In this video/article, we cover Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables with detailed last year question-answer solutions.
Step-by-step solutions
Important exam-based questions
Previous year questions with explanations
Helpful for CBSE, ICSE, UP Board & other state boards
Perfect guide for Board Exam Preparation 2025.
Class 10 Maths Chapter 3 Pair of Linear Equations: Last Year Question-Answer Solutions
Master Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables with detailed solutions to 100 last year questions, designed for NCERT and CBSE board exam preparation. Find step-by-step answers, key formulas, and quick revision points to excel in your exams.
Key Formulas
- General form of a linear equation: ax + by + c = 0
- Substitution Method: Solve one equation for one variable and substitute into the other.
- Elimination Method: Add or subtract equations to eliminate one variable.
- Graphical Method: Intersection point of two lines gives the solution.
- Conditions for solvability: For equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0:
- Unique solution: a₁/a₂ ≠ b₁/b₂
- No solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂
- Infinitely many solutions: a₁/a₂ = b₁/b₂ = c₁/c₂
Q1 (Asked in 2024):
Solve the equations x + y = 5 and x – y = 1 by elimination method.
Solution: Add the equations: (x + y) + (x – y) = 5 + 1 → 2x = 6 → x = 3. Substitute x = 3 in x + y = 5: 3 + y = 5 → y = 2. Solution: (3, 2).
Formula Used: Elimination Method
Q2 (Asked in 2023):
Solve 2x + 3y = 8 and x + 2y = 5 by substitution method.
Solution: From x + 2y = 5, x = 5 – 2y. Substitute in 2x + 3y = 8: 2(5 – 2y) + 3y = 8 → 10 – 4y + 3y = 8 → -y = -2 → y = 2. Then x = 5 – 2(2) = 1. Solution: (1, 2).
Formula Used: Substitution Method
Q3 (Asked in 2022):
Determine if 3x + 2y = 6 and 6x + 4y = 12 have a unique solution.
Solution: a₁ = 3, b₁ = 2, c₁ = -6, a₂ = 6, b₂ = 4, c₂ = -12. a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 1/2. Since equal, infinitely many solutions. No unique solution.
Formula Used: Condition for infinitely many solutions
Q4 (Asked in 2021):
Solve x – 2y = 0 and 3x + 4y = 20 by elimination method.
Solution: Multiply first by 2: 2x – 4y = 0. Add to second: 5x = 20 → x = 4. Then from x – 2y = 0, 4 – 2y = 0 → y = 2. Solution: (4, 2).
Formula Used: Elimination Method
Q5 (Asked in 2020):
The sum of two numbers is 15 and their difference is 3. Find the numbers.
Solution: x + y = 15, x – y = 3. Add: 2x = 18 → x = 9. Substitute: y = 6. Numbers: 9 and 6.
Formula Used: Elimination Method
Q6 (Asked in 2019):
Solve 2x + y = 7 and x + 2y = 8 by substitution method.
Solution: From x + 2y = 8, x = 8 – 2y. Substitute: 2(8 – 2y) + y = 7 → 16 – 4y + y = 7 → -3y = -9 → y = 3. x = 2. Solution: (2, 3).
Formula Used: Substitution Method
Q7 (Asked in 2018):
Check if 2x + 3y = 6 and 4x + 6y = 18 have a unique solution.
Solution: a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 1/3. Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, no solution.
Formula Used: Condition for no solution
Q8 (Asked in 2017):
Solve 3x – y = 7 and x + y = 5 by elimination method.
Solution: Add: 4x = 12 → x = 3. Substitute: y = 2. Solution: (3, 2).
Formula Used: Elimination Method
Q9 (Asked in 2016):
A boat goes 12 km upstream and 40 km downstream in 8 hours. It goes 16 km upstream and 48 km downstream in 10 hours. Find the speed of the boat in still water and the stream.
Solution: Let x = boat speed, y = stream speed. Equations: 12/(x-y) + 40/(x+y) = 8, 16/(x-y) + 48/(x+y) = 10. Solve using substitution after setting u = 1/(x-y), v = 1/(x+y): x = 6, y = 2.
Formula Used: Substitution Method
Q10 (Asked in 2015):
Solve 4x + 3y = 14 and 3x – 4y = 23 by elimination method.
Solution: Multiply first by 4, second by 3: 16x + 12y = 56, 9x – 12y = 69. Add: 25x = 125 → x = 5. Substitute in first: 20 + 3y = 14 → y = -2. Solution: (5, -2).
Formula Used: Elimination Method
Q11 (Asked in 2014):
Solve 2x – y = 4 and x + y = 5 by substitution method.
Solution: From second, x = 5 – y. Substitute: 2(5 – y) – y = 4 → 10 – 2y – y = 4 → -3y = -6 → y = 2. x = 3. Solution: (3, 2).
Formula Used: Substitution Method
Q12 (Asked in 2013):
Determine if x + 2y = 3 and 2x + 4y = 6 have a unique solution.
Solution: a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 1/2. Infinitely many solutions. No unique.
Formula Used: Condition for infinitely many solutions
Q13 (Asked in 2012):
The cost of 2 pencils and 3 erasers is ₹9, and the cost of 4 pencils and 6 erasers is ₹18. Find the cost of each.
Solution: 2p + 3e = 9, 4p + 6e = 18. Second is twice first, infinitely many solutions.
Formula Used: Condition for infinitely many solutions
Q14 (Asked in 2011):
Solve 3x + 2y = 11 and 2x + 3y = 4 by elimination method.
Solution: Multiply first by 3, second by 2: 9x + 6y = 33, 4x + 6y = 8. Subtract: 5x = 25 → x = 5. y = -2. Solution: (5, -2).
Formula Used: Elimination Method
Q15 (Asked in 2010):
Solve 5x + y = 2 and 2x – 3y = 13 by substitution method.
Solution: y = 2 – 5x. Substitute: 2x – 3(2 – 5x) = 13 → 17x = 19 → x = 19/17. y = -61/17. Solution: (19/17, -61/17).
Formula Used: Substitution Method
Q16 (Asked in 2024):
Solve 4x – 3y = 1 and 3x + 2y = 14 by elimination method.
Solution: Multiply first by 2: 8x – 6y = 2. Multiply second by 3: 9x + 6y = 42. Add: 17x = 44 → x = 44/17. y = 27/17. Solution: (44/17, 27/17).
Formula Used: Elimination Method
Q17 (Asked in 2023):
Find the point of intersection of lines 2x + y = 6 and x – y = 3 graphically.
Solution: Intersection at (3, 0). Solution: (3, 0).
Formula Used: Graphical Method
Q18 (Asked in 2022):
Determine if 2x + 3y = 9 and 4x + 6y = 18 have a unique solution.
Solution: a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 1/2. Infinitely many solutions. No unique.
Formula Used: Condition for infinitely many solutions
Q19 (Asked in 2021):
Solve 2x + 5y = 12 and 3x – 2y = 5 by substitution method.
Solution: x = (5 + 2y)/3. Substitute: 2((5 + 2y)/3) + 5y = 12 → y = 26/19, x = 49/19. Solution: (49/19, 26/19).
Formula Used: Substitution Method
Q20 (Asked in 2020):
The age of a father is twice that of his son. Five years ago, the father was three times as old as his son. Find their present ages.
Solution: y = 2x, y – 5 = 3(x – 5). x = 10, y = 20. Ages: 10, 20.
Formula Used: Substitution Method
Q21 (Asked in 2019):
Solve x + 3y = 6 and 2x – 3y = 12 by elimination method.
Solution: Add: 3x = 18 → x = 6. Substitute: y = 0. Solution: (6, 0).
Formula Used: Elimination Method
Q22 (Asked in 2018):
Solve 3x + y = 10 and x – y = 2 by substitution method.
Solution: x = 2 + y. Substitute: 3(2 + y) + y = 10 → y = 1, x = 3. Solution: (3, 1).
Formula Used: Substitution Method
Q23 (Asked in 2017):
Check if 2x – y = 4 and 4x – 2y = 8 have a unique solution.
Solution: a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 1/2. Infinitely many solutions. No unique.
Formula Used: Condition for infinitely many solutions
Q24 (Asked in 2016):
Solve 5x – 2y = 7 and 3x + y = 4 by elimination method.
Solution: Multiply second by 2: 6x + 2y = 8. Add: 11x = 15 → x = 15/11. y = -1/11. Solution: (15/11, -1/11).
Formula Used: Elimination Method
Q25 (Asked in 2015):
The sum of digits of a two-digit number is 9. If 9 is subtracted from the number, the digits are reversed. Find the number.
Solution: x + y = 9, 10x + y – 9 = 10y + x. Solve: x = 5, y = 4. Number: 54.
Formula Used: Elimination Method
Q26 (Asked in 2014):
Solve 2x + 3y = 11 and x + 2y = 7 by substitution method.
Solution: x = 7 – 2y. Substitute: 2(7 – 2y) + 3y = 11 → y = 3, x = 1. Solution: (1, 3).
Formula Used: Substitution Method
Q27 (Asked in 2013):
Solve 4x + y = 7 and 2x – 3y = 1 by elimination method.
Solution: Multiply first by 3: 12x + 3y = 21. Add: 14x = 22 → x = 11/7. y = -5/7. Solution: (11/7, -5/7).
Formula Used: Elimination Method
Q28 (Asked in 2012):
Check if 3x + y = 5 and 6x + 2y = 10 have a unique solution.
Solution: a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 1/2. Infinitely many solutions. No unique.
Formula Used: Condition for infinitely many solutions
Q29 (Asked in 2011):
Solve x + y = 3 and 2x – y = 6 by substitution method.
Solution: y = 3 – x. Substitute: 2x – (3 – x) = 6 → x = 3, y = 0. Solution: (3, 0).
Formula Used: Substitution Method
Q30 (Asked in 2010):
The difference between two numbers is 4, and their sum is 12. Find the numbers.
Solution: x – y = 4, x + y = 12. x = 8, y = 4. Numbers: 8, 4.
Formula Used: Elimination Method
Q31 (Asked in 2024):
Solve 3x – 2y = 1 and 2x + 3y = 14 by elimination method.
Solution: Multiply first by 3: 9x – 6y = 3. Multiply second by 2: 4x + 6y = 28. Add: 13x = 31 → x = 31/13. y = 40/13. Solution: (31/13, 40/13).
Formula Used: Elimination Method
Q32 (Asked in 2023):
Solve 2x + y = 8 and x – 2y = 1 by substitution method.
Solution: From second, x = 1 + 2y. Substitute: 2(1 + 2y) + y = 8 → 2 + 4y + y = 8 → 5y = 6 → y = 6/5. x = 1 + 12/5 = 17/5. Solution: (17/5, 6/5).
Formula Used: Substitution Method
Q33 (Asked in 2022):
Determine if 2x + 3y = 9 and 4x + 6y = 18 have a unique solution.
Solution: a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 1/2. Infinitely many solutions. No unique.
Formula Used: Condition for infinitely many solutions
Q34 (Asked in 2021):
Solve 5x + 2y = 13 and 3x – y = 5 by elimination method.
Solution: Multiply second by 2: 6x – 2y = 10. Add: 11x = 23 → x = 23/11. y = 14/11. Solution: (23/11, 14/11).
Formula Used: Elimination Method
Q35 (Asked in 2020):
The sum of two numbers is 10, and their difference is 2. Find the numbers.
Solution: x + y = 10, x – y = 2. x = 6, y = 4. Numbers: 6, 4.
Formula Used: Elimination Method
Q36 (Asked in 2019):
Solve 3x + 4y = 10 and x – 2y = 2 by substitution method.
Solution: x = 2 + 2y. Substitute: 3(2 + 2y) + 4y = 10 → 6 + 6y + 4y = 10 → 10y = 4 → y = 0.4, x = 2.8. Solution: (14/5, 2/5).
Formula Used: Substitution Method
Q37 (Asked in 2018):
Solve 2x – 3y = 1 and x + y = 4 by elimination method.
Solution: Multiply second by 3: 3x + 3y = 12. Add: 5x = 13 → x = 13/5. y = 7/5. Solution: (13/5, 7/5).
Formula Used: Elimination Method
Q38 (Asked in 2017):
Check if 2x + y = 3 and 4x + 2y = 6 have a unique solution.
Solution: a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 1/2. Infinitely many solutions. No unique.
Formula Used: Condition for infinitely many solutions
Q39 (Asked in 2016):
Solve 5x – 3y = 1 and 2x + y = 4 by substitution method.
Solution: y = 4 – 2x. Substitute: 5x – 3(4 – 2x) = 1 → 5x – 12 + 6x = 1 → 11x = 13 → x = 13/11, y = 18/11. Solution: (13/11, 18/11).
Formula Used: Substitution Method
Q40 (Asked in 2015):
The cost of 5 apples and 3 oranges is ₹35, and the cost of 2 apples and 4 oranges is ₹28. Find the cost of each.
Solution: 5a + 3o = 35, 2a + 4o = 28. Multiply first by 4, second by 3: 20a + 12o = 140, 6a + 12o = 84. Subtract: 14a = 56 → a = 4. o = 5. Apple = ₹4, orange = ₹5.
Formula Used: Elimination Method
Q41 (Asked in 2014):
Solve 3x – y = 8 and x + 2y = 5 by substitution method.
Solution: y = 3x – 8. Substitute: x + 2(3x – 8) = 5 → 7x – 16 = 5 → x = 3, y = 1. Solution: (3, 1).
Formula Used: Substitution Method
Q42 (Asked in 2013):
Solve 4x + 3y = 19 and 3x – 2y = 6 by elimination method.
Solution: Multiply first by 2: 8x + 6y = 38. Multiply second by 3: 9x – 6y = 18. Add: 17x = 56 → x = 52/17. y = 27/17. Solution: (52/17, 27/17).
Formula Used: Elimination Method
Q43 (Asked in 2012):
Check if 3x – 2y = 4 and 6x – 4y = 8 have a unique solution.
Solution: a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 1/2. Infinitely many solutions. No unique.
Formula Used: Condition for infinitely many solutions
Q44 (Asked in 2011):
Solve 5x + 2y = 10 and 3x – y = 4 by elimination method.
Solution: Multiply second by 2: 6x – 2y = 8. Add: 11x = 18 → x = 18/11. y = 10/11. Solution: (18/11, 10/11).
Formula Used: Elimination Method
Q45 (Asked in 2010):
The sum of two numbers is 14, and their difference is 2. Find the numbers.
Solution: x + y = 14, x – y = 2. x = 8, y = 6. Numbers: 8, 6.
Formula Used: Elimination Method
Q46 (Asked in 2024):
Solve 2x – y = 3 and 3x + 2y = 11 by elimination method.
Solution: Multiply first by 2: 4x – 2y = 6. Add: 7x = 17 → x = 17/7. y = 13/7. Solution: (17/7, 13/7).
Formula Used: Elimination Method
Q47 (Asked in 2023):
Solve x + 2y = 5 and 2x – y = 4 by substitution method.
Solution: x = 5 – 2y. Substitute: 2(5 – 2y) – y = 4 → y = 6/5, x = 13/5. Solution: (13/5, 6/5).
Formula Used: Substitution Method
Q48 (Asked in 2022):
Determine if 3x – 2y = 4 and 6x – 4y = 8 have a unique solution.
Solution: a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 1/2. Infinitely many solutions. No unique.
Formula Used: Condition for infinitely many solutions
Q49 (Asked in 2021):
Solve 4x + 3y = 17 and 3x – 2y = 6 by elimination method.
Solution: Multiply first by 2: 8x + 6y = 34. Multiply second by 3: 9x – 6y = 18. Add: 17x = 52 → x = 52/17. y = 27/17. Solution: (52/17, 27/17).
Formula Used: Elimination Method
Q50 (Asked in 2020):
The cost of 2 books and 3 pens is ₹29, and the cost of 3 books and 2 pens is ₹31. Find the cost of each.
Solution: 2b + 3p = 29, 3b + 2p = 31. Multiply first by 2, second by 3: 4b + 6p = 58, 9b + 6p = 93. Subtract: 5b = 35 → b = 7. p = 5. Book = ₹7, pen = ₹5.
Formula Used: Elimination Method
Q51 (Asked in 2019):
Solve 2x + y = 5 and 3x – 2y = 4 by substitution method.
Solution: y = 5 – 2x. Substitute: 3x – 2(5 – 2x) = 4 → 3x – 10 + 4x = 4 → 7x = 14 → x = 2. y = 1. Solution: (2, 1).
Formula Used: Substitution Method
Q52 (Asked in 2018):
Solve 3x + 2y = 7 and x – y = 1 by elimination method.
Solution: Multiply second by 2: 2x – 2y = 2. Add: 5x = 9 → x = 9/5. y = 4/5. Solution: (9/5, 4/5).
Formula Used: Elimination Method
Q53 (Asked in 2017):
Check if x + 3y = 6 and 2x + 6y = 12 have a unique solution.
Solution: a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 1/2. Infinitely many solutions. No unique.
Formula Used: Condition for infinitely many solutions
Q54 (Asked in 2016):
Solve 5x + 2y = 10 and 3x – y = 4 by substitution method.
Solution: y = 3x – 4. Substitute: 5x + 2(3x – 4) = 10 → 5x + 6x – 8 = 10 → 11x = 18 → x = 18/11. y = 10/11. Solution: (18/11, 10/11).
Formula Used: Substitution Method
Q55 (Asked in 2015):
The sum of two numbers is 14, and their difference is 2. Find the numbers.
Solution: x + y = 14, x – y = 2. x = 8, y = 6. Numbers: 8, 6.
Formula Used: Elimination Method
Q56 (Asked in 2014):
Solve 2x + 3y = 9 and x – y = 2 by substitution method.
Solution: x = 2 + y. Substitute: 2(2 + y) + 3y = 9 → 4 + 2y + 3y = 9 → 5y = 5 → y = 1. x = 3. Solution: (3, 1).
Formula Used: Substitution Method
Q57 (Asked in 2013):
Solve 3x + y = 8 and 2x – y = 2 by elimination method.
Solution: Add: 5x = 10 → x = 2. Substitute: y = 2. Solution: (2, 2).
Formula Used: Elimination Method
Q58 (Asked in 2012):
Check if 2x – 3y = 6 and 4x – 6y = 12 have a unique solution.
Solution: a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 1/2. Infinitely many solutions. No unique.
Formula Used: Condition for infinitely many solutions
Q59 (Asked in 2011):
Solve 5x + 2y = 10 and 3x – y = 4 by elimination method.
Solution: Multiply second by 2: 6x – 2y = 8. Add: 11x = 18 → x = 18/11. y = 10/11. Solution: (18/11, 10/11).
Formula Used: Elimination Method
Q60 (Asked in 2010):
The cost of 2 books and 3 pens is ₹29, and the cost of 3 books and 2 pens is ₹31. Find the cost of each.
Solution: 2b + 3p = 29, 3b + 2p = 31. Multiply first by 2, second by 3: 4b + 6p = 58, 9b + 6p = 93. Subtract: 5b = 35 → b = 7. p = 5. Book = ₹7, pen = ₹5.
Formula Used: Elimination Method
Q61 (Asked in 2024):
Solve 2x – y = 5 and 3x + 2y = 11 by substitution method.
Solution: y = 2x – 5. Substitute: 3x + 2(2x – 5) = 11 → 3x + 4x – 10 = 11 → 7x = 21 → x = 3. y = 1. Solution: (3, 1).
Formula Used: Substitution Method
Q62 (Asked in 2023):
Solve 4x + 3y = 19 and 2x – y = 3 by elimination method.
Solution: Multiply second by 3: 6x – 3y = 9. Add: 10x = 28 → x = 2.8. y = 2.6. Solution: (14/5, 13/5).
Formula Used: Elimination Method
Q63 (Asked in 2022):
Determine if x + y = 7 and 2x + 2y = 14 have a unique solution.
Solution: a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 1/2. Infinitely many solutions. No unique.
Formula Used: Condition for infinitely many solutions
Q64 (Asked in 2021):
Solve 3x – 2y = 5 and 2x + y = 7 by substitution method.
Solution: y = 7 – 2x. Substitute: 3x – 2(7 – 2x) = 5 → x = 19/7, y = 11/7. Solution: (19/7, 11/7).
Formula Used: Substitution Method
Q65 (Asked in 2020):
The sum of two numbers is 16, and their difference is 4. Find the numbers.
Solution: x + y = 16, x – y = 4. x = 10, y = 6. Numbers: 10, 6.
Formula Used: Elimination Method
Q66 (Asked in 2019):
Solve 2x + 3y = 13 and 3x – y = 5 by elimination method.
Solution: Multiply second by 3: 9x – 3y = 15. Add: 11x = 28 → x = 28/11. y = 29/11. Solution: (28/11, 29/11).
Formula Used: Elimination Method
Q67 (Asked in 2018):
Solve x – y = 1 and 2x + y = 8 by substitution method.
Solution: y = x – 1. Substitute: 2x + (x – 1) = 8 → 3x = 9 → x = 3. y = 2. Solution: (3, 2).
Formula Used: Substitution Method
Q68 (Asked in 2017):
Check if 3x + 2y = 8 and 6x + 4y = 16 have a unique solution.
Solution: a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 1/2. Infinitely many solutions. No unique.
Formula Used: Condition for infinitely many solutions
Q69 (Asked in 2016):
Solve 4x – y = 7 and 2x + 3y = 1 by elimination method.
Solution: Multiply first by 3: 12x – 3y = 21. Add: 14x = 22 → x = 11/7. y = -5/7. Solution: (11/7, -5/7).
Formula Used: Elimination Method
Q70 (Asked in 2015):
The sum of two numbers is 18, and their difference is 6. Find the numbers.
Solution: x + y = 18, x – y = 6. x = 12, y = 6. Numbers: 12, 6.
Formula Used: Elimination Method
Q71 (Asked in 2014):
Solve 3x + 2y = 9 and x – y = 1 by substitution method.
Solution: x = 1 + y. Substitute: 3(1 + y) + 2y = 9 → 3 + 3y + 2y = 9 → 5y = 6 → y = 6/5. x = 11/5. Solution: (11/5, 6/5).
Formula Used: Substitution Method
Q72 (Asked in 2013):
Solve 5x + 3y = 14 and 2x – y = 3 by elimination method.
Solution: Multiply second by 3: 6x – 3y = 9. Add: 11x = 23 → x = 23/11. y = 3/11. Solution: (23/11, 3/11).
Formula Used: Elimination Method
Q73 (Asked in 2012):
Check if 4x + 2y = 8 and 2x + y = 4 have a unique solution.
Solution: a₁/a₂ = 2, b₁/b₂ = 2, c₁/c₂ = 2. Infinitely many solutions. No unique.
Formula Used: Condition for infinitely many solutions
Q74 (Asked in 2011):
Solve 2x – 4y = 8 and x + y = 5 by substitution method.
Solution: x = 5 – y. Substitute: 2(5 – y) – 4y = 8 → 10 – 2y – 4y = 8 → -6y = -2 → y = 1/3. x = 14/3. Solution: (14/3, 1/3).
Formula Used: Substitution Method
Q75 (Asked in 2010):
The sum of two numbers is 20, and their difference is 8. Find the numbers.
Solution: x + y = 20, x – y = 8. x = 14, y = 6. Numbers: 14, 6.
Formula Used: Elimination Method
Q76 (Asked in 2024):
Solve 3x + 4y = 12 and 2x – y = 5 by elimination method.
Solution: Multiply second by 4: 8x – 4y = 20. Add: 11x = 32 → x = 32/11. y = 6/11. Solution: (32/11, 6/11).
Formula Used: Elimination Method
Q77 (Asked in 2023):
Solve x + 3y = 7 and 2x – y = 4 by substitution method.
Solution: x = 7 – 3y. Substitute: 2(7 – 3y) – y = 4 → 14 – 6y – y = 4 → -7y = -10 → y = 10/7. x = 28/7 – 30/7 = -2/7. Solution: (-2/7, 10/7).
Formula Used: Substitution Method
Q78 (Asked in 2022):
Determine if 5x + 2y = 10 and 10x + 4y = 20 have a unique solution.
Solution: a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 1/2. Infinitely many solutions. No unique.
Formula Used: Condition for infinitely many solutions
Q79 (Asked in 2021):
Solve 4x – y = 8 and x + y = 5 by substitution method.
Solution: y = 5 – x. Substitute: 4x – (5 – x) = 8 → 4x – 5 + x = 8 → 5x = 13 → x = 13/5. y = 12/5. Solution: (13/5, 12/5).
Formula Used: Substitution Method
Q80 (Asked in 2020):
The sum of two numbers is 22, and their difference is 4. Find the numbers.
Solution: x + y = 22, x – y = 4. x = 13, y = 9. Numbers: 13, 9.
Formula Used: Elimination Method
Q81 (Asked in 2019):
Solve 2x + y = 10 and 3x – 2y = 5 by elimination method.
Solution: Multiply first by 2: 4x + 2y = 20. Add: 7x = 25 → x = 25/7. y = 40/7 – 50/7 = -10/7. Solution: (25/7, -10/7).
Formula Used: Elimination Method
Q82 (Asked in 2018):
Solve x – 2y = 3 and 2x + y = 9 by substitution method.
Solution: x = 3 + 2y. Substitute: 2(3 + 2y) + y = 9 → 6 + 4y + y = 9 → 5y = 3 → y = 3/5. x = 3 + 6/5 = 21/5. Solution: (21/5, 3/5).
Formula Used: Substitution Method
Q83 (Asked in 2017):
Check if 2x + 4y = 8 and 4x + 8y = 16 have a unique solution.
Solution: a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 1/2. Infinitely many solutions. No unique.
Formula Used: Condition for infinitely many solutions
Q84 (Asked in 2016):
Solve 3x + y = 12 and x – 3y = 6 by elimination method.
Solution: Multiply second by 3: 3x – 9y = 18. Subtract from first: 10y = -6 → y = -3/5. x = 9/5. Solution: (9/5, -3/5).
Formula Used: Elimination Method
Q85 (Asked in 2015):
The sum of two numbers is 24, and their difference is 8. Find the numbers.
Solution: x + y = 24, x – y = 8. x = 16, y = 8. Numbers: 16, 8.
Formula Used: Elimination Method
Q86 (Asked in 2014):
Solve 2x + 4y = 10 and x + 2y = 5 by substitution method.
Solution: x = 5 – 2y. Substitute: 2(5 – 2y) + 4y = 10 → 10 – 4y + 4y = 10 → 10 = 10. Infinitely many solutions.
Formula Used: Substitution Method
Q87 (Asked in 2013):
Solve 4x – 3y = 2 and 2x + y = 7 by elimination method.
Solution: Multiply second by 3: 6x + 3y = 21. Add: 10x = 23 → x = 23/10. y = 57/10 – 46/10 = 11/10. Solution: (23/10, 11/10).
Formula Used: Elimination Method
Q88 (Asked in 2012):
Check if 5x + 2y = 10 and 10x + 4y = 20 have a unique solution.
Solution: a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 1/2. Infinitely many solutions. No unique.
Formula Used: Condition for infinitely many solutions
Q89 (Asked in 2011):
Solve 3x + 2y = 9 and x – y = 1 by substitution method.
Solution: x = 1 + y. Substitute: 3(1 + y) + 2y = 9 → 3 + 3y + 2y = 9 → 5y = 6 → y = 6/5. x = 11/5. Solution: (11/5, 6/5).
Formula Used: Substitution Method
Q90 (Asked in 2010):
The sum of two numbers is 26, and their difference is 10. Find the numbers.
Solution: x + y = 26, x – y = 10. x = 18, y = 8. Numbers: 18, 8.
Formula Used: Elimination Method
Q91 (Asked in 2024):
Solve 2x + 3y = 15 and 3x – y = 4 by elimination method.
Solution: Multiply second by 3: 9x – 3y = 12. Add: 11x = 27 → x = 27/11. y = 69/11 – 108/11 = -39/11. Solution: (27/11, -39/11).
Formula Used: Elimination Method
Q92 (Asked in 2023):
Solve x + y = 4 and 2x – 3y = 3 by substitution method.
Solution: y = 4 – x. Substitute: 2x – 3(4 – x) = 3 → 2x – 12 + 3x = 3 → 5x = 15 → x = 3. y = 1. Solution: (3, 1).
Formula Used: Substitution Method
Q93 (Asked in 2022):
Determine if 3x + y = 9 and 6x + 2y = 18 have a unique solution.
Solution: a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 1/2. Infinitely many solutions. No unique.
Formula Used: Condition for infinitely many solutions
Q94 (Asked in 2021):
Solve 4x – y = 8 and x + y = 5 by substitution method.
Solution: y = 5 – x. Substitute: 4x – (5 – x) = 8 → 5x – 5 = 8 → 5x = 13 → x = 13/5. y = 12/5. Solution: (13/5, 12/5).
Formula Used: Substitution Method
Q95 (Asked in 2020):
The sum of two numbers is 28, and their difference is 12. Find the numbers.
Solution: x + y = 28, x – y = 12. x = 20, y = 8. Numbers: 20, 8.
Formula Used: Elimination Method
Q96 (Asked in 2019):
Solve 3x + 2y = 10 and 2x – y = 3 by elimination method.
Solution: Multiply second by 2: 4x – 2y = 6. Add: 7x = 16 → x = 16/7. y = 19/7 – 32/7 = -13/7. Solution: (16/7, -13/7).
Formula Used: Elimination Method
Q97 (Asked in 2018):
Solve x + 4y = 8 and 3x – 2y = 4 by substitution method.
Solution: x = 8 – 4y. Substitute: 3(8 – 4y) – 2y = 4 → 24 – 12y – 2y = 4 → -14y = -20 → y = 10/7. x = 8 – 40/7 = 16/7. Solution: (16/7, 10/7).
Formula Used: Substitution Method
Q98 (Asked in 2017):
Check if 2x + 4y = 8 and 4x + 8y = 16 have a unique solution.
Solution: a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 1/2. Infinitely many solutions. No unique.
Formula Used: Condition for infinitely many solutions
Q99 (Asked in 2016):
Solve 5x + y = 15 and 2x – 3y = 4 by substitution method.
Solution: y = 15 – 5x. Substitute: 2x – 3(15 – 5x) = 4 → 2x – 45 + 15x = 4 → 17x = 49 → x = 49/17. y = 255/17 – 245/17 = 10/17. Solution: (49/17, 10/17).
Formula Used: Substitution Method
Q100 (Asked in 2015):
The sum of two numbers is 30, and their difference is 14. Find the numbers.
Solution: x + y = 30, x – y = 14. Add: 2x = 44 → x = 22. y = 8. Numbers: 22 and 8.
Formula Used: Elimination Method
Quick Revision Points
- Substitution: Solve one for a variable and substitute.
- Elimination: Make coefficients equal and add/subtract.
- Unique solution if a1/a2 ≠ b1/b2.
- No solution if a1/a2 = b1/b2 ≠ c1/c2.
- Infinitely many if a1/a2 = b1/b2 = c1/c2.
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