Linear Equations PYQ Bank | Class X Maths

Questions

2-Mark Qs

3-Mark Qs

Years Covered

📋 Key Topics Covered (As per CBSE 2025-26 Syllabus)

Substitution Method — solving pair of equations

Elimination Method — solving pair of equations

Consistency Conditions — unique / infinite / no solution

Word Problems — age, fraction, digit, speed, geometry

Reducible Equations — converting to linear form

Graphical Interpretation — intersecting / parallel / coincident

⚠️

DELETED for 2025-26: Cross-Multiplication Method is removed from the syllabus. Only Substitution and Elimination methods will be asked. Do NOT waste time practising cross-multiplication.

📝 SECTION A — 2-MARK QUESTIONS

15 Questions • Very Short Answer (VSA) Type • Board PYQs 2018–2025

CBSE 2025

Solve the following pair of equations algebraically: 101x + 102y = 304 ; 102x + 101y = 305

2 Marks Add & Subtract Trick ⭐ MOST RECENT

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✅ DETAILED SOLUTION

Step 1 Add both equations:

203x + 203y = 609 ⟹ x + y = 3 ... (iii)

Step 2 Subtract eq.(i) from eq.(ii):

x – y = 1 ... (iv)

Step 3 Adding (iii) and (iv): 2x = 4 ⟹ x = 2

Step 4 From (iii): y = 3 – 2 = 1

∴ x = 2, y = 1

CBSE 2025

In a pair of supplementary angles, the greater angle exceeds the smaller by 50°. Find the measure of each angle.

2 Marks Geometry Word Problem ⭐ MOST RECENT

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✅ DETAILED SOLUTION

Step 1 Let greater angle = x°, smaller angle = y°

Step 2 Supplementary: x + y = 180 ... (i)

Step 3 Given: x – y = 50 ... (ii)

Step 4 Adding (i) and (ii): 2x = 230 ⟹ x = 115°

Step 5 y = 180 – 115 = 65°

∴ The angles are 115° and 65°

CBSE 2024

If 2x + y = 13 and 4x – y = 17, find the value of (x – y).

2 Marks Elimination

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✅ DETAILED SOLUTION

Step 1 2x + y = 13 ... (i) ; 4x – y = 17 ... (ii)

Step 2 Add (i) and (ii): 6x = 30 ⟹ x = 5

Step 3 Put x = 5 in (i): 10 + y = 13 ⟹ y = 3

∴ x – y = 5 – 3 = 2

CBSE 2024

Solve the following pair of linear equations: x + 2y = 9 and y – 2x = 2

2 Marks Elimination

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✅ DETAILED SOLUTION

Step 1 x + 2y = 9 ... (i) ; –2x + y = 2 ... (ii)

Step 2 Multiply (i) by 2: 2x + 4y = 18 ... (iii)

Step 3 Add (ii) and (iii): 5y = 20 ⟹ y = 4

Step 4 Put y = 4 in (i): x + 8 = 9 ⟹ x = 1

∴ x = 1, y = 4

CBSE 2024

Find the value of k for which the pair of linear equations 5x + 2y – 7 = 0 and 2x + ky + 1 = 0 have no solution.

2 Marks Consistency / No Solution ⭐ REPEATED PATTERN

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✅ DETAILED SOLUTION

No solution condition: a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Step 1 a₁/a₂ = 5/2 ; b₁/b₂ = 2/k ; c₁/c₂ = –7/1 = –7

Step 2 For no solution: 5/2 = 2/k ⟹ k = 4/5

Verify c₁/c₂ = –7 ≠ 5/2 ✓ (condition satisfied)

∴ k = 4/5

CBSE 2024

Check whether the point (–4, 3) lies on both the lines represented by x + y + 1 = 0 and x – y = 1.

2 Marks Point Verification ⭐ COMPETENCY BASED

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✅ DETAILED SOLUTION

Step 1 Check in eq. (i): x + y + 1 = –4 + 3 + 1 = 0 ✓ (Satisfies)

Step 2 Check in eq. (ii): x – y = –4 – 3 = –7 ≠ 1 ✗ (Does NOT satisfy)

Alt The intersection point: Adding equations gives 2x + 1 = 1 ⟹ x = 0, y = –1. So intersection is (0, –1).

∴ The point (–4, 3) does NOT lie on both lines.

CBSE 2022

If 17x – 19y = 53 and 19x – 17y = 55, find the value of (x + y).

2 Marks Add & Subtract Trick

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✅ DETAILED SOLUTION

Step 1 Add both equations: 36x – 36y = 108 ⟹ x – y = 3 ... (iii)

Step 2 Subtract eq.(i) from eq.(ii): 2x + 2y = 2 ⟹ x + y = 1 ... (iv)

Quick method: Adding gives (x – y) = 3, subtracting gives (x + y) = 1. No need to find x, y individually!

∴ x + y = 1

CBSE 2021

Find the value of k for which the pair of linear equations x + y – 4 = 0 and 2x + ky – 3 = 0 have no solution.

2 Marks No Solution Condition

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✅ DETAILED SOLUTION

Step 1 a₁/a₂ = 1/2 ; b₁/b₂ = 1/k ; c₁/c₂ = –4/(–3) = 4/3

Step 2 For no solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂

1/2 = 1/k ⟹ k = 2

Verify 1/2 ≠ 4/3 ✓ (condition satisfied)

∴ k = 2

CBSE 2023

The pair of linear equations 2x = 5y + 6 and 15y = 6x – 18 represents two lines which are: (a) intersecting (b) parallel (c) coincident (d) either intersecting or parallel. Justify your answer.

2 Marks Consistency Check

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✅ DETAILED SOLUTION

Step 1 Rewrite in standard form:

2x – 5y – 6 = 0 ; 6x – 15y – 18 = 0

Step 2 a₁/a₂ = 2/6 = 1/3 ; b₁/b₂ = –5/(–15) = 1/3 ; c₁/c₂ = –6/(–18) = 1/3

Step 3 Since a₁/a₂ = b₁/b₂ = c₁/c₂ = 1/3

∴ The lines are coincident (infinitely many solutions). Answer: (c)

Q10

CBSE 2020

Solve by substitution method: x + y = 14 and x – y = 4

2 Marks Substitution ⭐ CLASSIC TYPE

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✅ DETAILED SOLUTION

Step 1 From (ii): x = y + 4 ... (iii)

Step 2 Substitute in (i): (y + 4) + y = 14 ⟹ 2y = 10 ⟹ y = 5

Step 3 x = 5 + 4 = 9

∴ x = 9, y = 5

Q11

CBSE 2019

Find the value of k for which the pair of equations 2x + 3y = 7 and (k+1)x + (2k–1)y = 4k+1 have infinitely many solutions.

2 Marks Infinite Solutions ⭐ IMPORTANT

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✅ DETAILED SOLUTION

Infinitely many solutions: a₁/a₂ = b₁/b₂ = c₁/c₂

Step 1 2/(k+1) = 3/(2k–1) = 7/(4k+1)

Step 2 From first two: 2(2k–1) = 3(k+1) ⟹ 4k – 2 = 3k + 3 ⟹ k = 5

Verify From last two: 3(4k+1) = 7(2k–1) ⟹ 12k+3 = 14k–7 ⟹ 2k = 10 ⟹ k = 5 ✓

∴ k = 5

Q12

CBSE 2022

Determine whether the pair of linear equations 3x + 2y = 7 and 4x + 8y – 11 = 0 are intersecting, parallel or coincident.

2 Marks Nature of Lines

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✅ DETAILED SOLUTION

Step 1 Standard form: 3x + 2y – 7 = 0 ; 4x + 8y – 11 = 0

Step 2 a₁/a₂ = 3/4 ; b₁/b₂ = 2/8 = 1/4

Step 3 Since a₁/a₂ ≠ b₁/b₂ (3/4 ≠ 1/4)

∴ The lines are intersecting (unique solution).

Q13

CBSE 2018

Find the two numbers whose sum is 75 and difference is 15.

2 Marks Simple Word Problem

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✅ DETAILED SOLUTION

Step 1 Let the numbers be x and y (x > y)

Step 2 x + y = 75 ... (i) ; x – y = 15 ... (ii)

Step 3 Adding: 2x = 90 ⟹ x = 45

Step 4 y = 75 – 45 = 30

∴ The two numbers are 45 and 30.

Q14

CBSE 2023

Find the value of k for which the system of equations 3x – 4y = 7 and 6x – 8y = k has infinitely many solutions.

2 Marks Infinite Solutions

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✅ DETAILED SOLUTION

Step 1 Standard form: 3x – 4y – 7 = 0 ; 6x – 8y – k = 0

Step 2 a₁/a₂ = 3/6 = 1/2 ; b₁/b₂ = –4/(–8) = 1/2 ; c₁/c₂ = –7/(–k) = 7/k

Step 3 For infinite solutions: 1/2 = 1/2 = 7/k ⟹ k = 14

∴ k = 14

Q15

CBSE 2020

Find the value of k for which the system of equations 3x + y = 1 and (2k–1)x + (k–1)y = 2k + 1 has no solution.

2 Marks No Solution

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✅ DETAILED SOLUTION

Step 1 a₁/a₂ = 3/(2k–1) ; b₁/b₂ = 1/(k–1) ; c₁/c₂ = –1/(–(2k+1)) = 1/(2k+1)

Step 2 For no solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂

3/(2k–1) = 1/(k–1) ⟹ 3(k–1) = 2k–1 ⟹ 3k – 3 = 2k – 1 ⟹ k = 2

Verify At k = 2: a₁/a₂ = 3/3 = 1, c₁/c₂ = 1/5. Since 1 ≠ 1/5 ✓

∴ k = 2

📝 SECTION B — 3-MARK QUESTIONS

15 Questions • Short Answer (SA) Type • Board PYQs 2018–2025

Q16

CBSE 2024

Solve the following system of linear equations and verify your answer: 7x – 2y = 5 and 8x + 7y = 15

3 Marks Elimination + Verify

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✅ DETAILED SOLUTION

Step 1 7x – 2y = 5 ... (i) ; 8x + 7y = 15 ... (ii)

Step 2 Multiply (i) by 7 and (ii) by 2:

49x – 14y = 35 ; 16x + 14y = 30

Step 3 Add: 65x = 65 ⟹ x = 1

Step 4 Put x = 1 in (i): 7 – 2y = 5 ⟹ y = 1

Verification in eq.(ii): 8(1) + 7(1) = 8 + 7 = 15 ✓

∴ x = 1, y = 1 (Verified)

Q17

CBSE 2024

Three years ago, Rashmi was thrice as old as Nazma. Ten years later, Rashmi will be twice as old as Nazma. How old are Rashmi and Nazma now?

3 Marks Age Problem ⭐ MOST REPEATED TYPE

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✅ DETAILED SOLUTION

Step 1 Let Rashmi's age = x, Nazma's age = y

Step 2 3 years ago: (x – 3) = 3(y – 3) ⟹ x – 3 = 3y – 9 ⟹ x = 3y – 6 ... (i)

Step 3 10 years later: (x + 10) = 2(y + 10) ⟹ x + 10 = 2y + 20 ⟹ x = 2y + 10 ... (ii)

Step 4 From (i) and (ii): 3y – 6 = 2y + 10 ⟹ y = 16

Step 5 x = 3(16) – 6 = 42

∴ Rashmi is 42 years old and Nazma is 16 years old.

Q18

CBSE 2023

5 students of Hockey and 4 students of Cricket together receive a prize amount of ₹9500. 4 students of Hockey and 3 students of Cricket together receive ₹7370. Find the prize amount given per student for each game.

3 Marks Real-Life Application ⭐ COMPETENCY BASED

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✅ DETAILED SOLUTION

Step 1 Let Hockey prize per student = ₹x, Cricket = ₹y

Step 2 5x + 4y = 9500 ... (i) ; 4x + 3y = 7370 ... (ii)

Step 3 Multiply (i) by 3 and (ii) by 4:

15x + 12y = 28500 ; 16x + 12y = 29480

Step 4 Subtract: –x = –980 ⟹ x = ₹980

Step 5 Put in (ii): 4(980) + 3y = 7370 ⟹ 3920 + 3y = 7370 ⟹ y = ₹1150

∴ Hockey prize = ₹980 per student, Cricket prize = ₹1150 per student

Q19

CBSE 2022

A fraction becomes 9/11 if 2 is added to both numerator and denominator. If 3 is added to both numerator and denominator, it becomes 5/6. Find the fraction.

3 Marks Fraction Problem ⭐ MOST REPEATED TYPE

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✅ DETAILED SOLUTION

Step 1 Let the fraction be x/y

Step 2 (x+2)/(y+2) = 9/11 ⟹ 11(x+2) = 9(y+2) ⟹ 11x – 9y = –4 ... (i)

Step 3 (x+3)/(y+3) = 5/6 ⟹ 6(x+3) = 5(y+3) ⟹ 6x – 5y = –3 ... (ii)

Step 4 Multiply (i) by 5 and (ii) by 9:

55x – 45y = –20 ; 54x – 45y = –27

Step 5 Subtract: x = 7. Put in (i): 77 – 9y = –4 ⟹ 9y = 81 ⟹ y = 9

∴ The fraction is 7/9

Q20

CBSE 2021

The sum of the digits of a two-digit number is 9. The number obtained by reversing the order of digits exceeds the given number by 27. Find the number.

3 Marks Digit Problem ⭐ CLASSIC

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✅ DETAILED SOLUTION

Step 1 Let tens digit = x, units digit = y

Step 2 Original number = 10x + y ; Reversed number = 10y + x

Step 3 x + y = 9 ... (i)

Step 4 (10y + x) – (10x + y) = 27 ⟹ 9y – 9x = 27 ⟹ y – x = 3 ... (ii)

Step 5 Adding (i) and (ii): 2y = 12 ⟹ y = 6 ; x = 9 – 6 = 3

∴ The number is 10(3) + 6 = 36

Q21

CBSE 2025

A part of monthly hostel charge is fixed and the remaining depends on the number of days one has taken food. When Swati takes food for 20 days, she pays ₹3000. Mansi, who takes food for 25 days, pays ₹3500. Find the fixed charges and cost of food per day.

3 Marks Real-Life Application ⭐ MOST RECENT

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✅ DETAILED SOLUTION

Step 1 Let fixed charge = ₹x, cost of food per day = ₹y

Step 2 x + 20y = 3000 ... (i) ; x + 25y = 3500 ... (ii)

Step 3 Subtract (i) from (ii): 5y = 500 ⟹ y = 100

Step 4 Put y = 100 in (i): x + 2000 = 3000 ⟹ x = 1000

∴ Fixed charge = ₹1000, Cost of food per day = ₹100

Q22

CBSE 2020

Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

3 Marks Age Problem

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✅ DETAILED SOLUTION

Step 1 Let Nuri's age = x, Sonu's age = y

Step 2 5 years ago: x – 5 = 3(y – 5) ⟹ x – 3y = –10 ... (i)

Step 3 10 years later: x + 10 = 2(y + 10) ⟹ x – 2y = 10 ... (ii)

Step 4 Subtract (i) from (ii): y = 20

Step 5 x = 10 + 2(20) = 50

∴ Nuri is 50 years old and Sonu is 20 years old.

Q23

CBSE 2019

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.

3 Marks Fraction Problem

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✅ DETAILED SOLUTION

Step 1 Let the fraction = x/y

Step 2 (x–1)/y = 1/3 ⟹ 3(x–1) = y ⟹ 3x – y = 3 ... (i)

Step 3 x/(y+8) = 1/4 ⟹ 4x = y + 8 ⟹ 4x – y = 8 ... (ii)

Step 4 Subtract (i) from (ii): x = 5

Step 5 y = 3(5) – 3 = 12

∴ The fraction is 5/12

Q24

CBSE 2023

Solve the following pair of equations by elimination method: 3x + 2y = 11 and 2x + 3y = 4

3 Marks Elimination

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✅ DETAILED SOLUTION

Step 1 3x + 2y = 11 ... (i) ; 2x + 3y = 4 ... (ii)

Step 2 Multiply (i) by 3 and (ii) by 2:

9x + 6y = 33 ... (iii) ; 4x + 6y = 8 ... (iv)

Step 3 Subtract (iv) from (iii): 5x = 25 ⟹ x = 5

Step 4 Put x = 5 in (i): 15 + 2y = 11 ⟹ 2y = –4 ⟹ y = –2

∴ x = 5, y = –2

Q25

CBSE 2021

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.

3 Marks Geometry Word Problem

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✅ DETAILED SOLUTION

Step 1 Let length = x m, width = y m

Step 2 Half perimeter: x + y = 36 ... (i)

Step 3 Length is 4 m more: x = y + 4 ⟹ x – y = 4 ... (ii)

Step 4 Adding (i) and (ii): 2x = 40 ⟹ x = 20

Step 5 y = 36 – 20 = 16

∴ Length = 20 m, Width = 16 m

Q26

CBSE 2022

The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes 1/2. Find the fraction.

3 Marks Fraction Problem

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✅ DETAILED SOLUTION

Step 1 Let fraction = x/y

Step 2 x + y = 12 ... (i)

Step 3 x/(y + 3) = 1/2 ⟹ 2x = y + 3 ⟹ 2x – y = 3 ... (ii)

Step 4 Adding (i) and (ii): 3x = 15 ⟹ x = 5

Step 5 y = 12 – 5 = 7

∴ The fraction is 5/7

Q27

CBSE 2019

Solve for x and y: 2/x + 3/y = 13 and 5/x – 4/y = –2 (where x ≠ 0, y ≠ 0)

3 Marks Reducible Equations ⭐ IMPORTANT TYPE

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✅ DETAILED SOLUTION

Step 1 Substitute: Let 1/x = a and 1/y = b

Step 2 2a + 3b = 13 ... (i) ; 5a – 4b = –2 ... (ii)

Step 3 Multiply (i) by 4 and (ii) by 3:

8a + 12b = 52 ; 15a – 12b = –6

Step 4 Add: 23a = 46 ⟹ a = 2, so 1/x = 2 ⟹ x = 1/2

Step 5 From (i): 4 + 3b = 13 ⟹ b = 3, so 1/y = 3 ⟹ y = 1/3

∴ x = 1/2, y = 1/3

Q28

CBSE 2018

The sum of a two-digit number and the number obtained by reversing the order of its digits is 99. If the digits differ by 3, find the number.

3 Marks Digit Problem

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✅ DETAILED SOLUTION

Step 1 Let tens digit = x, units digit = y

Step 2 Number + Reversed = (10x+y) + (10y+x) = 11(x+y) = 99 ⟹ x + y = 9 ... (i)

Step 3 |x – y| = 3 ⟹ x – y = 3 or x – y = –3

Case 1 x – y = 3 and x + y = 9 ⟹ 2x = 12 ⟹ x = 6, y = 3. Number = 63

Case 2 x – y = –3 and x + y = 9 ⟹ 2x = 6 ⟹ x = 3, y = 6. Number = 36

∴ The number is 63 or 36

Q29

CBSE 2020

The cost of 2 kg apples and 1 kg grapes is ₹160. The cost of 4 kg apples and 2 kg grapes is ₹300. Represent the situation as a pair of linear equations and determine whether the system is consistent.

3 Marks Consistency + Word Problem

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✅ DETAILED SOLUTION

Step 1 Let cost of 1 kg apples = ₹x, grapes = ₹y

Step 2 2x + y = 160 ... (i) ; 4x + 2y = 300 ... (ii)

Step 3 a₁/a₂ = 2/4 = 1/2 ; b₁/b₂ = 1/2 ; c₁/c₂ = –160/(–300) = 8/15

Step 4 Since a₁/a₂ = b₁/b₂ = 1/2 but ≠ c₁/c₂ = 8/15

Interpretation: The lines are parallel ⟹ no solution. This means the given prices are inconsistent (not possible simultaneously).

∴ The system is INCONSISTENT (no solution). The lines are parallel.

Q30

CBSE 2018

4 men and 6 boys can finish a piece of work in 5 days while 3 men and 4 boys can finish it in 7 days. Find the time taken by 1 man alone and by 1 boy alone to finish the work.

3 Marks Work-Time Problem ⭐ CHALLENGING

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✅ DETAILED SOLUTION

Step 1 Let 1 man's 1 day work = 1/x, 1 boy's 1 day work = 1/y

Step 2 4/x + 6/y = 1/5 ... (i) ; 3/x + 4/y = 1/7 ... (ii)

Step 3 Let 1/x = a, 1/y = b

4a + 6b = 1/5 ... (iii) ; 3a + 4b = 1/7 ... (iv)

Step 4 Multiply (iii) by 4 and (iv) by 6:

16a + 24b = 4/5 ; 18a + 24b = 6/7

Step 5 Subtract: –2a = 4/5 – 6/7 = (28 – 30)/35 = –2/35 ⟹ a = 1/35

Step 6 So x = 35. From (iv): 3/35 + 4b = 1/7 ⟹ 4b = 1/7 – 3/35 = (5–3)/35 = 2/35 ⟹ b = 1/70

Step 7 y = 70

∴ 1 man alone = 35 days, 1 boy alone = 70 days

📐 KEY CONDITIONS TO REMEMBER

For a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0:
• Unique Solution (Intersecting): a₁/a₂ ≠ b₁/b₂
• Infinite Solutions (Coincident): a₁/a₂ = b₁/b₂ = c₁/c₂
• No Solution (Parallel): a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Methods Allowed (CBSE 2025-26):
• Substitution Method: Express one variable, substitute in other equation
• Elimination Method: Make coefficients equal, add/subtract to eliminate
• Graphical Method: Plot both lines, intersection = solution
• ❌ Cross-Multiplication: DELETED — Not in syllabus 2025-26

💡 BOARD EXAM TIPS — LINEAR EQUATIONS

For word problems: Always define variables clearly ("Let x = ..., y = ...") — marks are given for this step!

Verify your answer by substituting back in BOTH equations — it helps catch errors and earns marks.

Add & Subtract trick: When coefficients are symmetric (like 101, 102), add/subtract equations first.

Reducible equations (1/x, 1/y type): Always substitute a = 1/x, b = 1/y first to convert to linear form.

Expected marks: 6–8 marks from this chapter (MCQ + SA + Word Problem).

Practice these 5 types: Age, Fraction, Digit, Speed-Distance, and Geometry word problems.