), the two lines intersect at the point (1300, 0), meaning each bat costs ₹1300 and each ball costs ₹0. However, note that both equations share the same x-intercept at (1300, 0), giving us the solution y = 0 and x = 1300.
⚠️ Note: Don't forget to simplify Equation (i) by dividing 3x + 6y = 3900 by 3 to get x + 2y = 1300. Working with simplified equations makes graphing and solving much easier.
3 Exercise 3.1 — Question 3
The cost of 2 kg of apples and 1 kg of grapes on a day was found to be ₹160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹300. Represent the situation algebraically and graphically.
✅ Solution
Step 1: Define variables
Let the cost of 1 kg of apples = ₹x
Let the cost of 1 kg of grapes = ₹y
Step 2: Form equations
Condition 1: 2x + y = 160 ... (i)
Condition 2: 4x + 2y = 300 → 2x + y = 150 ... (ii)
Step 3: Analyze the pair
Comparing: a₁/a₂ = 2/2 = 1, b₁/b₂ = 1/1 = 1, c₁/c₂ = 160/150 = 16/15
Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → The lines are parallel.
Step 4: Tables for graphing
For Equation (i): y = 160 − 2x
| x | 0 | 40 | 80 |
|---|---|---|---|
| y | 160 | 80 | 0 |
For Equation (ii): y = 150 − 2x
| x | 0 | 40 | 75 |
|---|---|---|---|
| y | 150 | 70 | 0 |
Fig: The two lines are parallel (same slope, different intercepts), indicating no solution exists
Answer: The algebraic representation is: 2x + y = 160 and 2x + y = 150. The graph shows two parallel lines that never intersect. This means the situation is inconsistent — the given data is contradictory and no solution exists.
🎯 Exam Tip: This question is a classic example of an inconsistent pair. In the board exam, when you see two equations that simplify to the same coefficients but different constants (like 2x + y = 160 and 2x + y = 150), immediately identify it as parallel lines with no solution.
Exercise 3.2 Solutions – Graphical Method of Solving
Exercise 3.2 focuses on using the graphical method to solve pairs of linear equations. You will plot equations on a graph, find their intersection, and determine whether the pair is consistent, inconsistent, or dependent. There are 7 questions in this exercise.
1 Exercise 3.2 — Question 1
Form the pair of linear equations in the following problems, and find their solutions graphically:
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and one pen.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and one pen.
✅ Solution — Part (i)
Step 1: Define variables and form equations
Let the number of boys = x and the number of girls = y
Total students: x + y = 10 ... (i)
Girls are 4 more than boys: y − x = 4 → y = x + 4 ... (ii)
Girls are 4 more than boys: y − x = 4 → y = x + 4 ... (ii)
Step 2: Create tables
For Equation (i): y = 10 − x
| x | 0 | 4 | 8 | 10 |
|---|---|---|---|---|
| y | 10 | 6 | 2 | 0 |
For Equation (ii): y = x + 4
| x | 0 | 2 | 4 | 6 |
|---|---|---|---|---|
| y | 4 | 6 | 8 | 10 |
Step 3: Find intersection point
From the tables, both equations give y = 7 when x = 3. The lines intersect at (3, 7).
Answer (i): Number of boys = 3, Number of girls = 7.
✅ Solution — Part (ii)
Step 1: Define variables and form equations
Let cost of one pencil = ₹x and cost of one pen = ₹y
5x + 7y = 50 ... (i)
7x + 5y = 46 ... (ii)
7x + 5y = 46 ... (ii)
Step 2: Create tables
For Equation (i): y = (50 − 5x) / 7
| x | 3 | 10 | −4 |
|---|---|---|---|
| y | 5 | 0 | 10 |
For Equation (ii): y = (46 − 7x) / 5
| x | 3 | 8 | −2 |
|---|---|---|---|
| y | 5 | −2 | 12 |
Step 3: Find intersection
Both equations give y = 5 when x = 3. The lines intersect at (3, 5).
Answer (ii): Cost of one pencil = ₹3, Cost of one pen = ₹5.
2 Exercise 3.2 — Question 2
On comparing the ratios a₁/a₂, b₁/b₂ and c₁/c₂, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel, or are coincident:
(i) 5x − 4y + 8 = 0 and 7x + 6y − 9 = 0
(ii) 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0
(iii) 6x − 3y + 10 = 0 and 2x − y + 9 = 0
(i) 5x − 4y + 8 = 0 and 7x + 6y − 9 = 0
(ii) 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0
(iii) 6x − 3y + 10 = 0 and 2x − y + 9 = 0
✅ Solution
Part (i): 5x − 4y + 8 = 0 and 7x + 6y − 9 = 0
a₁/a₂ = 5/7, b₁/b₂ = −4/6 = −2/3
Since 5/7 ≠ −2/3 → a₁/a₂ ≠ b₁/b₂
Answer: The lines intersect at a point (unique solution, consistent pair).
Part (ii): 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0
a₁/a₂ = 9/18 = 1/2, b₁/b₂ = 3/6 = 1/2, c₁/c₂ = 12/24 = 1/2
Since a₁/a₂ = b₁/b₂ = c₁/c₂ = 1/2
Answer: The lines are coincident (infinitely many solutions, dependent pair).
Part (iii): 6x − 3y + 10 = 0 and 2x − y + 9 = 0
a₁/a₂ = 6/2 = 3, b₁/b₂ = −3/−1 = 3, c₁/c₂ = 10/9
Since a₁/a₂ = b₁/b₂ = 3 but c₁/c₂ = 10/9 ≠ 3
Answer: The lines are parallel (no solution, inconsistent pair).
3 Exercise 3.2 — Question 3
On comparing the ratios a₁/a₂, b₁/b₂ and c₁/c₂, find out whether the following pair of linear equations are consistent, or inconsistent:
(i) 3x + 2y = 5 ; 2x − 3y = 7
(ii) 2x − 3y = 8 ; 4x − 6y = 9
(iii) 3/2 x + 5/3 y = 7 ; 9x − 10y = 14
(iv) 5x − 3y = 11 ; −10x + 6y = −22
(v) 4/3 x + 2y = 8 ; 2x + 3y = 12
(i) 3x + 2y = 5 ; 2x − 3y = 7
(ii) 2x − 3y = 8 ; 4x − 6y = 9
(iii) 3/2 x + 5/3 y = 7 ; 9x − 10y = 14
(iv) 5x − 3y = 11 ; −10x + 6y = −22
(v) 4/3 x + 2y = 8 ; 2x + 3y = 12
✅ Solution
Part (i): 3x + 2y = 5 and 2x − 3y = 7
a₁/a₂ = 3/2, b₁/b₂ = 2/(−3) = −2/3
Since 3/2 ≠ −2/3 → Lines intersect
Since 3/2 ≠ −2/3 → Lines intersect
Answer: Consistent (unique solution)
Part (ii): 2x − 3y = 8 and 4x − 6y = 9
Writing as: 2x − 3y − 8 = 0 and 4x − 6y − 9 = 0
a₁/a₂ = 2/4 = 1/2, b₁/b₂ = −3/−6 = 1/2, c₁/c₂ = −8/−9 = 8/9
Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → Parallel lines
a₁/a₂ = 2/4 = 1/2, b₁/b₂ = −3/−6 = 1/2, c₁/c₂ = −8/−9 = 8/9
Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → Parallel lines
Answer: Inconsistent (no solution)
Part (iii): 3/2 x + 5/3 y = 7 and 9x − 10y = 14
Multiply first equation by 6: 9x + 10y = 42
Now compare: 9x + 10y = 42 and 9x − 10y = 14
a₁/a₂ = 9/9 = 1, b₁/b₂ = 10/(−10) = −1
Since 1 ≠ −1 → Lines intersect
Now compare: 9x + 10y = 42 and 9x − 10y = 14
a₁/a₂ = 9/9 = 1, b₁/b₂ = 10/(−10) = −1
Since 1 ≠ −1 → Lines intersect
Answer: Consistent (unique solution)
Part (iv): 5x − 3y = 11 and −10x + 6y = −22
Writing as: 5x − 3y − 11 = 0 and −10x + 6y + 22 = 0
a₁/a₂ = 5/(−10) = −1/2, b₁/b₂ = −3/6 = −1/2, c₁/c₂ = −11/22 = −1/2
Since a₁/a₂ = b₁/b₂ = c₁/c₂ = −1/2 → Coincident lines
a₁/a₂ = 5/(−10) = −1/2, b₁/b₂ = −3/6 = −1/2, c₁/c₂ = −11/22 = −1/2
Since a₁/a₂ = b₁/b₂ = c₁/c₂ = −1/2 → Coincident lines
Answer: Consistent (dependent — infinitely many solutions)
Part (v): 4/3 x + 2y = 8 and 2x + 3y = 12
Multiply first equation by 3: 4x + 6y = 24 → 2x + 3y = 12
Now compare: 2x + 3y = 12 and 2x + 3y = 12
These are the same equation!
a₁/a₂ = b₁/b₂ = c₁/c₂ → Coincident lines
Now compare: 2x + 3y = 12 and 2x + 3y = 12
These are the same equation!
a₁/a₂ = b₁/b₂ = c₁/c₂ → Coincident lines
Answer: Consistent (dependent — infinitely many solutions)
4 Exercise 3.2 — Question 4
Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) x + y = 5, 2x + 2y = 10
(ii) x − y = 8, 3x − 3y = 16
(iii) 2x + y − 6 = 0, 4x − 2y − 4 = 0
(iv) 2x − 2y − 2 = 0, 4x − 4y − 5 = 0
(i) x + y = 5, 2x + 2y = 10
(ii) x − y = 8, 3x − 3y = 16
(iii) 2x + y − 6 = 0, 4x − 2y − 4 = 0
(iv) 2x − 2y − 2 = 0, 4x − 4y − 5 = 0
✅ Solution
Part (i): x + y = 5 and 2x + 2y = 10
a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 5/10 = 1/2
a₁/a₂ = b₁/b₂ = c₁/c₂ → Coincident lines
a₁/a₂ = b₁/b₂ = c₁/c₂ → Coincident lines
Answer: Consistent (dependent). The two equations are the same line. Every point on x + y = 5 is a solution. Infinitely many solutions.
Part (ii): x − y = 8 and 3x − 3y = 16
Writing as: x − y − 8 = 0 and 3x − 3y − 16 = 0
a₁/a₂ = 1/3, b₁/b₂ = −1/−3 = 1/3, c₁/c₂ = −8/−16 = 1/2
a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → Parallel lines
a₁/a₂ = 1/3, b₁/b₂ = −1/−3 = 1/3, c₁/c₂ = −8/−16 = 1/2
a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → Parallel lines
Answer: Inconsistent (no solution).
Part (iii): 2x + y − 6 = 0 and 4x − 2y − 4 = 0
a₁/a₂ = 2/4 = 1/2, b₁/b₂ = 1/(−2) = −1/2
Since 1/2 ≠ −1/2 → Lines intersect
Since 1/2 ≠ −1/2 → Lines intersect
Graphical solution:
For 2x + y = 6 → y = 6 − 2x
| x | 0 | 1 | 3 |
|---|---|---|---|
| y | 6 | 4 | 0 |
For 4x − 2y = 4 → y = 2x − 2
| x | 0 | 1 | 3 |
|---|---|---|---|
| y | −2 | 0 | 4 |
Intersection point: When x = 2, from Eq(i): y = 6 − 4 = 2. Check in Eq(ii): 4(2) − 2(2) = 4 ✓
Answer: Consistent. Solution: x = 2, y = 2. The lines intersect at (2, 2).
Part (iv): 2x − 2y − 2 = 0 and 4x − 4y − 5 = 0
a₁/a₂ = 2/4 = 1/2, b₁/b₂ = −2/−4 = 1/2, c₁/c₂ = −2/−5 = 2/5
a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → Parallel lines
a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → Parallel lines
Answer: Inconsistent (no solution).
5 Exercise 3.2 — Question 5
Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
✅ Solution
Step 1: Define variables
Let width = x metres and length = y metres
Step 2: Form equations
Length is 4 m more than width: y = x + 4 → y − x = 4 ... (i)
Half the perimeter = 36: x + y = 36 ... (ii)
Step 3: Tables for graphing
For Equation (i): y = x + 4
| x | 0 | 8 | 16 |
|---|---|---|---|
| y | 4 | 12 | 20 |
For Equation (ii): y = 36 − x
| x | 0 | 16 | 36 |
|---|---|---|---|
| y | 36 | 20 | 0 |
The lines intersect at (16, 20).
Answer: Width = 16 m, Length = 20 m.
6 Exercise 3.2 — Question 6
Given the linear equation 2x + 3y − 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines (ii) parallel lines (iii) coincident lines
(i) intersecting lines (ii) parallel lines (iii) coincident lines
✅ Solution
Given equation: 2x + 3y − 8 = 0 (here a₁ = 2, b₁ = 3, c₁ = −8)
Part (i): Intersecting lines
We need a₁/a₂ ≠ b₁/b₂. Choose any equation where the ratio of x-coefficient to y-coefficient is different from 2/3.
Example: 3x + 2y − 7 = 0
Check: a₁/a₂ = 2/3, b₁/b₂ = 3/2 → 2/3 ≠ 3/2 ✓
Part (ii): Parallel lines
We need a₁/a₂ = b₁/b₂ ≠ c₁/c₂. Keep the same coefficient ratio but change the constant.
Example: 4x + 6y − 10 = 0
Check: a₁/a₂ = 2/4 = 1/2, b₁/b₂ = 3/6 = 1/2, c₁/c₂ = −8/−10 = 4/5 → 1/2 = 1/2 ≠ 4/5 ✓
Part (iii): Coincident lines
We need a₁/a₂ = b₁/b₂ = c₁/c₂. Multiply the entire equation by any constant.
Example: 4x + 6y − 16 = 0
Check: a₁/a₂ = 2/4 = 1/2, b₁/b₂ = 3/6 = 1/2, c₁/c₂ = −8/−16 = 1/2 → All equal ✓
🎯 Exam Tip: There are infinitely many correct answers for this type of question. The key is understanding the conditions. For intersecting lines, just make sure the coefficient ratios differ. For parallel, keep ratios same but constant different. For coincident, multiply everything by the same number.
7 Exercise 3.2 — Question 7
Draw the graphs of the equations x − y + 1 = 0 and 3x + 2y − 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
✅ Solution
Step 1: Create tables
For x − y + 1 = 0 → y = x + 1
| x | −1 | 0 | 2 | 4 |
|---|---|---|---|---|
| y | 0 | 1 | 3 | 5 |
For 3x + 2y − 12 = 0 → y = (12 − 3x)/2
| x | 0 | 2 | 4 |
|---|---|---|---|
| y | 6 | 3 | 0 |
Step 2: Find intersection of the two lines
From both tables, when x = 2, y = 3. Check: 2 − 3 + 1 = 0 ✓ and 3(2) + 2(3) − 12 = 0 ✓
Intersection point: (2, 3)
Intersection point: (2, 3)
Step 3: Find where each line crosses the x-axis (y = 0)
Line 1 (y = x + 1): When y = 0, x = −1 → Point (−1, 0)
Line 2 (y = (12 − 3x)/2): When y = 0, x = 4 → Point (4, 0)
Line 2 (y = (12 − 3x)/2): When y = 0, x = 4 → Point (4, 0)
Fig: Triangle formed by the two lines and the x-axis, with vertices at (−1, 0), (4, 0), and (2, 3)
Answer: The vertices of the triangle are (−1, 0), (4, 0), and (2, 3).
Exercise 3.3 Solutions – Algebraic Methods (Substitution & Elimination)
Exercise 3.3 teaches the two most powerful algebraic methods for solving linear equations: the Substitution Method and the Elimination Method. These are far more reliable than graphical methods when answers involve decimals or fractions. There are 2 questions, but they contain multiple parts covering different real-world applications.
📝 Substitution Method (Steps)
1. From one equation, express one variable in terms of the other (e.g., y = ...)
2. Substitute this expression into the second equation
3. Solve the resulting single-variable equation
4. Substitute back to find the other variable
📝 Elimination Method (Steps)
1. Multiply one or both equations so that the coefficients of one variable become equal (or opposite)
2. Add or subtract the equations to eliminate that variable
3. Solve for the remaining variable
4. Substitute back to find the other variable
1 Exercise 3.3 — Question 1
Solve the following pair of linear equations by the substitution and elimination methods:
(i) x + y = 14 ; x − y = 4
(ii) 3x + 4y = 10 ; 2x − 2y = 2 (i.e., s − t = 3 ; s/3 + t/2 = 6)
(iii) 3x − 5y − 4 = 0 ; 9x = 2y + 7
(iv) x/2 + 2y/3 = −1 ; x − y/3 = 3
(i) x + y = 14 ; x − y = 4
(ii) 3x + 4y = 10 ; 2x − 2y = 2 (i.e., s − t = 3 ; s/3 + t/2 = 6)
(iii) 3x − 5y − 4 = 0 ; 9x = 2y + 7
(iv) x/2 + 2y/3 = −1 ; x − y/3 = 3
✅ Solution — Part (i): x + y = 14 and x − y = 4
By Substitution Method:
Step 1: From Equation (i): x = 14 − y
Step 2: Substitute in Equation (ii):
(14 − y) − y = 4
14 − 2y = 4
−2y = 4 − 14 = −10
y = 5
14 − 2y = 4
−2y = 4 − 14 = −10
y = 5
Step 3: Substitute y = 5 in x = 14 − y:
x = 14 − 5 = 9
By Elimination Method:
Step 1: Add both equations (coefficients of y are already opposite):
x + y = 14
+ x − y = 4
─────────────
2x = 18
x = 9
+ x − y = 4
─────────────
2x = 18
x = 9
Step 2: Substitute x = 9 in Equation (i): 9 + y = 14 → y = 5
Answer: x = 9, y = 5
✅ Solution — Part (ii): s − t = 3 and s/3 + t/2 = 6
By Substitution Method:
Step 1: From Equation (i): s = t + 3
Step 2: Substitute in Equation (ii):
(t + 3)/3 + t/2 = 6
Multiply by 6 (LCM of 3 and 2):
2(t + 3) + 3t = 36
2t + 6 + 3t = 36
5t = 30
t = 6
Multiply by 6 (LCM of 3 and 2):
2(t + 3) + 3t = 36
2t + 6 + 3t = 36
5t = 30
t = 6
Step 3: s = t + 3 = 6 + 3 = 9
By Elimination Method:
Step 1: First, clear fractions in Equation (ii). Multiply by 6:
Eq (i): s − t = 3
Eq (ii): 2s + 3t = 36
Eq (ii): 2s + 3t = 36
Step 2: Multiply Eq (i) by 3:
3s − 3t = 9 ... (iii)
2s + 3t = 36 ... (ii)
2s + 3t = 36 ... (ii)
Step 3: Add (iii) and (ii):
5s = 45 → s = 9
From Eq(i): 9 − t = 3 → t = 6
From Eq(i): 9 − t = 3 → t = 6
Answer: s = 9, t = 6
✅ Solution — Part (iii): 3x − 5y − 4 = 0 and 9x = 2y + 7
By Substitution Method:
Step 1: Rewrite:
3x − 5y = 4 ... (i)
9x − 2y = 7 ... (ii)
9x − 2y = 7 ... (ii)
Step 2: From Eq(i): x = (4 + 5y)/3. Substitute in Eq(ii):
9 × (4 + 5y)/3 − 2y = 7
3(4 + 5y) − 2y = 7
12 + 15y − 2y = 7
13y = −5
y = −5/13
3(4 + 5y) − 2y = 7
12 + 15y − 2y = 7
13y = −5
y = −5/13
Step 3: x = (4 + 5(−5/13))/3 = (4 − 25/13)/3 = (52/13 − 25/13)/3 = (27/13)/3 = 9/13
By Elimination Method:
Step 1: Multiply Eq(i) by 3:
9x − 15y = 12 ... (iii)
9x − 2y = 7 ... (ii)
9x − 2y = 7 ... (ii)
Step 2: Subtract (ii) from (iii):
−15y − (−2y) = 12 − 7
−13y = 5
y = −5/13
x = (4 + 5(−5/13))/3 = 9/13
−13y = 5
y = −5/13
x = (4 + 5(−5/13))/3 = 9/13
Answer: x = 9/13, y = −5/13
✅ Solution — Part (iv): x/2 + 2y/3 = −1 and x − y/3 = 3
By Elimination Method:
Step 1: Clear fractions
Eq(i) × 6: 3x + 4y = −6 ... (iii)
Eq(ii) × 3: 3x − y = 9 ... (iv)
Eq(ii) × 3: 3x − y = 9 ... (iv)
Step 2: Subtract (iv) from (iii):
(3x + 4y) − (3x − y) = −6 − 9
5y = −15
y = −3
5y = −15
y = −3
Step 3: Substitute in Eq(ii): x − (−3)/3 = 3 → x + 1 = 3 → x = 2
Answer: x = 2, y = −3
2 Exercise 3.3 — Question 2
Form the pair of linear equations for the following problems and find their solution by substitution/elimination method:
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹105 and for a journey of 15 km, the charge paid is ₹155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹105 and for a journey of 15 km, the charge paid is ₹155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.
✅ Solution — Part (i)
Step 1: Let the two numbers be x and y (x > y)
x − y = 26 ... (i)
x = 3y ... (ii)
x = 3y ... (ii)
Step 2: Substitute Eq(ii) in Eq(i):
3y − y = 26
2y = 26
y = 13
x = 3 × 13 = 39
2y = 26
y = 13
x = 3 × 13 = 39
Answer: The two numbers are 39 and 13.
✅ Solution — Part (ii)
Step 1: Let the larger angle = x° and smaller angle = y°
Supplementary: x + y = 180 ... (i)
Larger exceeds smaller by 18°: x − y = 18 ... (ii)
Larger exceeds smaller by 18°: x − y = 18 ... (ii)
Step 2: Add both equations:
2x = 198
x = 99°
y = 180 − 99 = 81°
x = 99°
y = 180 − 99 = 81°
Answer: The angles are 99° and 81°.
✅ Solution — Part (iii)
Step 1: Let cost of one bat = ₹x and cost of one ball = ₹y
7x + 6y = 3800 ... (i)
3x + 5y = 1750 ... (ii)
3x + 5y = 1750 ... (ii)
Step 2 (Elimination): Multiply Eq(i) by 5 and Eq(ii) by 6:
35x + 30y = 19000 ... (iii)
18x + 30y = 10500 ... (iv)
18x + 30y = 10500 ... (iv)
Step 3: Subtract (iv) from (iii):
17x = 8500
x = 500
x = 500
Step 4: Substitute in Eq(ii): 3(500) + 5y = 1750 → 5y = 250 → y = 50
Answer: Cost of each bat = ₹500, Cost of each ball = ₹50.
✅ Solution — Part (iv)
Step 1: Let the fixed charge = ₹x and charge per km = ₹y
x + 10y = 105 ... (i)
x + 15y = 155 ... (ii)
x + 15y = 155 ... (ii)
Step 2: Subtract (i) from (ii):
5y = 50
y = 10
y = 10
Step 3: Substitute in Eq(i): x + 10(10) = 105 → x = 105 − 100 = 5
Step 4: Charge for 25 km = 5 + 25 × 10 = ₹255
Answer: Fixed charge = ₹5, Charge per km = ₹10. Cost for 25 km = ₹255.
✅ Solution — Part (v)
Step 1: Let the fraction be x/y
(x + 2)/(y + 2) = 9/11 → 11(x + 2) = 9(y + 2) → 11x − 9y = −4 ... (i)
(x + 3)/(y + 3) = 5/6 → 6(x + 3) = 5(y + 3) → 6x − 5y = −3 ... (ii)
Step 2 (Elimination): Multiply Eq(i) by 5 and Eq(ii) by 9:
55x − 45y = −20 ... (iii)
54x − 45y = −27 ... (iv)
54x − 45y = −27 ... (iv)
Step 3: Subtract (iv) from (iii):
x = −20 − (−27) = 7
Step 4: Substitute in Eq(ii): 6(7) − 5y = −3 → 42 − 5y = −3 → 5y = 45 → y = 9
Answer: The fraction is 7/9.
🎯 Board Exam Alert: Word problems from Q2 are the most frequently asked questions from this chapter. Part (iii) — the bat and ball problem — and Part (v) — the fraction problem — have appeared multiple times in CBSE board exams. Practice forming equations from word problems until it becomes second nature.
Exam Tips & Common Mistakes to Avoid
🏆 How to Score Full Marks in Chapter 3
1. Always verify your answer. Substitute the values back into BOTH original equations. If they don't satisfy both, recheck your work.
2. Show all steps. CBSE awards step marks. Even if you make a calculation error, you can get 2-3 marks out of 4 if your method is correct.
3. Label your graphs clearly. For graphical questions, always label axes, scale, equations, and the intersection point.
4. Simplify before solving. If equations have fractions, multiply through by the LCM first. This reduces calculation errors dramatically.
5. Read the question carefully. Some questions ask for graphical method specifically — don't use algebraic methods for those, and vice versa.
⚠️ Common Mistake #1: When using the elimination method, students forget to multiply ALL terms in the equation (including the constant on the right side). If you multiply the left side by 3, multiply the right side by 3 too!
⚠️ Common Mistake #2: Sign errors during subtraction. When subtracting equations, remember: subtracting a negative becomes adding. Write it out carefully: (−3y) − (−5y) = −3y + 5y = 2y, not −8y.
⚠️ Common Mistake #3: In word problems, students sometimes set up the equation incorrectly. Always define your variables clearly and re-read the problem after forming equations to make sure they match the conditions given.
Fig: Comparison of the three methods taught in Chapter 3
Frequently Asked Questions
Q1: How many exercises are there in NCERT Class 10 Maths Chapter 3?
As per the 2025-26 NCERT syllabus, Chapter 3 (Pair of Linear Equations in Two Variables) has 3 exercises: Exercise 3.1 (3 questions on graphical representation), Exercise 3.2 (7 questions on graphical method of solving), and Exercise 3.3 (2 questions with multiple parts on algebraic methods — substitution and elimination).
Q2: Is the Cross-Multiplication method still in the syllabus?
No. The Cross-Multiplication method has been removed from the NCERT textbook for the 2025-26 session. Only the Graphical Method, Substitution Method, and Elimination Method are part of the current syllabus.
Q3: What is the difference between consistent and inconsistent pair of linear equations?
A consistent pair has at least one solution — either a unique solution (intersecting lines) or infinitely many solutions (coincident lines). An inconsistent pair has no solution at all, meaning the lines are parallel and never meet.
Q4: Which method should I use in the board exam?
Use whichever method the question specifies. If it says "solve graphically," draw the graph. If it says "solve algebraically," use either substitution or elimination — the elimination method is usually faster. If no method is specified, elimination is recommended as it's quickest and least error-prone for most problems.
Q5: When should I use the substitution method vs the elimination method?
Use Substitution when one equation already has a variable with coefficient 1 (like y = 2x + 3 or x = 5 − y), making it easy to substitute directly. Use Elimination when both equations are in standard form (ax + by = c) and coefficients can be made equal through multiplication. Both give the same answer — pick whichever feels simpler for the given pair.
Q6: How important is Chapter 3 for board exams?
Chapter 3 carries significant weightage in CBSE board exams. You can expect 1-2 questions worth 3-5 marks from this chapter. Word problems from Exercise 3.3 (especially the bat-ball problem and fraction problems) are asked almost every year.
Q7: What are the conditions for a pair of linear equations to have a unique solution?
For the pair a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, the condition for a unique solution is: a₁/a₂ ≠ b₁/b₂. This means the lines intersect at exactly one point. Geometrically, the two lines have different slopes.
Conclusion
Chapter 3 — Pair of Linear Equations in Two Variables — is a fundamental chapter that builds your equation-solving skills for higher classes. The key takeaways are: understand the three possible cases (intersecting, parallel, coincident), master the condition ratios (a₁/a₂, b₁/b₂, c₁/c₂), and practice word problems until forming equations from verbal descriptions becomes automatic.
We have covered every question from Exercise 3.1, Exercise 3.2, and Exercise 3.3 with detailed step-by-step solutions and visual diagrams. Bookmark this page for quick revision before your exams. For more NCERT solutions, explore our guides for Chapter 2 (Polynomials) and Chapter 4 (Quadratic Equations).
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