Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Notes

📅 Published on Friday, April 17, 2026 👤 By Rathna CBSE Tutor
Student Mathematics Blog · Grade 10

Mastering Pairs of Linear Equations in Two Variables

Pair of Linear Equations Overview: To solve a pair of linear equations in two variables, first compare coefficient ratios (a₁/a₂, b₁/b₂, c₁/c₂). This determines if lines intersect (consistent), coincide (infinite solutions), or are parallel (no solution).

A professor-guided walkthrough with 15 progressively challenging problems across every method — graphical, substitution, and elimination.

Sections in this guide
§1 — The Graphical Method & Consistency Ratios §2 — The Substitution Method §3 — The Elimination Method
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Section 1

The Graphical Method & Consistency Ratios

Professor's Commentary: Before you ever pick up a pencil to draw a graph, check the ratios of coefficients. This simple comparison tells you the "destiny" of the two lines — whether they will intersect (one unique answer), coincide (infinite answers), or run parallel forever (no answer). Master this diagnostic step, and you will save yourself enormous effort.
1
Determine graphically whether the system x - 2y = 0 and 3x + 4y - 20 = 0 is consistent. If so, find its solution.
Foundational ▾
Step 1
Identify the coefficients from the general form a₁x + b₁y + c₁ = 0:
Eq. 1: a₁=1, b₁=-2, c₁=0  |  Eq. 2: a₂=3, b₂=4, c₂=-20
Step 2
Compare the coefficient ratios:
a₁/a₂ = 1/3    and    b₁/b₂ = -2/4 = -1/2
Since 1/3 ≠ -1/2, the lines must intersect — the system is consistent with a unique solution.
Step 3
Build a two-point table for each line:
Eq. 1 (y = x/2):   (0, 0) and (4, 2) Eq. 2 (y = (20-3x)/4): (0, 5) and (4, 2)
Step 4
Plot both lines. They share the point (4, 2) — this is the unique intersection.
Verification ✓ Plug (4, 2) into both equations:
Eq. 1: 4 - 2(2) = 4 - 4 = 0 ✓
Eq. 2: 3(4) + 4(2) - 20 = 12 + 8 - 20 = 0 ✓
Answer Consistent. Unique solution: x = 4, y = 2
2
Without graphing, classify: 5x - 4y + 8 = 0 and 7x + 6y - 9 = 0.
Basic ▾
Step 1
Extract coefficients: a₁=5, b₁=-4 / a₂=7, b₂=6
Step 2
Compare ratios:
a₁/a₂ = 5/7    and    b₁/b₂ = -4/6 = -2/3
Step 3
Since 5/7 ≠ -2/3, the ratios differ → the lines intersect.
Answer Intersecting lines — consistent (unique solution).
3
A rectangular garden has half-perimeter 36 m, and its length is 4 m more than its width. Set up and solve the system graphically.
Intermediate ▾
Step 1
Let length = x, width = y. Translate the conditions:
Condition 1: x = y + 4 → x - y = 4 Condition 2: x + y = 36
Step 2
Check ratios: a₁/a₂ = 1/1 = 1, b₁/b₂ = -1/1 = -1. Since 1 ≠ -1 → intersecting, consistent.
Step 3
Two points for each line:
x - y = 4: (4, 0) and (0, -4) x + y = 36: (0, 36) and (20, 16)
Step 4
The lines cross at (20, 16).
Answer Length = 20 m, Width = 16 m.
4
Determine whether the rails x + 2y - 4 = 0 and 2x + 4y - 12 = 0 will ever cross each other.
Intermediate ▾
Step 1
Identify: a₁=1, b₁=2, c₁=-4 / a₂=2, b₂=4, c₂=-12
Step 2
Compute all three ratios:
a₁/a₂ = 1/2 b₁/b₂ = 2/4 = 1/2 c₁/c₂ = -4/-12 = 1/3
Step 3
We find a₁/a₂ = b₁/b₂ ≠ c₁/c₂, i.e., 1/2 = 1/2 ≠ 1/3. This is the parallel condition — the lines never meet.
Answer No solution — the rails are parallel and will never cross.
5
5 pencils and 7 pens cost ₹ 50; 7 pencils and 5 pens cost ₹ 46. Verify graphically that the system is consistent and find both costs.
Challenging ▾
Step 1
Let pencil cost = x (₹), pen cost = y (₹).
Equation 1: 5x + 7y = 50 Equation 2: 7x + 5y = 46
Step 2
Check ratios: a₁/a₂ = 5/7, b₁/b₂ = 7/5. Since 5/7 ≠ 7/5 → consistent, intersecting.
Step 3
Two points per line:
Eq. 1: (10, 0) and (3, 5) Eq. 2: (8, -2) and (3, 5)
Step 4
Both lines pass through (3, 5). Check: 5(3)+7(5)=15+35=50 ✓ and 7(3)+5(5)=21+25=46 ✓
Answer Pencil = ₹ 3, Pen = ₹ 5.
Section 2

The Substitution Method

Professor's Insight: When solutions involve messy non-integer coordinates like 49/29 or -1.75, graphs become unreliable. That's where substitution excels — it is algebraically exact. The logic is simple: Isolate → Substitute → Solve → Back-substitute.
1
Solve: x + y = 14 and x - y = 4.
Foundational ▾
Step 1
Isolate x from the simpler Equation 2:
x - y = 4 → x = y + 4 ... (3)
Step 2
Substitute (3) into Equation 1:
(y + 4) + y = 14 2y + 4 = 14 2y = 10 y = 5
Step 3
Back-substitute y=5 into (3): x = 5 + 4 = 9
Verification ✓ Eq. 1: 9 + 5 = 14 ✓  |  Eq. 2: 9 - 5 = 4 ✓
Answer x = 9, y = 5
2
Solve: 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3.
Basic ▾
Step 1
Clear decimals — multiply both equations by 10:
2x + 3y = 13 ... (1) 4x + 5y = 23 ... (2)
Step 2
Isolate x from (1):
x = (13 - 3y) / 2 ... (3)
Step 3
Substitute (3) into (2):
4 × (13 - 3y)/2 + 5y = 23 2(13 - 3y) + 5y = 23 26 - 6y + 5y = 23 -y = -3 → y = 3
Step 4
Back-substitute: x = (13 - 9)/2 = 4/2 = 2
Answer x = 2, y = 3
3
Aftab tells his daughter: "Seven years ago, I was seven times your age. Three years from now, I will be three times your age." Find their present ages.
Intermediate ▾
Step 1
Define variables: Aftab's age = a, daughter's age = d. Translate conditions:
Seven years ago: a - 7 = 7(d - 7) → a - 7d + 42 = 0 ... (1) Three years hence: a + 3 = 3(d + 3) → a - 3d - 6 = 0 ... (2)
Step 2
Isolate a from Equation (2):
a = 3d + 6 ... (3)
Step 3
Substitute (3) into (1):
(3d + 6) - 7d + 42 = 0 -4d + 48 = 0 4d = 48 d = 12
Step 4
Back-substitute: a = 3(12) + 6 = 42
Answer Aftab = 42 years, Daughter = 12 years.
4
A fraction becomes 9/11 when 2 is added to both numerator and denominator. It becomes 5/6 when 3 is added to both. Find the fraction.
Intermediate ▾
Step 1
Let numerator = x, denominator = y. Set up equations by cross-multiplying:
(x+2)/(y+2) = 9/11 → 11x + 22 = 9y + 18 → 11x - 9y + 4 = 0 ... (1) (x+3)/(y+3) = 5/6 → 6x + 18 = 5y + 15 → 6x - 5y + 3 = 0 ... (2)
Step 2
Isolate x from (1):
x = (9y - 4) / 11 ... (3)
Step 3
Substitute into (2):
6(9y - 4)/11 - 5y = -3 (54y - 24)/11 - 5y = -3 54y - 24 - 55y = -33 -y = -9 → y = 9
Step 4
Back-substitute: x = (81 - 4)/11 = 77/11 = 7
Answer The fraction is 7/9.
5
Use substitution to solve 7x - 15y = 2 and x + 2y = 3. Confirm the solution is not a neat integer.
Challenging ▾
Step 1
Isolate x from the simpler Equation 2:
x = 3 - 2y ... (3)
Step 2
Substitute into Equation 1:
7(3 - 2y) - 15y = 2 21 - 14y - 15y = 2 21 - 29y = 2 -29y = 2 - 21 = -19 y = 19/29
Step 3
Back-substitute:
x = 3 - 2(19/29) x = 3 - 38/29 x = 87/29 - 38/29 x = 49/29
Step 4
This is exactly why the graphical method would fail here — 49/29 ≈ 1.69 is nearly impossible to read off a graph accurately.
Answer x = 49/29, y = 19/29. A non-integer solution — only algebra can find this reliably.
Section 3

The Elimination Method

Professor's Commentary: I find elimination particularly elegant. Rather than isolating a variable, you neutralize it — multiply the equations by well-chosen constants, then add or subtract to make one variable vanish. It is especially powerful when both equations have large coefficients, where substitution becomes algebraically messy.
1
Solve by elimination: x + y = 5 and 2x - 3y = 4.
Foundational ▾
Step 1
Multiply Equation 1 by 3 to match the y-coefficient:
3(x + y) = 3(5) → 3x + 3y = 15 ... (3) Keep Eq. 2: 2x - 3y = 4 ... (2)
Step 2
Add (3) and (2) — the y terms cancel:
(3x + 3y) + (2x - 3y) = 15 + 4 5x = 19 x = 19/5 = 3.8
Step 3
Substitute x=3.8 into Equation 1:
3.8 + y = 5 → y = 1.2
Verification ✓ Eq. 1: 3.8 + 1.2 = 5 ✓
Eq. 2: 2(3.8) - 3(1.2) = 7.6 - 3.6 = 4 ✓
Answer x = 3.8, y = 1.2
2
Determine all solutions of 2x + 3y = 8 and 4x + 6y = 7 using elimination.
Basic ▾
Step 1
Multiply Equation 1 by 2 to match x-coefficients:
4x + 6y = 16 ... (3) 4x + 6y = 7 ... (2)
Step 2
Subtract (2) from (3):
(4x - 4x) + (6y - 6y) = 16 - 7 0 = 9 — a FALSE statement
Step 3
A false statement with no variables means the lines are parallel — there is no point of intersection.
Answer No solution — the system is inconsistent (parallel lines).
3
The sum of digits of a two-digit number is 9. Nine times the number equals twice its digit-reversal. Find the number.
Intermediate ▾
Step 1
Let ten's digit = x, unit's digit = y. Original number = 10x + y. Reversed = 10y + x.
Condition 1: x + y = 9 ... (1) Condition 2: 9(10x + y) = 2(10y + x) 90x + 9y = 20y + 2x 88x - 11y = 0 8x - y = 0 ... (2)
Step 2
Add (1) and (2) — y cancels:
(x + y) + (8x - y) = 9 + 0 9x = 9 → x = 1
Step 3
Substitute x=1 into (1): 1 + y = 9 → y = 8
Answer The number is 18. Check: 9 × 18 = 162 = 2 × 81 ✓
4
Two persons have income ratio 9:7 and expenditure ratio 4:3. Each saves ₹ 2000 per month. Find their monthly incomes.
Intermediate ▾
Step 1
Let incomes = 9x and 7x; expenditures = 4y and 3y. Income − Expenditure = Savings:
9x - 4y = 2000 ... (1) 7x - 3y = 2000 ... (2)
Step 2
Multiply (1) by 3 and (2) by 4 to equate y-coefficients:
27x - 12y = 6000 ... (3) 28x - 12y = 8000 ... (4)
Step 3
Subtract (3) from (4):
(28x - 27x) - (12y - 12y) = 8000 - 6000 x = 2000
Step 4
Substitute x=2000 into (1): 9(2000) - 4y = 2000 → 4y = 16000 → y = 4000
Answer Person A earns ₹ 18,000 (= 9×2000), Person B earns ₹ 14,000 (= 7×2000).
5
A library charges a fixed fee for the first 3 days, then a daily rate. Saritha paid ₹ 27 for 7 days; Susy paid ₹ 21 for 5 days. Find both charges.
Challenging ▾
Step 1
Let fixed charge = x (₹), extra daily charge = y (₹ per day). Extra days = total days − 3.
Saritha (7 days): x + (7-3)y = 27 → x + 4y = 27 ... (1) Susy (5 days): x + (5-3)y = 21 → x + 2y = 21 ... (2)
Step 2
The x-coefficients are already equal (both 1). Subtract (2) from (1):
(x + 4y) - (x + 2y) = 27 - 21 2y = 6 → y = 3
Step 3
Substitute y=3 into (2):
x + 2(3) = 21 → x + 6 = 21 → x = 15
Step 4
Sense-check: Saritha: 15 + 4(3) = 15 + 12 = 27 ✓
Answer Fixed charge = ₹ 15, Extra daily charge = ₹ 3.

Click any problem number to expand its full step-by-step solution. Difficulty increases from Foundational → Basic → Intermediate → Challenging within each section.