CBSE Class 10 Mathematics: NCERT Solutions Class 10 Maths Chapter 3 Pair of Linear Equations (2025–2026 Guide)

For students tackling CBSE Class 10 Mathematics, Chapter 3 on the Pair of Linear Equations in Two Variables is a cornerstone of algebraic problem-solving in the 2025–2026 curriculum. Mastering NCERT Solutions Class 10 Maths Chapter 3 Pair of Linear Equations equips students with analytical tools to model real-world scenarios, verify consistency, and solve simultaneous equations using both graphical and algebraic techniques. Whether preparing for basic or standard CBSE Mathematics, this comprehensive guide delivers complete conceptual explanations, step-by-step methods, and high-frequency questions tailored for board exam 10 success.

Key Concepts

A linear equation in two variables represents a straight line on the Cartesian plane. When two such equations are considered simultaneously, they form a pair of linear equations in two variables (or a system of simultaneous linear equations).

1. General Algebraic Form

The standard form of a pair of linear equations in two variables x and y is expressed as:

a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0

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

2. Geometric Representation and Consistency Criteria

The solution to a pair of linear equations corresponds to the point(s) of intersection of their respective lines. Comparing the ratios of their coefficients allows us to determine the nature of their solutions without plotting graphs:

3. Algebraic Methods of Solution

While the graphical method is useful for visual understanding, algebraic methods provide direct and exact numerical solutions:

• Substitution Method: Express one variable in terms of the other from one equation (e.g., y = f(x)) and substitute this expression into the second equation to obtain a single-variable linear equation.
• Elimination Method: Multiply one or both equations by non-zero constants so that the absolute values of the coefficients of one variable become equal. Then add or subtract the equations to eliminate that variable.

4. Translating Real-World Problems into Linear Equations

Mathematical modeling requires identifying unknowns, assigning variables (usually x and y), and formulating two independent relations based on the given conditions. Typical applications in CBSE Class 10 include:

• Speed, Distance, and Time: Relative speed problems involving upstream (x − y) and downstream (x + y) motion.
• Numbers and Digits: Reversing two-digit numbers where the original number is 10x + y and the reversed number is 10y + x.
• Fixed and Variable Charges: Taxi fares, hostel fees, and library charges with a base fee plus a daily/per-kilometer rate.
• Fraction Problems: Numerator-denominator adjustments where the fraction is represented as x/y.

Important CBSE Questions with Answers

Below are essential questions representative of official CBSE board exam patterns, categorized by question type and solved with complete step-by-step working.

Question 1 (Short Answer - 2 Marks: Condition for Infinite Solutions)

Problem: Find the value of k for which the following pair of linear equations has infinitely many solutions:
kx + 3y = k − 3
12x + ky = k

Solution:

1. Write both equations in standard form ax + by + c = 0:
   • kx + 3y − (k − 3) = 0 ⇒ a₁ = k, b₁ = 3, c₁ = −(k − 3)
   • 12x + ky − k = 0 ⇒ a₂ = 12, b₂ = k, c₂ = −k
2. For infinitely many solutions (coincident lines), the condition is:

   a₁/a₂ = b₁/b₂ = c₁/c₂

   k / 12 = 3 / k = (k − 3) / k

3. Equating the first two ratios:

   k / 12 = 3 / k ⇒ k² = 36 ⇒ k = ±6

4. Equating the second and third ratios (where k ≠ 0):

   3 / k = (k − 3) / k ⇒ 3 = k − 3 ⇒ k = 6

5. Verify consistency across all ratios:
   • If k = 6: 6/12 = 3/6 = 3/6 = 1/2 (Satisfied).
   • If k = −6: −6/12 = −1/2, but 3/(−6) = −1/2 ≠ (−9)/(−6) = 3/2 (Not satisfied).

Answer: The value of k = 6.

Question 2 (Algebraic Solution - 3 Marks: Elimination Method)

Problem: Solve the following pair of linear equations using the elimination method:
3x + 4y = 10
2x − 2y = 2

Solution:

1. Given equations:

   (1) 3x + 4y = 10

   (2) 2x − 2y = 2

2. Multiply equation (2) by 2 so that the coefficient of y matches equation (1) in magnitude:

   2 × (2x − 2y = 2) ⇒ 4x − 4y = 4   --- (3)

3. Add equation (1) and equation (3) to eliminate y:

   (3x + 4y) + (4x − 4y) = 10 + 4

   7x = 14 ⇒ x = 2

4. Substitute x = 2 into equation (1):

   3(2) + 4y = 10

   6 + 4y = 10 ⇒ 4y = 4 ⇒ y = 1

5. Verification: In equation (2), 2(2) − 2(1) = 4 − 2 = 2 (LHS = RHS).

Answer: x = 2, y = 1.

Question 3 (Word Problem - 3 Marks: Fraction Relations)

Problem: A fraction becomes 1/3 when 1 is subtracted from the numerator, and it becomes 1/4 when 8 is added to its denominator. Find the fraction.

Solution:

1. Let the numerator be x and the denominator be y (where y ≠ 0). The required fraction is x/y.
2. From the first condition:

   (x − 1) / y = 1/3

   3(x − 1) = y ⇒ 3x − 3 = y ⇒ 3x − y = 3   --- (1)

3. From the second condition:

   x / (y + 8) = 1/4

   4x = y + 8 ⇒ 4x − y = 8   --- (2)

4. Subtract equation (1) from equation (2):

   (4x − y) − (3x − y) = 8 − 3

   x = 5

5. Substitute x = 5 into equation (1):

   3(5) − y = 3 ⇒ 15 − y = 3 ⇒ y = 12

Answer: The required fraction is 5/12.

Question 4 (Long Answer - 5 Marks: Upstream & Downstream Speed Problem)

Problem: A motorboat covers 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can cover 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.

Solution:

1. Let the speed of the boat in still water be x km/h and the speed of the stream be y km/h (x > y).
   • Speed upstream = (x − y) km/h
   • Speed downstream = (x + y) km/h
2. Using Time = Distance / Speed, establish the equations:

   Case 1: 30 / (x − y) + 44 / (x + y) = 10   --- (1)

   Case 2: 40 / (x − y) + 55 / (x + y) = 13   --- (2)

3. Substitute u = 1 / (x − y) and v = 1 / (x + y):

   30u + 44v = 10   --- (3)

   40u + 55v = 13   --- (4)

4. Multiply equation (3) by 4 and equation (4) by 3 to eliminate u:

   120u + 176v = 40   --- (5)

   120u + 165v = 39   --- (6)

   Subtracting (6) from (5): 11v = 1 ⇒ v = 1/11

5. Substitute v = 1/11 into equation (3):

   30u + 44(1/11) = 10 ⇒ 30u + 4 = 10 ⇒ 30u = 6 ⇒ u = 1/5

6. Re-substitute u and v to find x and y:

   1 / (x − y) = 1/5 ⇒ x − y = 5   --- (7)

   1 / (x + y) = 1/11 ⇒ x + y = 11   --- (8)

7. Add equations (7) and (8):

   2x = 16 ⇒ x = 8 km/h

   Substitute x = 8 into (8): 8 + y = 11 ⇒ y = 3 km/h

Answer: The speed of the boat in still water is 8 km/h, and the speed of the stream is 3 km/h.

Question 5 (Case-Based / Application - 4 Marks: Fixed and Variable Rates)

Problem: A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for a book kept for seven days, while Susy paid ₹21 for a book kept for five days. Formulate the linear equations and determine the fixed charge and the daily extra charge.

Solution:

1. Let the fixed charge for the first 3 days be ₹x and the charge for each additional day be ₹y.
2. Saritha kept the book for 7 days (3 fixed days + 4 extra days):

   x + 4y = 27   --- (1)

3. Susy kept the book for 5 days (3 fixed days + 2 extra days):

   x + 2y = 21   --- (2)

4. Subtract equation (2) from equation (1):

   (x + 4y) − (x + 2y) = 27 − 21

   2y = 6 ⇒ y = 3

5. Substitute y = 3 into equation (2):

   x + 2(3) = 21 ⇒ x + 6 = 21 ⇒ x = 15

Answer: The fixed charge for the first three days is ₹15 and the charge for each additional day is ₹3.

How to Prepare for This Topic

Scoring full marks in Chapter 3 in board exam 10 requires structured practice and systematic habits. Follow these targeted preparation guidelines:

1. Check Consistency First: Before solving any unfamiliar pair of equations algebraically, evaluate a₁/a₂, b₁/b₂, and c₁/c₂. This prevents wasted time on inconsistent systems that yield no solution.
2. Standardize Equation Formats: Always shift all constant terms to the LHS (ax + by + c = 0) or all to the RHS (ax + by = c) consistently before identifying coefficients. A mismatch in signs between c₁ and c₂ is the most common source of calculation error.
3. Master Variable Elimination: When using the elimination method, select the variable that requires multiplying only one equation by an integer, reducing arithmetic workload and potential sign errors.
4. Always Verify by Back-Substitution: After obtaining the values of x and y, substitute both into the second original equation. If LHS equals RHS, your solution is verified.
5. Focus on Statement Interpretation: In word problems (ages, digits, speed, geometry), write down explicit variable declarations with proper units (e.g., "Let the present age of father be x years") to secure mandatory step marks in the CBSE marking scheme.

Where to Practice More

To secure top scores in CBSE Class 10 Mathematics, regular test practice with authentic board-style question papers is essential. Visit Theorify QPTool at qptool.theorify.in to generate chapter-wise practice tests, access curated CBSE question banks with detailed marking schemes, and download customized model question papers for the 2025–2026 board examinations.