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Complete NCERT Solutions for CBSE Class 10 Maths Chapter 3 — Pair of Linear Equations in Two Variables, covering consistency by ratios, the substitution and elimination methods, and word problems.
Key Concepts (Quick Recap)
For a pair a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0:
| Condition | Lines | Solutions | Nature |
|---|---|---|---|
| a₁/a₂ ≠ b₁/b₂ | Intersecting | Unique | Consistent |
| a₁/a₂ = b₁/b₂ ≠ c₁/c₂ | Parallel | None | Inconsistent |
| a₁/a₂ = b₁/b₂ = c₁/c₂ | Coincident | Infinite | Dependent (consistent) |
Exercise 3.1 — Consistency by Ratios
Check whether each pair is consistent, and the nature of the lines.
(a) x + 2y = 5; 2x + 4y = 10 a₁/a₂ = 1/2, b₁/b₂ = 2/4 = 1/2, c₁/c₂ = 5/10 = 1/2. All equal → coincident lines, infinitely many solutions (consistent/dependent).
(b) x − 2y = 8; 2x − 4y = 5 1/2 = (−2)/(−4) = 1/2, but c: 8/5 ≠ 1/2. So a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → parallel lines, no solution (inconsistent).
(c) 3x + 2y = 12; 5x − 2y = 4 3/5 ≠ 2/(−2). a₁/a₂ ≠ b₁/b₂ → intersecting lines, unique solution (consistent).
Exercise 3.2 — Substitution Method
Solve the following pairs by substitution.
(i) x + y = 14; x − y = 4 From (2): x = y + 4. Substitute: (y + 4) + y = 14 → 2y = 10 → y = 5, x = 9.
(ii) s − t = 3; s/3 + t/2 = 6 s = t + 3. Substitute: (t+3)/3 + t/2 = 6 → multiply by 6: 2(t+3) + 3t = 36 → 5t + 6 = 36 → t = 6, s = 9, t = 6.
(iii) 3x − y = 3; 9x − 3y = 9 The second is 3 × the first → same line → infinitely many solutions.
(iv) 0.2x + 0.3y = 1.3; 0.4x + 0.5y = 2.3 ×10: 2x + 3y = 13; 4x + 5y = 23. From first x = (13 − 3y)/2. Substitute into second: 4(13−3y)/2 + 5y = 23 → 2(13−3y) + 5y = 23 → 26 − y = 23 → y = 3, x = 2, y = 3.
(v) √2 x + √3 y = 0; √3 x − √8 y = 0 Both pass through the origin and are not multiples → x = 0, y = 0.
(vi) (3/2)x − (5/3)y = −2; x/3 + y/2 = 13/6 ×6 (first): 9x − 10y = −12. ×6 (second): 2x + 3y = 13 → x = (13 − 3y)/2. Substitute: 9(13−3y)/2 − 10y = −12 → 9(13−3y) − 20y = −24 → 117 − 47y = −24 → y = 3, x = 2, y = 3.
Exercise 3.3 — Elimination Method
Solve by elimination.
(i) x + y = 5; 2x − 3y = 4 ×2 (first): 2x + 2y = 10. Subtract second: (2x+2y) − (2x−3y) = 10 − 4 → 5y = 6 → y = 6/5, x = 5 − 6/5 = 19/5.
(ii) 3x + 4y = 10; 2x − 2y = 2 Second ÷2: x − y = 1 → x = y + 1. Substitute: 3(y+1) + 4y = 10 → 7y = 7 → y = 1, x = 2.
(iii) 3x − 5y − 4 = 0; 9x = 2y + 7 3x − 5y = 4 … (1); 9x − 2y = 7 … (2). ×3 (1): 9x − 15y = 12. Subtract (2): −13y = 5 → y = −5/13. From (1): 3x = 4 + 5(−5/13) … 3x = 4 − 25/13 = 27/13 → x = 9/13.
(iv) x/2 + 2y/3 = −1; x − y/3 = 3 ×6 (first): 3x + 4y = −6. ×3 (second): 3x − y = 9. Subtract: 5y = −15 → y = −3. Then 3x − (−3) = 9 → 3x = 6 → x = 2.
Word Problems (Application)
Q. The sum of the digits of a two-digit number is 9. Nine times this number is twice the number obtained by reversing the order of the digits. Find the number. Let the number be 10x + y. Then x + y = 9 … (1). Reversed number = 10y + x. Given 9(10x + y) = 2(10y + x) → 90x + 9y = 20y + 2x → 88x = 11y → y = 8x. With (1): x + 8x = 9 → x = 1, y = 8. Number = 18.
Q. Five years ago, A was thrice as old as B. Ten years later, A will be twice as old as B. Find their present ages. Let present ages be a and b. (a − 5) = 3(b − 5) → a − 3b = −10 … (1). (a + 10) = 2(b + 10) → a − 2b = 10 … (2). Subtract (1) from (2): b = 20. Then a = 10 + 2(20) = 50. A = 50, B = 20.
Common Mistakes
- Mixing up the consistency rule — parallel needs a₁/a₂ = b₁/b₂ ≠ c₁/c₂.
- Arithmetic slips when clearing fractions — multiply the whole equation by the LCM.
- Forgetting to back-substitute to find the second variable.
Exam Tips
- Choose the method that fits: substitution when a variable has coefficient 1; elimination when coefficients align easily.
- Always verify your answer in both original equations.
- For word problems, define variables clearly and write the two equations before solving.