Curriculum
Mathematics Grade 10
Grade 10 Mathematics
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Chapter 1. Real Numbers
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Chapter 2. Polynomials
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Chapter 3: PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
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Chapter 4: QUADRATIC EQUATIONS
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Chapter 5: ARITHMETIC PROGRESSIONS
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Chapter 6: TRIANGLES
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Chapter 7: COORDINATE GEOMETRY
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Chapter 8: INTRODUCTION TO TRIGONOMETRY
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Chapter 9: SOME APPLICATIONS OF TRIGONOMETRY
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Chapter 10: CIRCLES
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Chapter 11: AREAS RELATED TO CIRCLES
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Chapter 12: SURFACE AREAS AND VOLUMES
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Chapter 13: STATISTICS
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Chapter 14: PROBABILITY
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Interactive Worksheets
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Chapter Wise Worksheets
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Solved Problems and solutions
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Chapter 3: PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
Grade 10 · Mathematics · Chapter 3
Pair of Linear Equations
in Two Variables
Master graphical, substitution and elimination methods with 14 worked examples and real-life word problems.
| 📊 Graphical | 🔄 Substitution | ➕ Elimination | ⚖️ Ratio Test |
📚 What You Will Learn
| ✦ Introduction to pairs of linear equations | ✦ Three types of line behaviour on a graph |
| ✦ Ratio test — classify without solving | ✦ Graphical method with coordinate diagram |
| ✦ Substitution method step by step | ✦ Elimination method step by step |
| ✦ Worked Examples A – N (14 solved) | ✦ Real-life word problems |
| ✦ Practice Sets A – D (24 questions) | ✦ Chapter summary and exam quick-check |
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📐
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Section 1 What Is a Pair of Linear Equations? |
A linear equation in two variables has the form ax + by + c = 0. Its graph is always a straight line. Two such equations considered together form a pair of linear equations:
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
A solution is a pair (x, y) satisfying both equations — on a graph, it is the intersection point of the two lines.
💡 Real-Life Example
Riya visits a fair. Each ride costs ₹3 and each game costs ₹4. She spends ₹20 in total, and the number of games equals half the rides. Writing two equations and solving them simultaneously — that is exactly what this chapter is about.
📊 Three Ways Two Lines Can Behave on a Graph
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Case 1 ● Meeting point = Solution ✅ One unique solution Consistent |
Case 2 Both lines lie exactly ⚠️ Infinitely many solutions Dependent |
Case 3 Lines run side by side ❌ No solution Inconsistent |
Intersecting → one solution | Coincident → infinite solutions | Parallel → no solution
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⚖️
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Section 2 The Ratio Test — Classify Without Solving |
For a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, compare three ratios to instantly know the type of pair — no solving needed.
| Ratio Condition | Lines | Solutions | Pair Type |
|---|---|---|---|
| a₁/a₂ ≠ b₁/b₂ | Intersecting | Exactly one ✅ | Consistent |
| a₁/a₂ = b₁/b₂ = c₁/c₂ | Coincident | Infinite ⚠️ | Dependent |
| a₁/a₂ = b₁/b₂ ≠ c₁/c₂ | Parallel | None ❌ | Inconsistent |
WORKED EXAMPLE A
Use the ratio test to classify these pairs — without solving.
a₁/a₂ = 2, b₁/b₂ = 2, c₁/c₂ = 2 — all equal
→ Coincident — Infinitely many solutions ⚠️
(ii) 3x + 2y − 5 = 0 and 4x − y − 6 = 0
a₁/a₂ = 3/4, b₁/b₂ = −2 — not equal
→ Intersecting — Unique solution ✅
(iii) x + 4y − 6 = 0 and 2x + 8y − 18 = 0
a₁/a₂ = 1/2 = b₁/b₂ (equal), c₁/c₂ = 6/18 = 1/3 (different)
→ Parallel — No solution ❌
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📈
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Section 3 Graphical Method |
Draw both equations as lines on the same graph. Their intersection point is the solution.
📋 Four Steps
Step 1: Find two (x, y) points that satisfy each equation.
Step 2: Plot and draw both straight lines on the same grid.
Step 3: Intersect once = unique; same line = infinite; never meet = none.
Step 4: Read the intersection and verify in both original equations.
Graph of x + 3y = 6 and 2x − 3y = 12 — Meeting at B(6, 0)
| y=3 | |||||||
| y=2 ◀A | ● | ||||||
| y=1 | |||||||
| y=0 | x=0 | x=1 | x=2 | x=3 | x=4 | x=5 | ● B(6,0) |
| y=−1 | Q(3,−2)→ |
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━━ x + 3y = 6 Points: A(0, 2) and B(6, 0) |
━━ 2x − 3y = 12 Points: P(0, −4) and B(6, 0) |
WORKED EXAMPLE B
Solve graphically: x + 3y = 6 and 2x − 3y = 12
2x − 3y = 12: x=0 → y=−4 (Point P) | x=3 → y=−2 (Point Q)
Both lines pass through B(6, 0). Verify: 6+3(0)=6 ✓ 2(6)−3(0)=12 ✓
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x = 6
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y = 0
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WORKED EXAMPLE C — Infinitely Many Solutions
Show that 5x − 8y + 1 = 0 and 15x − 24y + 3 = 0 have infinitely many solutions.
This is identical to the first equation. Both are the same line.
Every point on the line satisfies both equations — infinitely many solutions.
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🔄
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Section 4 Substitution Method |
Express one variable from one equation, substitute it into the other, then solve.
📋 Steps
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1
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Isolate one variable from either equation (prefer coefficient = 1). |
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2
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Substitute into the other equation — now one variable only. |
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3
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Solve for that variable. If 0=0 → infinite; 0=number → no solution. |
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4
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Back-substitute to find the other variable. Verify in both equations. |
WORKED EXAMPLE D
Solve: 7x − 15y = 2 and x + 2y = 3
Substitute: 7(3−2y) − 15y = 2 → 21 − 14y − 15y = 2 → −29y = −19 → y = 19/29
Back-sub: x = 3 − 2(19/29) = 87/29 − 38/29 = x = 49/29
Verify: 7(49/29) − 15(19/29) = 343/29 − 285/29 = 58/29 = 2 ✓
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x = 49/29
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y = 19/29
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WORKED EXAMPLE E — Ages
Seven years ago a father was 7× his son’s age. In 3 years he will be 3× his son’s age. Find present ages.
s − 7 = 7(t − 7) → s − 7t = −42 …(1)
s + 3 = 3(t + 3) → s − 3t = 6 …(2)
From (2): s = 3t + 6. Sub in (1): 3t + 6 − 7t = −42 → −4t = −48 → t = 12
s = 3(12)+6 = 42 Verify: 35=7×5 ✓ 45=3×15 ✓
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Father = 42 years
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Son = 12 years
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WORKED EXAMPLE F — Infinite Solutions
2 pencils + 3 erasers = ₹9 and 4 pencils + 6 erasers = ₹18. Find individual costs.
From (1): x = (9−3y)/2. Sub in (2): 4·(9−3y)/2 + 6y = 18 → 18−6y+6y = 18 → 18 = 18
Always true — equations are identical. No unique cost can be determined.
WORKED EXAMPLE G — No Solution
Two roads: x + 2y − 4 = 0 and 2x + 4y − 12 = 0. Will they ever meet?
Contradiction — the roads are parallel and will never meet.
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➕
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Section 5 Elimination Method |
Multiply one or both equations to make one variable’s coefficient equal, then add or subtract to remove it.
📋 Steps
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1
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Multiply to make one variable’s coefficients equal in both equations. |
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2
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Add or subtract to eliminate that variable completely. |
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3
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Solve the remaining one-variable equation. 0=0 → infinite; 0=c → no solution. |
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4
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Substitute back to find the other variable. Verify in both equations. |
Elimination Flow
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📋 Two Equations a₁x+b₁y=c₁ |
→ |
✖️ Multiply Match one |
→ |
➕ Add/Subtract Eliminate |
→ |
✅ Solve & Verify Find (x, y) |
WORKED EXAMPLE H — Incomes
Two friends earn in ratio 9:7 and spend in ratio 4:3. Each saves ₹2000/month. Find their incomes.
9x − 4y = 2000 …(1) 7x − 3y = 2000 …(2)
(1)×3: 27x−12y=6000 | (2)×4: 28x−12y=8000 → Subtract: x = 2000
Sub in (1): 18000−4y=2000 → y = 4000
Incomes: 9×2000 = ₹18,000 and 7×2000 = ₹14,000
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Person A = ₹18,000
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Person B = ₹14,000
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WORKED EXAMPLE I — No Solution
Show that 2x + 3y = 8 and 4x + 6y = 7 have no solution.
Subtract eq(2) from (3): 0 = 9 — always false. No solution.
WORKED EXAMPLE J — Number Puzzle
A two-digit number reversed and added to itself gives 66. Digits differ by 2. Find the number.
(10x+y)+(10y+x) = 66 → 11(x+y)=66 → x+y=6 …(1)
x−y=2 …(2)
Add: 2x=8 → x=4 | Sub: 2y=4 → y=2 → Number = 42
Verify: 42+24=66 ✓ 4−2=2 ✓
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Number: 42
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Reversed: 24
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🌍
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Section 6 Word Problems — Step by Step |
🔑 Golden Rules
1. Assign x and y to the two unknowns clearly.
2. Each number fact in the question gives one equation.
3. Write neatly, solve, then verify using the original words.
WORKED EXAMPLE K — Shopping
7 bats + 6 balls = ₹3800 and 3 bats + 5 balls = ₹1750. Find cost of each.
7x+6y=3800 …(1) 3x+5y=1750 …(2)
(1)×5: 35x+30y=19000 | (2)×6: 18x+30y=10500 → Subtract: 17x=8500 → x=500
Sub: 6y=300 → y=50
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1 Bat = ₹500
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1 Ball = ₹50
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WORKED EXAMPLE L — Taxi Fare
10 km costs ₹105. 15 km costs ₹155. Find fixed charge, per-km rate, and cost for 25 km.
a+10b=105 …(1) a+15b=155 …(2)
Subtract: 5b=50 → b=10 Sub: a=5
25 km: 5+25×10 = ₹255
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Fixed = ₹5
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Per km = ₹10
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25 km = ₹255
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WORKED EXAMPLE M — Fraction
Add 2 to numerator and denominator → 9/11. Add 3 to both → 5/6. Find the original fraction.
11(p+2)=9(q+2) → 11p−9q=−4 …(1)
6(p+3)=5(q+3) → 6p−5q=−3 …(2)
(1)×5: 55p−45q=−20 | (2)×9: 54p−45q=−27 → Subtract: p=7
9q=77+4=81 → q=9 Verify: 9/11 ✓ 10/12=5/6 ✓
WORKED EXAMPLE N — Bank Withdrawal
Meena withdraws ₹2000 using exactly 25 notes of ₹50 and ₹100. How many of each?
x+y=25 …(1) 50x+100y=2000 → x+2y=40 …(2)
Subtract (1) from (2): y=15 → x=10
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₹50 notes: 10
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₹100 notes: 15
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🏋️
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Section 7 Practice Exercises with Answers |
Set A — Classify Using the Ratio Test
Without solving, determine if each pair is consistent, inconsistent, or dependent.
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(1) 5x − 4y + 8 = 0 and 7x + 6y − 9 = 0
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(2) 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0
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(3) 6x − 3y + 10 = 0 and 2x − y + 9 = 0
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(4) 3x + 2y = 5 and 2x − 3y = 7
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(5) 2x − 3y = 8 and 4x − 6y = 9
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(6) 5x − 3y = 11 and −10x + 6y = −22
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📋 Reveal Answers ▼
(4) Consistent — unique | (5) Inconsistent — no solution | (6) Dependent — infinite
Set B — Solve by Substitution
Show all working. Verify in both equations.
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(1) x + y = 14 and x − y = 4
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(2) s − t = 3 and s/3 + t/2 = 6
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(3) 3x − y = 3 and 9x − 3y = 9
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(4) 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3
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(5) √2·x + √3·y = 0 and √3·x − √8·y = 0
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(6) 2x + 3y = 11 and 2x − 4y = −24
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📋 Reveal Answers ▼
(4) x=2, y=3 | (5) x=0, y=0 | (6) x=−2, y=5
Set C — Solve by Elimination
State if the pair is inconsistent or dependent where applicable.
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(1) x + y = 5 and 2x − 3y = 4
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(2) 3x + 4y = 10 and 2x − 2y = 2
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(3) 3x − 5y − 4 = 0 and 9x = 2y + 7
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(4) x/2 + 2y/3 = −1 and x − y/3 = 3
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(5) 2x + 3y = 8 and 4x + 6y = 7
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(6) 5x + 2y = 16 and 3x − 4y = −2
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📋 Reveal Answers ▼
(4) x=2, y=−3 | (5) No solution | (6) x=2, y=3
Set D — Word Problems
Form two equations and solve. Always verify your answer.
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(1) A class of 30 students has twice as many girls as boys. How many boys and girls?
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(2) 5 pens + 3 notebooks = ₹70 and 3 pens + 5 notebooks = ₹74. Find cost of each.
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(3) Two numbers add to 48 and their difference is 16. Find both numbers.
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(4) Two complementary angles differ by 22°. Find both angles.
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(5) A library charges a fixed fee for the first 3 days then ₹3/day extra. Siya paid ₹27 for 7 days; Monu paid ₹21 for 5 days. Find the fixed fee and daily rate.
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(6) A rectangle has half-perimeter 36 m and its length is 4 m more than its width. Find both dimensions.
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📋 Reveal Answers ▼
(4) 56° and 34° | (5) Fixed fee=₹15, Extra=₹3/day | (6) Length=20m, Width=16m
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📌
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Section 8 Chapter Summary |
🔑 Everything on One Page
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📐 What Is a Pair Two linear equations together. A solution (x, y) satisfies both simultaneously. |
📊 Graphical Draw both lines. Cross = unique. Same line = infinite. Never cross = none. |
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⚖️ Ratio Test a₁/a₂ ≠ b₁/b₂ → unique | all equal → infinite | two equal, third ≠ → none |
🔄 Substitution Isolate → substitute → solve → back-sub → verify. |
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➕ Elimination Match coefficient → add/subtract to remove variable → solve → verify. |
📝 Word Problems Name unknowns → 2 conditions → 2 equations → solve → check in words. |
✅ 8-Point Quick-Check Before Your Exam
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1A solution must satisfy both equations — check in both every time.
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2Graphically, the solution is the intersection point of both lines.
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3Use the ratio test to classify instantly — compare a₁/a₂, b₁/b₂, c₁/c₂.
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4Consistent = solution exists. Inconsistent = none. Dependent = infinite.
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5In substitution, isolate the variable with coefficient 1 first to keep it simple.
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6In elimination: 0 = 0 → infinite solutions; 0 = number → no solution.
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7For word problems, define both unknowns clearly before writing any equation.
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8Graphical method works best for integer answers; algebra works for all cases.
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Grade 10 Maths Chapter 3 — Pair of Linear Equations in Two Variables
Topics covered: consistent and inconsistent pairs, graphical method, substitution method, elimination method, ratio test, classification of linear equation pairs, word problems, worked examples A to N. Ideal for CBSE Class 10 board exam revision and preparation.