SEBA Class 10 Maths Chapter 3 Exercise 3.1 Solutions – Linear Equations in two variables
By Jamal Ali (M.Sc Physics, 5+ years teaching experience) · Reviewed by Editorial Board
SEBA Class 10 Maths Chapter 3 Exercise 3.1 Solutions focuses on the basic concepts of pair of linear equations in two variables, especially their graphical representation and general form. As per the latest ASSEB Division 1 syllabus (March 2026), this exercise helps students understand how two linear equations behave when plotted on a graph and how solutions are determined visually.
Students can build a strong foundation by referring to complete chapter 3 solutions and practicing step-by-step methods through chapterwise maths solutions. These resources ensure better clarity and structured preparation.
Practicing SEBA Class 10 Maths chapter 3 important questions exercise 3.1 along with SEBA HSLC Maths linear equations ex 3.1 solved questions helps improve accuracy in interpreting graphs and solving MCQs. Students can also revise using SEBA Class 10 Maths linear equations exercise 3.1 solutions pdf, based on the updated 2026 textbook.
SEBA Class 10 Maths Chapter 3 Exercise 3.1 Solutions with Step-by-Step Answers & Important Questions
Exercise 3.1 – Pair of Linear Equations in Two Variables
Q1. Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls.
Answer:
Let number of boys be x and number of girls be y.
Total students = 10
\[
i.e x + y = 10 \quad ...(1)
\]
also,
Girls are 4 more than boys:
\[
so, y = x + 4 \quad ...(2)
\]
Table for equation (1):
| x | y |
|---|---|
| 2 | 8 |
| 6 | 4 |
Table for equation (2):
| x | y |
|---|---|
| 1 | 5 |
| 3 | 7 |
Graph of x + y = 10 and y = x + 4
From the graph, the lines intersect at (3, 7).
Therefore,
Number of boys = 3
Number of girls = 7
(ii) 5 pens and 7 pencils together cost Rs 50, whereas 7 pens and 5 pencils together cost Rs 46. Find the cost of one pencil and one pen.
Answer:
Let the cost of one pencil be x and cost of one pen be y.
\[so, as per the question
7x + 5y = 50 \quad ...(1) and
\]
\[
5x + 7y = 46 \quad ...(2)
\]
Table for equation (1):
| x | y |
|---|---|
| 5 | 3 |
| 0 | 10 |
Table for equation (2):
| x | y |
|---|---|
| 5 | 3 |
| 2 | 4 |
Graph of 7x + 5y = 50 and 5x + 7y = 46:
From the graph, the lines intersect at (5, 3).
Therefore,
Cost of pencil = Rs 5
Cost of pen = Rs 3
Q2. On comparing the ratios \( \frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2} \), find whether the lines intersect, are parallel or coincident.
(i) \(5x - 4y + 8 = 0\) and \(7x + 6y - 9 = 0\)
\[
\frac{a_1}{a_2} = \frac{5}{7}, \quad \frac{b_1}{b_2} = \frac{-4}{6} = -\frac{2}{3}
\]
Since,
\[
\frac{a_1}{a_2} \ne \frac{b_1}{b_2}
\]
Therefore, the lines are intersecting.
(ii) \(9x + 3y + 12 = 0\) and \(18x + 6y + 24 = 0\)
\[
\frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}, \quad
\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}, \quad
\frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2}
\]
Since,
\[
\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}
\]
Therefore, the lines are coincident.
(iii) \(6x - 3y + 10 = 0\) and \(2x - y + 9 = 0\)
\[
\frac{a_1}{a_2} = \frac{6}{2} = 3, \quad
\frac{b_1}{b_2} = \frac{-3}{-1} = 3, \quad
\frac{c_1}{c_2} = \frac{10}{9}
\]
Since,
\[
\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}
\]
Therefore, the lines are parallel.
(iv) \(6x - y + 9 = 0\) and \(2x - 3y + 10 = 0\)
\[
\frac{a_1}{a_2} = \frac{6}{2} = 3, \quad
\frac{b_1}{b_2} = \frac{-1}{-3} = \frac{1}{3}
\]
Since,
\[
\frac{a_1}{a_2} \ne \frac{b_1}{b_2}
\]
Therefore, the lines are intersecting.
Q3. On comparing the ratios \( \frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2} \), find out whether the following pair of linear equations is consistent, or inconsistent.
(i) \(3x + 2y = 5\) and \(2x - 3y = 7\)
\[
\frac{a_1}{a_2} = \frac{3}{2}, \quad
\frac{b_1}{b_2} = \frac{2}{-3}
\]
Since,
\[
\frac{a_1}{a_2} \ne \frac{b_1}{b_2}
\]
Therefore, the lines are intersecting.
Hence, the system is consistent (unique solution).
(ii) \(2x - 3y = 8\) and \(4x - 6y = 9\)
\[
\frac{a_1}{a_2} = \frac{2}{4}, \quad
\frac{b_1}{b_2} = \frac{-3}{-6}, \quad
\frac{c_1}{c_2} = \frac{8}{9}
\]
Since,
\[
\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}
\]
Therefore, the lines are parallel.
Hence, the system is inconsistent (no solution).
(iii) \( \frac{2}{3}x + \frac{3}{7}y = 7\) and \(9x - 10y = 14\)
Convert:
\[
14x + 9y = 147,\quad 9x - 10y = 14
\]
\[
\frac{a_1}{a_2} = \frac{14}{9}, \quad
\frac{b_1}{b_2} = \frac{9}{-10}
\]
Since,
\[
\frac{a_1}{a_2} \ne \frac{b_1}{b_2}
\]
Therefore, the lines are intersecting.
Hence, the system is consistent (unique solution).
(iv) \(5x - 3y = 11\) and \(-10x + 6y = -22\)
\[
\frac{a_1}{a_2} = \frac{5}{-10}, \quad
\frac{b_1}{b_2} = \frac{-3}{6}, \quad
\frac{c_1}{c_2} = \frac{11}{-22}
\]
Since,
\[
\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}
\]
Therefore, the lines are coincident.
Hence, the system is consistent (infinitely many solutions).
(v) \( \frac{3}{4}x + 2y = 8\) and \(2x + 3y = 12\)
Convert:
\[
3x + 8y = 32,\quad 2x + 3y = 12
\]
\[
\frac{a_1}{a_2} = \frac{3}{2}, \quad
\frac{b_1}{b_2} = \frac{8}{3}
\]
Since,
\[
\frac{a_1}{a_2} \ne \frac{b_1}{b_2}
\]
Therefore, the lines are intersecting.
Hence, the system is consistent (unique solution).
Q4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically.
(i) \(x + y = 5\) and \(2x + 2y = 10\)
\[
\frac{a_1}{a_2} = \frac{1}{2}, \quad
\frac{b_1}{b_2} = \frac{1}{2}, \quad
\frac{c_1}{c_2} = \frac{5}{10} = \frac{1}{2}
\]
Since,
\[
\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}
\]
Therefore, the lines are coincident.
Hence, the system is consistent (infinitely many solutions).
Table for \(x + y = 5\)
| x | y |
|---|---|
| 1 | 4 |
| 3 | 2 |
Graph of \(x + y = 5\) and \(2x + 2y = 10\)
(ii) \(x - 2y = 4\) and \(2x - 4y = 9\)
\[
\frac{a_1}{a_2} = \frac{1}{2}, \quad
\frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2}, \quad
\frac{c_1}{c_2} = \frac{4}{9}
\]
Since,
\[
\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}
\]
Therefore, the lines are parallel.
Hence, the system is inconsistent (no solution).
(iii) \(2x - y - 6 = 0\) and \(4x - 2y - 5 = 0\)
Convert:
\[
2x - y = 6,\quad 4x - 2y = 5
\]
\[
\frac{a_1}{a_2} = \frac{2}{4}, \quad
\frac{b_1}{b_2} = \frac{-1}{-2} = \frac{1}{2}, \quad
\frac{c_1}{c_2} = \frac{6}{5}
\]
Since,
\[
\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}
\]
Therefore, the lines are parallel.
Hence, the system is inconsistent (no solution).
(iv) \(2x - y + 2 = 0\) and \(4x - y - 5 = 0\)
Convert:
\[
2x - y = -2,\quad 4x - y = 5
\]
\[
\frac{a_1}{a_2} = \frac{2}{4}, \quad
\frac{b_1}{b_2} = \frac{-1}{-1} = 1
\]
Since,
\[
\frac{a_1}{a_2} \ne \frac{b_1}{b_2}
\]
Therefore, the lines are intersecting.
Hence, the system is consistent (unique solution).
Table for \(2x - y = -2\)
| x | y |
|---|---|
| 0 | 2 |
| 1 | 4 |
Table for \(4x - y = 5\)
| x | y |
|---|---|
| 1 | -1 |
| 2 | 3 |
Graph of \(2x - y + 2 = 0\) and \(4x - y - 5 = 0\)
Solution point: \( \left(\frac{7}{2}, 9\right) \)
Q5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Answer:
Let width of the garden be x m and length be y m.
Given, half of the perimeter = 36 m
\[
i.e\; x + y = 36 \quad ...(1)
\]
also,
Length is 4 m more than width:
\[
so,\; y = x + 4 \quad ...(2)
\]
Substituting (2) in (1):
\[
x + (x + 4) = 36
\]
\[
2x + 4 = 36
\]
\[
2x = 32
\Rightarrow x = 16
\]
Now, from (2):
\[
y = 16 + 4 = 20
\]
Therefore,
Width of the garden = 16 m
Length of the garden = 20 m
Q6. Given the linear equation \(2x + 3y - 8 = 0\), write another linear equation such that the pair of equations formed is:
(i) Intersecting lines
Answer:
Given equation:
\[
2x + 3y - 8 = 0
\]
For intersecting lines,
\[
since\; \frac{a_1}{a_2} \ne \frac{b_1}{b_2}
\]
Choose another equation such that ratios are not equal.
\[
so,\; x - y + 1 = 0
\]
Therefore, the given pair represents intersecting lines.
(ii) Parallel lines
Answer:
Given equation:
\[
2x + 3y - 8 = 0
\]
For parallel lines,
\[
since\; \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}
\]
Multiply first equation by 2:
\[
4x + 6y - 16 = 0
\]
Change constant term:
\[
so,\; 4x + 6y + 5 = 0
\]
Therefore, the lines are parallel.
(iii) Coincident lines
Answer:
Given equation:
\[
2x + 3y - 8 = 0
\]
For coincident lines,
\[
since\; \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}
\]
Multiply by 2:
\[
so,\; 4x + 6y - 16 = 0
\]
Therefore, both equations represent the same line (coincident lines).
Q7. Draw the graphs of the equations \(x - y + 1 = 0\) and \(3x + 2y - 12 = 0\). Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Answer:
Given equations:
\[
x - y + 1 = 0 \quad ...(1)
\]
\[
3x + 2y - 12 = 0 \quad ...(2)
\]
From (1):
\[
x = y - 1 \; \text{or} \; y = x + 1
\]
From (2):
\[
x = \frac{12 - 2y}{3}
\]
Table for equation (1):
| x | 0 | 1 | 2 |
|---|---|---|---|
| y | 1 | 2 | 3 |
Table for equation (2):
| x | 4 | 2 | 0 |
|---|---|---|---|
| y | 0 | 3 | 6 |
Graph of \(x - y + 1 = 0\) and \(3x + 2y - 12 = 0\)
Now, find intersection points with x-axis:
For (1):
\[
y = 0 \Rightarrow x = -1
\Rightarrow (-1,0)
\]
For (2):
\[
y = 0 \Rightarrow 3x = 12 \Rightarrow x = 4
\Rightarrow (4,0)
\]
Now, intersection of both lines:
\[
y = x + 1
\]
Put in (2):
\[
3x + 2(x+1) = 12
\]
\[
5x + 2 = 12
\Rightarrow x = 2,\; y = 3
\]
Therefore, vertices of triangle are:
\[
(-1,0),\; (4,0),\; (2,3)
\]
Q8. For what value of p does the pair of equations have a unique solution?
\[ 4x + py + 8 = 0 \] \[ 2x + 2y + 2 = 0 \]
(a) \(p = 4\)
(b) \(p \ne 4\)
(c) \(p = -4\)
(d) \(p \ne -4\)
Answer: (b) \(p \ne 4\)
For unique solution,
\[ since\; \frac{a_1}{a_2} \ne \frac{b_1}{b_2} \] \[ \frac{4}{2} = \frac{p}{2} \] \[ 2 = \frac{p}{2} \Rightarrow p = 4 \] So, for unique solution:
\[ p \ne 4 \]
Q9. Which of the following has infinitely many solutions?
(a) \(2x + 3y = 6\) and \(4x + 6y - 12 = 0\)
(b) \(x + y = 4\) and \(x - y - 2 = 0\)
(c) \(x - y = 3\) and \(x + y = 7\)
(d) \(x + y = 5\) and \(2x - y = 3\)
Answer: (a)
(a): \[ 4x + 6y - 12 = 0 \Rightarrow 4x + 6y = 12 \] Multiply first equation by 2: \[ 4x + 6y = 12 \] Since, \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \] Therefore, lines are coincident → infinitely many solutions.
Q10. In the following question a statement of assertion (A) is followed by a statement of reason (R).
Assertion (A): The equation \(2x - 3y = 0\) has an infinite number of solutions.
Reason (R): A linear equation in two variables represents a straight line on the coordinate plane.
(a) Both A and R are true and R is correct explanation
(b) Both A and R are true but R is not correct explanation
(c) A is true but R is false
(d) A is false but R is true
Answer: (a)
Assertion is true because a linear equation has infinitely many solutions.
Reason is also true because such equations represent a straight line.
Since all points on the line satisfy the equation,
therefore it has infinitely many solutions.
Hence, both are true and R explains A.
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SEBA Class 10 Maths Chapter 3 Exercise 3.1 Solutions – Complete Graphical Approach Guide | Assam Eduverse
Understanding SEBA Class 10 Maths Chapter 3 Exercise 3.1 Solutions is the first step toward mastering linear equations in two variables. This exercise introduces students to the graphical representation of equations and helps them visualize how solutions are formed when two lines intersect, coincide, or remain parallel.
As per the latest ASSEB guidelines, questions from this exercise are asked in both MCQ and descriptive formats. Graph-based MCQs test quick understanding, while descriptive questions require proper explanation of graphical methods. Therefore, students must focus on both conceptual clarity and regular practice.
By solving SEBA HSLC Maths linear equations ex 3.1 solved questions, students can easily understand how to plot equations and identify solutions from graphs. These concepts are essential for building a strong base for algebraic methods covered in later exercises. To strengthen overall understanding, students can also explore Exercise 3.2 solutions, which focus on algebraic techniques.
Regular practice of SEBA Class 10 Maths chapter 3 important questions exercise 3.1 helps students gain confidence in solving graph-based problems. These questions are often designed based on exam patterns and improve both speed and accuracy. For a structured study plan, referring to chapterwise maths solutions ensures complete topic coverage.
For quick revision, students can use SEBA Class 10 Maths linear equations exercise 3.1 solutions pdf, which provides easy access to all important questions. However, it is important to combine revision with practical graph plotting to fully understand the concepts. Students can also refer to Class 9 and 10 study materials for additional practice.
It is important to follow the updated Mathematics textbook introduced in March 2026, as the old book has been discontinued. The new book includes a variety of MCQs and real-life word problems, making it essential to practice different types of questions. Assam Eduverse provides updated, expert-reviewed solutions aligned with the latest syllabus.
To further improve performance, students can practice additional problems from chapterwise question answers and understand exam trends using SEBA blueprint 2026-27.
In conclusion, mastering SEBA Class 10 Maths Chapter 3 Exercise 3.1 Solutions requires consistent practice, clear understanding of graphs, and the ability to interpret results correctly. With proper preparation and the right resources, students can confidently solve any question from this exercise in the HSLC examination.
FAQs – Exercise 3.1 Linear Equations (Class 10 Maths)
1. How do you solve linear equations graphically in Exercise 3.1?
To solve graphically, both equations are plotted on a coordinate plane. The point where the two lines intersect represents the solution. If the lines do not intersect or overlap completely, it indicates no solution or infinitely many solutions.
2. What do parallel and intersecting lines represent in linear equations?
Intersecting lines indicate a unique solution, parallel lines show no solution, and coincident lines represent infinitely many solutions. Understanding this helps in quickly identifying answers in MCQs.
3. Why is graphical representation important in this exercise?
Graphical representation helps students visualize how equations behave and makes it easier to understand solutions. It also builds the base for solving equations using algebraic methods in later exercises.
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