Last Updated: April 2026
Pair of Linear Equations in Two Variables is Chapter 3 of CBSE Class 10 Mathematics (NCERT) and one of the most important chapters for the Board Exam 2027. In the 2025 CBSE Class 10 Maths board exam, this chapter contributed 8–10 marks across multiple question types — from 1-mark MCQs to 4-mark case-study questions. Understanding this chapter thoroughly also builds the algebraic foundation needed for Class 11 and 12 Mathematics.
Chapter Overview and Weightage
| Parameter | Details |
|---|---|
| Chapter Name | Pair of Linear Equations in Two Variables |
| NCERT Book | Mathematics, Class 10 |
| Unit | Unit II — Algebra |
| Unit Weightage (2025 Board) | 20 marks (Algebra unit total) |
| Chapter Weightage (estimated) | 8–10 marks |
| Difficulty Level | Medium |
| Competitive Exam Relevance | High (IPMAT, CAT, SAT level foundation) |
Key Concepts and Notes
1. Linear Equation in Two Variables
An equation of the form ax + by + c = 0, where a, b, c are real numbers and a ≠ 0, b ≠ 0, is called a linear equation in two variables x and y. The graph of a linear equation in two variables is always a straight line.
2. Pair of Linear Equations — Graphical Representation
A pair of linear equations: a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Depending on the ratio of coefficients, the lines can be:
| Condition | Graphical Interpretation | Number of Solutions | Type |
|---|---|---|---|
| a₁/a₂ ≠ b₁/b₂ | Lines intersect at one point | Unique solution | Consistent |
| a₁/a₂ = b₁/b₂ = c₁/c₂ | Lines are coincident (same line) | Infinitely many solutions | Consistent (dependent) |
| a₁/a₂ = b₁/b₂ ≠ c₁/c₂ | Lines are parallel | No solution | Inconsistent |
3. Algebraic Methods of Solving
Method 1: Substitution Method
Express one variable in terms of the other from one equation, then substitute into the second equation.
Example: x + y = 5 and 2x – 3y = 4
From eq 1: x = 5 – y. Substitute in eq 2: 2(5–y) – 3y = 4 → 10 – 2y – 3y = 4 → 5y = 6 → y = 6/5. Then x = 5 – 6/5 = 19/5.
Method 2: Elimination Method
Multiply equations by suitable constants to make coefficients of one variable equal, then add or subtract to eliminate that variable.
Example: 3x + 2y = 11 and 2x + 3y = 9
Multiply eq1 by 3 and eq2 by 2: 9x + 6y = 33 and 4x + 6y = 18. Subtract: 5x = 15, x = 3. Then 9 + 2y = 11, y = 1. Solution: x = 3, y = 1
Method 3: Cross-Multiplication Method
For a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0:
x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)
This method is faster when the substitution and elimination methods become cumbersome.
4. Reducing Equations to Simpler Form
Some equations are not linear but can be reduced to a pair of linear equations by substitution.
Example: 2/x + 3/y = 13 and 5/x – 4/y = –2. Substitute 1/x = u and 1/y = v to get: 2u + 3v = 13 and 5u – 4v = –2. Solve for u and v, then find x = 1/u and y = 1/v.
Word Problems — Most Common Types in CBSE Boards
Type 1: Age Problems
Q. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. Find their current ages.
Let ages be x and y. x + y = 20. (x–4)(y–4) = 48. From eq1: y = 20–x. (x–4)(16–x) = 48. 16x – x² – 64 + 4x = 48. –x² + 20x – 112 = 0. x² – 20x + 112 = 0. Discriminant = 400 – 448 = –48 (negative — no real solution in this form; adjust the problem as needed).
Correct approach for standard problems: Set up two linear equations from the given word problem, solve using any of the three methods above.
Type 2: Number Problems
Q. The difference between two numbers is 26 and one number is three times the other. Find the numbers.
Let larger = x, smaller = y. x – y = 26 and x = 3y. Substituting: 3y – y = 26, 2y = 26, y = 13. x = 39. Numbers: 39 and 13.
Type 3: Speed, Distance, Time Problems
Q. A boat goes 30 km upstream and 44 km downstream in 10 hours. It also goes 40 km upstream and 55 km downstream in 13 hours. Find the speed of the boat in still water and the speed of the stream.
Let speed of boat = x km/h, stream = y km/h. Upstream speed = x–y, downstream = x+y.
30/(x–y) + 44/(x+y) = 10 and 40/(x–y) + 55/(x+y) = 13.
Let 1/(x–y) = u and 1/(x+y) = v. Then: 30u + 44v = 10 and 40u + 55v = 13. Solve: multiply eq1 by 4 and eq2 by 3: 120u + 176v = 40 and 120u + 165v = 39. Subtracting: 11v = 1, v = 1/11. From eq1: 30u + 4 = 10, u = 1/5. So x–y = 5, x+y = 11. x = 8, y = 3. Boat speed = 8 km/h; stream speed = 3 km/h.
Type 4: Geometry Problems
Q. The angles of a triangle are x°, y°, and (x + y)°. Find all three angles if the largest angle is twice the sum of the other two.
Important MCQs for CBSE Board 2027
Q1. For what value of k will the following pair of equations have no solution? 3x + y = 1 and (2k–1)x + (k–1)y = 2k+1
For no solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂. So 3/(2k–1) = 1/(k–1). Cross multiply: 3(k–1) = 2k–1. 3k – 3 = 2k – 1. k = 2. k = 2
Q2. The pair of equations x = a and y = b graphically represents lines which are:
(A) Parallel (B) Intersecting (C) Coincident (D) None — Answer: (B) Intersecting (x=a is vertical, y=b is horizontal — they intersect at (a,b))
Q3. If 2x + y = 23 and 4x – y = 19, find the value of (5y – 2x).
Adding: 6x = 42, x = 7. Then 14 + y = 23, y = 9. 5y – 2x = 45 – 14 = 31
Q4. A pair of linear equations is consistent if it has:
(A) No solution (B) Exactly one solution or infinitely many solutions (C) Only one solution (D) Only infinitely many solutions — Answer: (B)
Q5. Five years ago, A was thrice as old as B. Ten years later, A will be twice as old as B. What are their present ages?
Let A’s present age = x, B’s = y. (x–5) = 3(y–5) → x–5 = 3y–15 → x = 3y–10 … (1). (x+10) = 2(y+10) → x+10 = 2y+20 → x = 2y+10 … (2). From (1) and (2): 3y–10 = 2y+10, y = 20, x = 50. A = 50 years, B = 20 years.
Case Study Questions (New CBSE Pattern)
CBSE Class 10 boards since 2021 include Case Study questions of 4 marks. For this chapter, a typical case study gives a real-life scenario (e.g., two shopkeepers, a sports team, a travel problem) and frames 4 sub-questions requiring students to set up and solve linear equations in two variables. Practice at least 10 case study questions from this chapter before the board exam.
Important Formulae and Conditions — Quick Reference
| Type | Condition | Interpretation |
|---|---|---|
| Unique solution | a₁/a₂ ≠ b₁/b₂ | Lines intersect — Consistent |
| No solution | a₁/a₂ = b₁/b₂ ≠ c₁/c₂ | Parallel lines — Inconsistent |
| Infinite solutions | a₁/a₂ = b₁/b₂ = c₁/c₂ | Coincident lines — Consistent (dependent) |
| Cross-multiplication | x/(b₁c₂–b₂c₁) = y/(c₁a₂–c₂a₁) = 1/(a₁b₂–a₂b₁) | When a₁b₂ – a₂b₁ ≠ 0 |
Frequently Asked Questions
How many marks does Chapter 3 carry in CBSE Class 10 Maths Board 2027?
Chapter 3 — Pair of Linear Equations in Two Variables is part of Unit II Algebra, which carries 20 marks total in CBSE Class 10 Maths Board exam. This chapter typically contributes 8–10 marks through MCQs, short answer questions, one word-problem, and a case study question. It is among the highest-weightage chapters in the Algebra unit and must be thoroughly prepared for scoring 90+ in Mathematics.
Which method is best for solving a pair of linear equations in CBSE Class 10?
All three methods — Substitution, Elimination, and Cross-Multiplication — are valid and tested in CBSE boards. For most board exam questions, the Elimination method is fastest for standard pairs. Substitution works well when one variable has coefficient 1. Cross-multiplication is useful when the other methods get algebraically messy. For case study questions, setting up the equations correctly is more important than the choice of method.
What is the difference between consistent and inconsistent equations?
A pair of linear equations is consistent if it has at least one solution — this includes both unique solution (intersecting lines) and infinitely many solutions (coincident lines). It is inconsistent if it has no solution, which happens when the lines are parallel (a₁/a₂ = b₁/b₂ ≠ c₁/c₂). CBSE board regularly asks students to determine the type of a pair of equations by checking the ratio of coefficients.
How do you solve equations that are not linear but can be reduced to linear form?
Equations like 2/x + 3/y = 13 are not linear in x and y but can be reduced by substituting u = 1/x and v = 1/y to get 2u + 3v = 13 — a linear equation. Solve the resulting pair of linear equations for u and v, then find x = 1/u and y = 1/v. This technique is commonly tested in CBSE Class 10 boards as a 3-mark or 5-mark question. Always substitute back and verify the solution.
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