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NCERT Class 10 Mathematics — Chapter 3

Pair of Linear Equations in Two Variables

Graphical representation, consistency conditions (intersecting, parallel, coincident), algebraic methods of elimination and substitution, and word problems on speed, upstream-downstream, and age.

Quick Key Takeaways:
General Standard Form: a1x+b1y+c1=0a_1 x + b_1 y + c_1 = 0 and a2x+b2y+c2=0a_2 x + b_2 y + c_2 = 0, where a12+b12≠0a_1^2 + b_1^2 \neq 0 and a22+b22≠0a_2^2 + b_2^2 \neq 0.
Consistency Matrix: (1) a1a2≠b1b2  ⟹  \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \implies Intersecting lines, unique solution, consistent; (2) a1a2=b1b2=c1c2  ⟹  \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \implies Coincident lines, infinitely many solutions, dependent consistent; (3) a1a2=b1b2≠c1c2  ⟹  \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \implies Parallel lines, no solution, inconsistent.
Algebraic Methods: Elimination by equating coefficients, and Substitution method.
Upstream-Downstream Physics: Speed upstream =(u−v)= (u - v) km/h, Speed downstream =(u+v)= (u + v) km/h, where uu is speed in still water and vv is stream current speed.
Graph & Coordinate VisualizerPlot pairs of linear equations, test consistency criteria, and observe intersection points
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Jump to Concept:1. 1. Linear Systems, Graphical Consistency Conditions & Algebraic Foundations2. 2. Methods of Solution & Word Problem Physics Formulations3. 3. Important Solved Board Examination Questions (3-Mark & 5-Mark)4. 4. CBSE Case-Study Modeling: Taxi Fare Structure Modeling in a Metro City5. 5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric
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1. Linear Systems, Graphical Consistency Conditions & Algebraic Foundations

Axioms & Foundational Theory

Comprehensive mathematical foundations, definitions, formulas, and visual conceptual model for Pair of Linear Equations in Two Variables.

• The 3 Geometric Possibilities & Algebraic Ratio Test
•
1. Intersecting Lines (Unique Solution):
a1a2≠b1b2\frac{a_1}{a_2} \neq \frac{b_1}{b_2}
Lines intersect at a single unique coordinate (x,y)(x, y). System is Consistent.
•
2. Coincident Lines (Infinitely Many Solutions):
a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}
Both equations represent the exact same line. System is Consistent and Dependent.
•
3. Parallel Lines (No Solution):
a1a2=b1b2≠c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}
Lines run parallel and never intersect. System is Inconsistent.
📊 Linear Systems: Ratio Consistency & Intersecting LinesVisual Model
1. Intersectinga₁/a₂ ≠ b₁/b₂Unique · Consistent1 Point (x, y)2. Coincidenta₁/a₂ = b₁/b₂ = c₁/c₂Infinite · DependentOverlapping Line3. Parallela₁/a₂ = b₁/b₂ ≠ c₁/c₂No Solution · InconsistentLines Never Meet

Visual schematic mapping the 3 geometric configurations (intersecting, parallel, coincident) with their algebraic ratio criteria.

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2. Methods of Solution & Word Problem Physics Formulations

Theorem Proofs & Derivations

Step-by-step rigorous geometric and algebraic theorem proofs and formula derivations for Pair of Linear Equations in Two Variables.

• Standard Algebraic Solution Algorithms
•
Method of Elimination: Multiply equations by non-zero constants so that the coefficients of one variable become numerically equal; then add or subtract to eliminate that variable.
•
Method of Substitution: Express one variable in terms of the other from one equation and substitute into the second equation.
•
Reducible Equations: In equations like ax+by=c\frac{a}{x} + \frac{b}{y} = c, substitute 1x=u\frac{1}{x} = u and 1y=v\frac{1}{y} = v to convert into standard linear form.
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3. Important Solved Board Examination Questions (3-Mark & 5-Mark)

Topper Step Solutions

Standard CBSE board exam numericals and riders with complete step-by-step mathematical working and justifications.

• 3-Mark Standard Board Question: Solve the pair of linear equations by elimination: 2x+3y=112x + 3y = 11 and 2x−4y=−242x - 4y = -24. Hence, find the value of mm for which y=mx+3y = mx + 3.
•
Step 1 (Equations):
2x+3y=11— (1)2x + 3y = 11 \quad \text{--- (1)}
2x−4y=−24— (2)2x - 4y = -24 \quad \text{--- (2)}
•
Step 2 (Elimination): Subtracting Equation (2) from Equation (1):
(2x−2x)+(3y−(−4y))=11−(−24)(2x - 2x) + (3y - (-4y)) = 11 - (-24)
7y=35  ⟹  y=57y = 35 \implies \mathbf{y = 5}
•
Step 3 (Finding xx): Substituting y=5y = 5 into Equation (1):
2x+3(5)=11  ⟹  2x+15=11  ⟹  2x=−4  ⟹  x=−22x + 3(5) = 11 \implies 2x + 15 = 11 \implies 2x = -4 \implies \mathbf{x = -2}
•
Step 4 (Finding mm): Substitute x=−2,y=5x = -2, y = 5 into y=mx+3y = mx + 3:
5=m(−2)+3  ⟹  −2m=2  ⟹  m=−15 = m(-2) + 3 \implies -2m = 2 \implies \mathbf{m = -1}
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Final Boxed Answer: x=−2,y=5,m=−1\mathbf{x = -2, \quad y = 5, \quad m = -1}
• 5-Mark Heavyweight Board Problem / Rider: A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.
•
Step 1 (Variables Setup):
Let speed of boat in still water =x= x km/h.
Let speed of stream =y= y km/h (x>yx > y).
Speed upstream =(x−y)= (x - y) km/h; Speed downstream =(x+y)= (x + y) km/h.
•
Step 2 (Formulating Equations):
30x−y+44x+y=10— (1)\frac{30}{x-y} + \frac{44}{x+y} = 10 \quad \text{--- (1)}
40x−y+55x+y=13— (2)\frac{40}{x-y} + \frac{55}{x+y} = 13 \quad \text{--- (2)}
•
Step 3 (Substitution u=1x−y,v=1x+yu = \frac{1}{x-y}, v = \frac{1}{x+y}):
30u+44v=10  ⟹  15u+22v=5— (3)30u + 44v = 10 \implies 15u + 22v = 5 \quad \text{--- (3)}
40u+55v=13— (4)40u + 55v = 13 \quad \text{--- (4)}
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Step 4 (Solving for u,vu, v):
Multiply (3) by 8: 120u+176v=40120u + 176v = 40.
Multiply (4) by 3: 120u+165v=39120u + 165v = 39.
Subtracting: 11v=1  ⟹  v=11111v = 1 \implies v = \frac{1}{11}.
From (3): 15u+22(111)=5  ⟹  15u+2=5  ⟹  15u=3  ⟹  u=1515u + 22(\frac{1}{11}) = 5 \implies 15u + 2 = 5 \implies 15u = 3 \implies u = \frac{1}{5}.
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Step 5 (Finding x,yx, y):
x−y=5— (5)x - y = 5 \quad \text{--- (5)}
x+y=11— (6)x + y = 11 \quad \text{--- (6)}
Adding (5) and (6): 2x=16  ⟹  x=8 km/h2x = 16 \implies \mathbf{x = 8\text{ km/h}}.
Subtracting (5) from (6): 2y=6  ⟹  y=3 km/h2y = 6 \implies \mathbf{y = 3\text{ km/h}}.
•
Final Boxed Answer: Speed of boat in still water=8 km/h,Speed of stream=3 km/h\mathbf{\text{Speed of boat in still water} = 8\text{ km/h}, \quad \text{Speed of stream} = 3\text{ km/h}}
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4. CBSE Case-Study Modeling: Taxi Fare Structure Modeling in a Metro City

Case Study Mastery (4 Marks)

Real-world mathematical modeling scenario with structured multi-part questions and full step solutions.

• Practical Application Context: Taxi Fare Structure Modeling in a Metro City
The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155.
•
Q1: Formulate the pair of linear equations representing this situation. →\rightarrow Let fixed charge =x= x and per-km charge =y= y. Equations: x+10y=105x + 10y = 105 and x+15y=155x + 15y = 155.
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Q2: Find the fixed charge and the charge per kilometer. →\rightarrow Subtracting: 5y=50  ⟹  y=105y = 50 \implies y = 10 Rs/km. From first: x+10(10)=105  ⟹  x=5x + 10(10) = 105 \implies x = 5 Rs. Fixed charge =Rs 5= \mathbf{\text{Rs } 5}, per-km charge =Rs 10= \mathbf{\text{Rs } 10}.
•
Q3: How much does a person have to pay for travelling a distance of 25 km? →\rightarrow Total fare =x+25y=5+25(10)=5+250=Rs 255= x + 25y = 5 + 25(10) = 5 + 250 = \mathbf{\text{Rs } 255}.
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5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

Important Solved Board Questions

Examiner step-marking allocations, common algebraic/arithmetic deduction traps, and presentation guidelines.

• Step-by-Step Marking Rubric & Presentation Guidelines
•
1 Mark: Defining variables and setting up correct algebraic equations.
•
1 Mark: Applying substitution u=1/(x−y),v=1/(x+y)u = 1/(x-y), v = 1/(x+y) cleanly.
•
2 Marks: Step-by-step solving to obtain u,vu, v and subsequently x,yx, y.
•
1 Mark: Stating the final result with appropriate physical units (km/h).
• Common Calculation Traps & Verification Checklist
•
Trap 1: Forgetting to convert equations into standard form ax+by+c=0ax + by + c = 0 before taking ratios a1/a2,b1/b2,c1/c2a_1/a_2, b_1/b_2, c_1/c_2.
•
Trap 2: In speed problems, assuming upstream speed is y−xy - x instead of x−yx - y (boat speed must exceed stream speed).
•
Trap 3: Forgetting to find the final variable requested (e.g. mm in y=mx+3y = mx + 3).
Authentic Board Question (3 Marks)Topic: Pair of Linear Equations in Two Variables Algebraic Formulations & Exact Methods
Solve the algebraic equation, find unknown coefficients, and apply standard theorem methods for Pair of Linear Equations in Two Variables.

Official CBSE Step-by-Step Marking Breakdown:

Step 1: Standard Form Equation & Algebraic Setup: Convert given conditions into standard algebraic form (ax2+bx+c=0ax^2+bx+c=0, an=a+(n−1)da_n = a+(n-1)d, or system of equations).
1 Mark
Step 2: Step-by-Step Factorisation / Reduction Method: Apply formal algebraic method (splitting the middle term, quadratic formula, elimination, or AP summation).
1 Mark
Step 3: Final Solution & Domain Verification: State the final values of x,n,dx, n, d, verifying against real-life boundary conditions (e.g. n>0n > 0, positive speeds).
1 Mark
Model Student Answer (Target: Full 3/3 Marks):
To score full 3 marks on Pair of Linear Equations in Two Variables in CBSE Mathematics:

1. Algebraic Setup: Express the given problem into standard algebraic form.
2. Solving Technique: Carry out step-by-step algebraic manipulation showing all factorization or formula substitution lines.
3. Boxed Result: State and box the final root values, discarding any non-viable negative or fractional answers where context dictates.
Examiner Mark Deduction Traps:
•Always check boundary conditions (e.g., number of terms nn in an AP must be a positive natural number).
•Show all factorization steps clearly—do not jump directly from equation to roots.

High-Frequency Conceptual Doubts & FAQs

Curated answers to the most common questions asked by Class 10 students.
For a1x+b1y+c1=0a_1x + b_1y + c_1 = 0 and a2x+b2y+c2=0a_2x + b_2y + c_2 = 0: (1) Unique solution (intersecting lines, consistent): a1a2≠b1b2\frac{a_1}{a_2} \neq \frac{b_1}{b_2}. (2) Infinitely many solutions (coincident lines, consistent/dependent): a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}. (3) No solution (parallel lines, inconsistent): a1a2=b1b2≠c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}.

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