Chapter 3: Pair of Linear Equations in Two Variables

Overview

Study pairs of linear equations in two variables. Master graphical consistency ratios (\(\frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2}\)), algebraic elimination and substitution techniques, and high-yield real-world modeling questions.

Graphical Method • Consistency Conditions • Substitution • Elimination • Word Problems

Exam Weightage & Blueprint

Total: 5-6 Marks

This chapter falls under Unit II: Algebra (20 marks total) in the 2026–27 syllabus. Focus areas include graphical solutions, consistency conditions, and algebraic methods (substitution and elimination) for solving situational problems.

Question Type Marks Frequency Focus Topic
MCQ 1 High Consistency Conditions ($a_1/a_2$...)
Short Answer 2 or 3 Medium Substitution/Elimination Method
Word Problem 3 or 5 High Ages, Digits, Fractions

CBSE 2026–27 Syllabus Status: Pair of Linear Equations in Two Variables

Unit II: Algebra • 5–6 Marks

This chapter is strictly structured according to the CBSE Rationalised Mathematics Syllabus. Focus your board preparation only on prescribed topics and avoid studying deleted sections:

In 2026–27 Board Syllabus:

✓ Graphical Representation & Intersecting/Parallel/Coincident Lines ✓ Consistency Conditions (a₁/a₂ ≠ b₁/b₂, etc.) ✓ Substitution Method ✓ Elimination Method ✓ Word Problems (Speed, Age, Fixed Charges)

Deleted / Not in Board Exam:

✕ Cross-Multiplication Method ✕ Equations Reducible to Linear (1/x, 1/y)

⏰ Last 24-Hour Checklist

  • General Form: $a_1x + b_1y + c_1 = 0$.
  • Consistency Table: Memorize unique, no, and infinite solution conditions.
  • Elimination Method: Equate coefficients and subtract.
  • Substitution Method: Express $x$ in terms of $y$.
  • Digit Problems: Number $= 10x + y$. Reverse $= 10y + x$.
  • Fraction Problems: Assume fraction as $x/y$.

📊 Consistency & Graph Nature

For equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

Ratio Condition Graphical Representation Algebraic Interpretation Consistency
$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ Intersecting Lines Unique Solution Consistent
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ Coincident Lines Infinitely Many Solutions Dependent
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ Parallel Lines No Solution Inconsistent
(x, y) 1. Intersecting Lines Unique Solution (Consistent) 2. Coincident Lines ∞ Solutions (Dependent) 3. Parallel Lines No Solution (Inconsistent) Geometric Behavior: Intersecting ($a_1/a_2 \neq b_1/b_2$), Coincident ($=b_1/b_2 = c_1/c_2$), Parallel ($=b_1/b_2 \neq c_1/c_2$)
⚡ High-Yield Conceptual Rules (Board MCQs & A-R):
  • Standard Form Alignment Rule: Always rewrite both equations in the same form ($a_1x + b_1y + c_1 = 0$ or $a_1x + b_1y = c_1$). Comparing ratios with constants on opposite sides (e.g. $x+2y=3$ and $2x+4y+6=0$) flips the sign of $c_1/c_2$!
  • Equations of Coordinate Axes: Equation of $x$-axis is $y = 0$; equation of $y$-axis is $x = 0$. A line parallel to $x$-axis is $y = k$; parallel to $y$-axis is $x = k$.
  • Homogeneous System ($c_1 = c_2 = 0$): A pair like $a_1x + b_1y = 0$ and $a_2x + b_2y = 0$ always passes through the origin $(0, 0)$, so $(0, 0)$ is always a guaranteed solution.
  • Triangular Region Area: For the triangle formed by lines with an axis, $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$, where base is the distance between the two axis intercepts and height is the coordinate perpendicular to that axis.

Algebraic Methods for Solving

1. Substitution Method

  1. Find value of one variable ($y$) in terms of other ($x$) from Eq 1.
  2. Substitute this into Eq 2 to get equation in one variable.
  3. Solve for $x$.
  4. Put $x$ back in Step 1 to find $y$.

2. Elimination Method

  1. Multiply equations by constants to make coefficients of one variable equal.
  2. Add or Subtract equations to eliminate that variable.
  3. Solve for the remaining variable.
  4. Substitute back to find the eliminated variable.

⚡ Board Special Shortcut: Interchanged Large Coefficients

For equations of the form: $ax + by = c$ and $bx + ay = d$ (e.g., $152x - 378y = -74$ and $-378x + 152y = -604$):

  • Step 1 (Add): Add both equations → $(a+b)x + (a+b)y = c+d \implies \mathbf{x + y = p}$
  • Step 2 (Subtract): Subtract both equations → $(a-b)x - (a-b)y = c-d \implies \mathbf{x - y = q}$
  • Step 3 (Solve): Simply add and subtract these two tiny equations to find $x$ and $y$ in seconds!
🚀 Word Problem Equation Models (CBSE Blueprint):
  • Two-digit number: Number $= 10x + y$, Reversed $= 10y + x$.
  • Fractions: Fraction $= \frac{x}{y}$ (where $x =$ numerator, $y =$ denominator).
  • Fixed + Variable charges: Total Cost $= Fixed Charge (x) + (days/km \times y)$.
  • Upstream & Downstream: Let boat speed $= x km/h$, stream speed $= y km/h$.
    $Upstream Speed = x - y$,   $Downstream Speed = x + y$.   $Time = \frac{Distance}{Speed}$.

Solved Examples (Board Marking Scheme)

Q1. The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits differ by 2, find the number. (3 Marks)

Step 1: Forming Equations 1 Mark

Let tens digit be $x$ and units digit be $y$. Number $= 10x + y$.

Reverse Number $= 10y + x$.

Case 1: Sum is 66 $\Rightarrow (10x+y) + (10y+x) = 66 \Rightarrow 11(x+y)=66 \Rightarrow x+y=6$ ...(1)

Case 2: Digits differ by 2 $\Rightarrow x-y=2$ ...(2) OR $y-x=2$ ...(3)

Step 2: Solving 1 Mark

Add (1) and (2): $2x = 8 \Rightarrow x=4$. Then $y=2$. Number is 42.

Add (1) and (3): $2y = 8 \Rightarrow y=4$. Then $x=2$. Number is 24.

Step 3: Conclusion 1 Mark

There are two such numbers: 42 and 24.

Q2. Check if $x-2y=0$ and $3x+4y-20=0$ are consistent. (2 Marks)

Step 1: Compare Ratios 1 Mark

$a_1=1, b_1=-2, c_1=0$

$a_2=3, b_2=4, c_2=-20$

$\frac{a_1}{a_2} = \frac{1}{3}$ and $\frac{b_1}{b_2} = \frac{-2}{4} = -\frac{1}{2}$.

Step 2: Conclusion 1 Mark

Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at one point.

Therefore, the pair of equations is consistent with a unique solution.

Previous Year Questions (PYQs)

2023 (1 Mark): Find the value of $k$ for which equations $x+2y=3$ and $5x+ky+7=0$ are inconsistent.
Hint: Condition for inconsistent (parallel): $$\frac{1}{5}$ = $\frac{2}{k}$ \neq $\frac{3}{-7}$ \Rightarrow k=10$.
2020 (3 Marks): 5 pencils and 7 pens cost ?50, whereas 7 pencils and 5 pens cost ?46. Find cost of one pencil and one pen.
Ans: Pencil (x) = ₹3, Pen (y) = ₹5. (Solve $5x+7y=50, 7x+5y=46$).
2019 (3 Marks): A fraction becomes 9/11 if 2 is added to both numerator and denominator. If 3 is added to both, it becomes 5/6. Find the fraction.
Ans: Fraction is 7/9. (Equations: $11x-9y=-4$ and $6x-5y=-3$).

Exam Strategy & Mistake Bank

⚠️ Mistake Bank

Sign Errors: When standard form is $ax+by=c$, $c_1/c_2$ ratio sign can be tricky. Always bring to $ax+by+c=0$.
Digit Problem: Writing number as $xy$ instead of $10x+y$.
Unit Conversions: In speed/distance problems, ensure km/h and mins are converted correctly.

💡 Scoring Tips

Let Statement: Always start word problems with "Let the number be x..."
Verify: If time permits, put your values of x and y back into the question to check.
Nature of Lines: Memorize the table. 1 Mark is guaranteed from there.

Self-Assessment Mock Test (10 Marks)

Q1 (1M): Write the condition for a pair of linear equations to have infinitely many solutions.


Q2 (2M): Solve for x and y: $x+y=14$, $x-y=4$.


Q3 (3M): Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. Find their ages.


Q4 (4M): Meena went to bank to withdraw ₹2000. She received only ₹50 and ₹100 notes. Total notes are 25. Find number of notes of each type.

📈 Graphical Method of Solution

Each linear equation $ax + by + c = 0$ represents a straight line on the graph. The solution to the pair is the point of intersection of the two lines.

Steps to Solve Graphically:
  1. Express each equation as $y = \frac{-ax - c}{b}$
  2. Find 2–3 ordered pairs $(x, y)$ satisfying each equation
  3. Plot points and draw both lines on the same graph
  4. The intersection point (if any) is the solution
✅ Intersecting Lines
One common point → Unique Solution
Consistent pair
♾️ Coincident Lines
Infinite points → Infinite Solutions
Dependent/Consistent
⚫ Parallel Lines
No common point → No Solution → Inconsistent pair
⚠️ Board Tip: In graphical questions, always find at least 3 points per line for accuracy, and clearly label the intersection point as the solution.

📝 More Solved Board Questions

Q3. Solve by Elimination: $3x + 2y = 11$ and $2x + 3y = 4$. 3 Marks

Sol. Multiply Eq 1 by 3 and Eq 2 by 2:

$9x + 6y = 33$    ...(3)

$4x + 6y = 8$    ...(4)

Subtract (4) from (3): $5x = 25 \Rightarrow x = 5$

Substitute in Eq 1: $15 + 2y = 11 \Rightarrow y = -2$

Answer: $x = 5, y = -2$

Q4. Solve for $x$ and $y$: $152x - 378y = -74$ and $-378x + 152y = -604$. 3 Marks (Frequent Standard Paper)

Sol. Coefficients of $x$ and $y$ are interchanged. Use the Add & Subtract method.

Adding both equations:
$(152 - 378)x + (-378 + 152)y = -74 - 604$
$\Rightarrow -226x - 226y = -678 \implies -226(x + y) = -678 \implies \mathbf{x + y = 3}$   ...(3)

Subtracting second from first:
$(152 - (-378))x + (-378 - 152)y = -74 - (-604)$
$\Rightarrow 530x - 530y = 530 \implies \mathbf{x - y = 1}$   ...(4)

Adding (3) and (4): $2x = 4 \Rightarrow \mathbf{x = 2}$.
Subtracting (4) from (3): $2y = 2 \Rightarrow \mathbf{y = 1}$.
Answer: $x = 2, y = 1$

Q5. 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. 5 Marks (Long Answer)

Sol. Let speed of boat in still water $= x km/h$ and speed of stream $= y km/h$.

Speed upstream $= (x - y) km/h$, Speed downstream $= (x + y) km/h$.

$\frac{30}{x-y} + \frac{44}{x+y} = 10$   ...(1)     and     $\frac{40}{x-y} + \frac{55}{x+y} = 13$   ...(2)

Let $\frac{1}{x-y} = u$ and $\frac{1}{x+y} = v$:

$30u + 44v = 10 \Rightarrow 15u + 22v = 5$   ...(3)
$40u + 55v = 13$   ...(4)

Multiplying (3) by 8 and (4) by 3:
$120u + 176v = 40$
$120u + 165v = 39$
Subtracting gives $11v = 1 \Rightarrow v = \frac{1}{11} \implies x + y = 11$.
Substituting $v$ into (3): $15u + 2 = 5 \Rightarrow 15u = 3 \Rightarrow u = \frac{1}{5} \implies x - y = 5$.

Adding $x+y=11$ and $x-y=5$: $2x = 16 \Rightarrow \mathbf{x = 8 km/h}$.
Then $y = 11 - 8 = \mathbf{3 km/h}$.
Speed of boat = 8 km/h, Speed of stream = 3 km/h.

Q6. Find the value of $k$ for which equations $2x + 3y = 7$ and $(k-1)x + (k+2)y = 3k$ have infinitely many solutions. 3 Marks

Sol. For infinitely many solutions: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

$\frac{2}{k-1} = \frac{3}{k+2} = \frac{7}{3k}$

From first two: $2(k+2) = 3(k-1) \Rightarrow 2k + 4 = 3k - 3 \Rightarrow k = 7$

Verify with third ratio: $\frac{7}{21} = \frac{1}{3}$ and $\frac{2}{6} = \frac{1}{3}$ ✅

$k = 7$

🎯 Board Pattern (2018–2025): Word problems (ages, money, fractions, upstream/downstream) appear almost every year as 3-mark or 5-mark questions. Setting up equations with clear “Let statements” guarantees partial marks even if an arithmetic slip happens later.
CBSE Section E • 4 Marks

Case Study: Eco-Cab Electric Taxi Tariff Model

Scenario: A municipal smart city launches an eco-friendly electric cab service. The fare in the city consists of a fixed base charge together with a charge for the distance travelled in kilometers. For a journey of 10 km, the charge paid is ₹105, and for a journey of 15 km, the charge paid is ₹155.

Q1: Formulate the pair of linear equations representing the given scenario. 1 Mark

Q2: Find the fixed base charge ($x$) and the running charge per kilometer ($y$). 1 Mark

Q3: How much will a commuter pay for travelling a distance of 25 km? (OR) If another cab company charges fixed ₹25 and ₹9/km, for what distance do both companies charge exactly the same total fare? 2 Marks

📋 Board Revision Checklist

  • ✅ General form: $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$
  • ✅ $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ → Intersecting → Unique solution (Consistent)
  • ✅ $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ → Coincident → Infinite solutions (Consistent/Dependent)
  • ✅ $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ → Parallel → No solution (Inconsistent)
  • ✅ Interchanged coefficients ($ax+by=c, bx+ay=d$) → Add once ($x+y=p$), Subtract once ($x-y=q$)
  • ✅ Upstream speed $= (x - y)$, Downstream speed $= (x + y)$
  • ✅ Substitution: express one variable, substitute into other equation
  • ✅ Elimination: multiply to equate coefficients, then add/subtract
  • ✅ Two-digit number: tens digit $x$, units digit $y$ → number $= 10x + y$, reverse $= 10y + x$
  • ✅ Always verify answer by substituting back in BOTH equations
💡 Exam Tip:
For the consistency MCQ, memorize: “Different ratios → Unique → Consistent; Same all three → Infinite; Same first two, different third → No solution.”

Concept Mastery Quiz 🎯

Test your readiness for the board exam.

1. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the pair of equations has:

2. The graphical representation of two coincident lines means:

3. For $x + 2y = 5$ and $3x + 6y = 15$, the pair is:

4. In a two-digit number, if tens digit is $x$ and units digit is $y$, the number reversed is:

5. The value of $k$ for which $x + ky = 2$ and $2x + 4y = 6$ has no solution is:

Frequently Asked Questions: Class 10 Maths Chapter 3

What is the marks distribution for Chapter 3 Linear Equations?

Chapter 3 contributes 4 to 6 marks in the CBSE Board exam, featuring 1-mark consistency questions, 2-mark/3-mark algebraic solutions, or 4-mark case study questions.

Which methods of solving linear equations are included in the CBSE syllabus?

The current syllabus includes Graphical Method, Substitution Method, and Elimination Method. The Cross-Multiplication Method has been removed.

How can we determine whether a system of linear equations is consistent or inconsistent?

For a1*x + b1*y + c1 = 0 and a2*x + b2*y + c2 = 0: if a1/a2 != b1/b2, lines intersect (unique solution, consistent); if a1/a2 = b1/b2 = c1/c2, lines coincide (infinitely many solutions, consistent/dependent); if a1/a2 = b1/b2 != c1/c2, lines are parallel (no solution, inconsistent).

What types of word problems are most frequent in CBSE exams?

High-frequency word problem types include speed-distance-time (upstream and downstream boat problems), fixed and per-day charge models, age problems, and fraction problems.

Chapter 3 Study Modules & Practice Resources

Complete Notes & Formulas • NCERT Solutions • Previous Year Questions • Chapter Tests & Quizzes • Practice Worksheets