ASSEB Class 10 Updated Syllabus 2026-27 | Maths Chapter 3 Exercise 3.1 Solutions
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With the recent transition to the ASSEB (Assam State School Education Board), the syllabus and exam patterns for the 2026–2027 academic session have been streamlined. To help you stay ahead, we are providing the most accurate, step-by-step solutions for Class 10 Mathematics, Chapter 3: Pair of Linear Equations in Two Variables.
This post covers the complete Exercise 3.1 according to the latest ASSEB guidelines.
📌 Why This Guide Matters for 2026-27
New Board Alignment: Tailored specifically for the ASSEB Class 10 framework.
Concept Clarity: Focuses on the "Algebraic to Graphical" conversion process.
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🔢 Exercise 3.1 – Solved for ASSEB Students
Let there be \(x\) number of girls and \(y\) number of boys. As per the given question, the algebraic expression can be represented as follows:
\[ x + y = 10 \] \[ x - y = 4 \]Now, for \(x + y = 10\) or \(x = 10 - y\), the solutions are:
| x | 5 | 4 |
| y | 5 | 6 |
For \(x - y = 4\) or \(x = 4 + y\), the solutions are:
| x | 4 | 5 |
| y | 0 | 1 |
The graphical representation is as follows:
From the graph, it can be seen that the given lines cross each other at point \((7, 3)\).
If 1 pencil costs Rs. \(x\) and 1 pen costs Rs. \(y\).
According to the question, the algebraic expressions can be represented as:
\[ 5x + 7y = 50 \] \[ 7x + 5y = 46 \]For \(5x + 7y = 50 \Rightarrow x = \frac{50 - 7y}{5}\), the solutions are:
| x | 3 | 10 |
| y | 5 | 0 |
For \(7x + 5y = 46 \Rightarrow x = \frac{46 - 5y}{7}\), the solutions are:
| x | 8 | 3 |
| y | -2 | 5 |
Hence, the graphical representation is as follows:
From the graph, it can be seen that the given lines intersect at point \((3, 5)\).
(i) Given expressions:
\[ 5x - 4y + 8 = 0 \] \[ 7x + 6y - 9 = 0 \]Comparing these equations with \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\), we get—
\(a_1 = 5,\; b_1 = -4,\; c_1 = 8\)
\(a_2 = 7,\; b_2 = 6,\; c_2 = -9\)
Since \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\), the given pair of equations has a unique solution and the lines intersect at exactly one point.
(ii) Given expressions:
\[ 9x + 3y + 12 = 0 \] \[ 18x + 6y + 24 = 0 \]Comparing with \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\), we get—
\(a_1 = 9,\; b_1 = 3,\; c_1 = 12\)
\(a_2 = 18,\; b_2 = 6,\; c_2 = 24\)
Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\), the given pair of equations has infinitely many solutions and the lines are coincident.
(iii) Given expressions:
\[ 6x - 3y + 10 = 0 \] \[ 2x - y + 9 = 0 \]Comparing with \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\), we get—
\(a_1 = 6,\; b_1 = -3,\; c_1 = 10\)
\(a_2 = 2,\; b_2 = -1,\; c_2 = 9\)
Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}\), the given pair of equations represents parallel lines. The lines never intersect, so there is no solution.
Solution (i)
Given: \(3x + 2y = 5 \Rightarrow 3x + 2y - 5 = 0\)
\(2x - 3y = 7 \Rightarrow 2x - 3y - 7 = 0\)
Comparing with \(a_1x+b_1y+c_1=0\) and \(a_2x+b_2y+c_2=0\):
\[ a_1=3,\; b_1=2,\; c_1=-5 \] \[ a_2=2,\; b_2=-3,\; c_2=-7 \] \[ \frac{a_1}{a_2}=\frac{3}{2}, \quad \frac{b_1}{b_2}=\frac{2}{-3}=-\frac{2}{3} \]Since \( \frac{a_1}{a_2} \ne \frac{b_1}{b_2} \), the lines intersect at one point ⇒ Consistent.
(ii)
Given: \(2x - 3y = 8 \Rightarrow 2x - 3y - 8 = 0\)
\(4x - 6y = 9 \Rightarrow 4x - 6y - 9 = 0\)
\[ a_1=2,\; b_1=-3,\; c_1=-8 \] \[ a_2=4,\; b_2=-6,\; c_2=-9 \] \[ \frac{a_1}{a_2}=\frac{1}{2}, \quad \frac{b_1}{b_2}=\frac{1}{2}, \quad \frac{c_1}{c_2}=\frac{8}{9} \]Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2} \), the lines are parallel ⇒ Inconsistent.
(iii)
Given: \(\frac{3}{2}x + \frac{5}{3}y = 7 \Rightarrow \frac{3}{2}x + \frac{5}{3}y - 7 = 0\)
\(9x - 10y = 14 \Rightarrow 9x - 10y - 14 = 0\)
\[ a_1=\frac{3}{2},\; b_1=\frac{5}{3},\; c_1=-7 \] \[ a_2=9,\; b_2=-10,\; c_2=-14 \] \[ \frac{a_1}{a_2}=\frac{3}{2\times9}=\frac{1}{6} \] \[ \frac{b_1}{b_2}=\frac{5}{3\times(-10)}=-\frac{1}{6} \]Since \( \frac{a_1}{a_2} \ne \frac{b_1}{b_2} \), the lines intersect ⇒ Consistent.
(iv)
Given: \(5x - 3y = 11 \Rightarrow 5x - 3y - 11 = 0\)
\(-10x + 6y = -22 \Rightarrow -10x + 6y + 22 = 0\)
\[ a_1=5,\; b_1=-3,\; c_1=-11 \] \[ a_2=-10,\; b_2=6,\; c_2=22 \] \[ \frac{a_1}{a_2}=-\frac{1}{2}, \quad \frac{b_1}{b_2}=-\frac{1}{2}, \quad \frac{c_1}{c_2}=-\frac{1}{2} \]Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), lines are coincident ⇒ Consistent (infinitely many solutions).
(v)
Given: \(\frac{4}{3}x + 2y = 8 \Rightarrow \frac{4}{3}x + 2y - 8 = 0\)
\(2x + 3y = 12 \Rightarrow 2x + 3y - 12 = 0\)
\[ a_1=\frac{4}{3},\; b_1=2,\; c_1=-8 \] \[ a_2=2,\; b_2=3,\; c_2=-12 \] \[ \frac{a_1}{a_2}=\frac{2}{3}, \quad \frac{b_1}{b_2}=\frac{2}{3}, \quad \frac{c_1}{c_2}=\frac{2}{3} \]Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), lines are coincident ⇒ Consistent.
Given:
\[ x + y = 5 \] \[ 2x + 2y = 10 \]Comparing with \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\), we get:
\[ a_1 = 1,\; b_1 = 1,\; c_1 = -5 \] \[ a_2 = 2,\; b_2 = 2,\; c_2 = -10 \] \[ \frac{a_1}{a_2} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{1}{2} \]Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the equations are coincident.
Therefore, the equations have infinitely many solutions and are consistent.
Table of values:
For \(x + y = 5 \Rightarrow x = 5 - y\)
| x | 4 | 3 |
| y | 1 | 2 |
For \(2x + 2y = 10 \Rightarrow x = \frac{10 - 2y}{2}\)
| x | 4 | 3 |
| y | 1 | 2 |
The graphical representation is as follows:
From the graph, we can see that the lines overlap each other.
Given Equations:
\[ x - y = 8 \implies x - y - 8 = 0 \] \[ 3x - 3y = 16 \implies 3x - 3y - 16 = 0 \]Comparing with the standard forms \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\):
\(a_2 = 3,\; b_2 = -3,\; c_2 = -16\)
Calculating the Ratios:
\[ \frac{a_1}{a_2} = \frac{1}{3} \] \[ \frac{b_1}{b_2} = \frac{-1}{-3} = \frac{1}{3} \] \[ \frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2} \]
From the above calculations, we observe that:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]
This condition indicates that the lines represented by these equations are parallel to each other.
Given:
\[ 2x + y - 6 = 0 \] \[ 4x - 2y - 4 = 0 \]Comparing with \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\):
\[ a_1=2,\; b_1=1,\; c_1=-6 \] \[ a_2=4,\; b_2=-2,\; c_2=-4 \] \[ \frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \quad \frac{b_1}{b_2}=\frac{1}{-2}=-\frac{1}{2} \]Since \( \frac{a_1}{a_2} \ne \frac{b_1}{b_2} \), the lines intersect at one point.
Hence, the pair of linear equations is consistent and has a unique solution.
Table of values:
For \(2x + y - 6 = 0 \Rightarrow y = 6 - 2x\)
| x | 0 | 2 |
| y | 6 | 2 |
For \(4x - 2y - 4 = 0 \Rightarrow y = 2x - 2\)
| x | 1 | 2 |
| y | 0 | 2 |
The graphical representation is as follows:
From the graph, the lines intersect at the point \((2, 2)\).
Given Equations:
\[ 2x - 2y - 2 = 0 \] \[ 4x - 4y - 5 = 0 \]Comparing with the standard forms \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\):
\(a_2 = 4,\; b_2 = -4,\; c_2 = -5\)
Calculating the Ratios:
\[ \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \] \[ \frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2} \] \[ \frac{c_1}{c_2} = \frac{-2}{-5} = \frac{2}{5} \]
From the above results, we observe that:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]
This mathematical condition confirms that the lines represented by these linear equations are parallel to each other.
Let the width of the rectangular garden be \(x\) m. ⇒ Length of the garden \(= x + 4\) m
According to the question, half the perimeter is 36 m.
\[ \frac{1}{2} \times 2(\text{Length} + \text{Width}) = 36 \]
\[ \Rightarrow \text{Length} + \text{Width} = 36 \]
\[ \Rightarrow (x + 4) + x = 36 \]
\[ \Rightarrow 2x + 4 = 36 \]
\[ \Rightarrow 2x = 32 \]
\[ \Rightarrow x = 16 \]
\[ \Rightarrow \text{Width} = 16 \text{ m} \]
\[ \Rightarrow \text{Length} = x + 4 = 16 + 4 = 20 \text{ m} \]
Given linear equation: \( 2x + 3y - 8 = 0 \)
Here, \( a_1 = 2, b_1 = 3, c_1 = -8 \)
(i) Intersecting lines: For the lines to intersect at a single point, the condition is:
Here, \( \frac{a_1}{a_2} = \frac{2}{3} \) and \( \frac{b_1}{b_2} = \frac{3}{2} \)
Since \( \frac{2}{3} \neq \frac{3}{2} \), the condition is satisfied. (ii) Parallel lines: For the lines to be parallel, the condition is:
Here, \( \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2} \) and \( \frac{c_1}{c_2} = \frac{-8}{-12} = \frac{2}{3} \)
Since \( \frac{1}{2} = \frac{1}{2} \neq \frac{2}{3} \), the condition is satisfied. (iii) Coincident lines: For the lines to be coincident (overlapping), the condition is:
Here, \( \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2} \) and \( \frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2} \)
Since \( \frac{1}{2} = \frac{1}{2} = \frac{1}{2} \), the condition is satisfied.
| x | 1 | 2 |
| y | 2 | 3 |
| x | 4 | 2 |
| y | 0 | 3 |
\( 2x + 2y + 2 = 0 \)
Given equations:
\[ 4x + py + 8 = 0 \] \[ 2x + 2y + 2 = 0 \]Comparing with \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\), we get:
\[ a_1 = 4,\; b_1 = p,\; c_1 = 8 \] \[ a_2 = 2,\; b_2 = 2,\; c_2 = 2 \]For a unique solution, we must have:
\[ \frac{a_1}{a_2} \ne \frac{b_1}{b_2} \] \[ \frac{4}{2} \ne \frac{p}{2} \] \[ 2 \ne \frac{p}{2} \] \[ \Rightarrow p \ne 4 \]Condition for infinitely many solutions:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \](a)
\[ 2x + 3y = 6 \Rightarrow 2x + 3y - 6 = 0 \] \[ 4x + 6y - 12 = 0 \] \[ a_1=2,\; b_1=3,\; c_1=-6 \] \[ a_2=4,\; b_2=6,\; c_2=-12 \] \[ \frac{2}{4}=\frac{1}{2}, \quad \frac{3}{6}=\frac{1}{2}, \quad \frac{-6}{-12}=\frac{1}{2} \]Since all ratios are equal, the equations are coincident.
(b)
\[ x + y = 4,\quad x - y - 2 = 0 \] Not proportional ⇒ Not infinite solutions.(c)
\[ x - y = 3,\quad x + y = 7 \] Not proportional ⇒ Not infinite solutions.(d)
\[ x + y = 5,\quad 2x - y = 3 \] Not proportional ⇒ Not infinite solutions.Step 1: Analyzing Statement (A)
A linear equation in two variables has infinitely many solutions because for every value of \(x\), there is a corresponding value of \(y\) that satisfies the equation. Geometrically, every point on the line is a solution.
Therefore, Statement (A) is True.
Step 2: Analyzing Reason (R)
This is a fundamental mathematical fact. Since a straight line consists of an infinite number of points, it explains why the equation has infinitely many solutions.
Therefore, Reason (R) is True.
Conclusion:
Since a linear equation represents a straight line, and a line is made of infinite points, the Reason (R) correctly explains why Statement (A) has infinitely many solutions.
📌 Important Note
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🚀 Join Our WhatsApp Channel📊 Graphical Analysis Table
Under the updated ASSEB syllabus, understanding the ratio of coefficients is crucial for multiple-choice questions (MCQs).
| Ratio Comparison | Nature of Lines | Number of Solutions |
| `\frac{a_1}{a_2} \neq \frac{b_1}{b_2}` | Intersecting Lines | Unique Solution |
| `\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}` | Parallel Lines | No Solution |
| `\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}` | Coincident Lines | Infinite Solutions |
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❓ Frequently Asked Questions (FAQ)
Q1: How has the ASSEB syllabus changed for Chapter 3?
Ans: While the core concepts remain aligned with NCERT, the ASSEB Updated Syllabus 2026-27 emphasizes more on application-based word problems and graphical interpretation.
Q2: Will the 2027 Board Exams be harder?
Ans: Not necessarily! With the right resources like Digital Pipal Academy, you will find the new ASSEB pattern much more logical and scoring.
Q3: Where can I find the rest of the exercises?
Ans: All solutions for Chapter 3 (3.2, 3.3, etc.) are being uploaded daily. Bookmark our site www.pipalacademy.com.
🔑Tags
ASSEB Class 10 Updated Syllabus 2026-27
Class 10 Maths Chapter 3 Exercise 3.1 Solution
Assam Board Class 10 Maths English Medium
Pair of Linear Equations in Two Variables ASSEB
Digital Pipal Academy ASSEB Notes
📝 Final Thoughts
The transition to the ASSEB board is an exciting step for education in Assam. By mastering the fundamentals in Exercise 3.1 today, you are building a strong foundation for your 2027 finals.
Digital Pipal Academy – Navigating the New ASSEB Syllabus Together.
📢 Description
Looking for the ASSEB Class 10 Updated Syllabus 2026-27 solutions? Get the complete English Medium guide for Maths Chapter 3 Exercise 3.1 at Digital Pipal Academy. Clear, accurate, and exam-ready!
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