School Assignments for Class 10 Mathematics: Chapter 03 Pair of Linear Equations in Two Variables
Explore structured practice materials through the CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Assignment Set 18. Tailored for Class 10 learners, utilizing these Mathematics assignments ensures thorough preparation and strengthens foundational knowledge before final CBSE evaluations.
Practice Class 10 Mathematics Assignments: Chapter 03 Pair of Linear Equations in Two Variables
View or download the dedicated CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Assignment Set 18 resource below. Engaging with these assignments under focused study conditions ensures continuous academic progress and mastery of the 2026-27 curriculum.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 2 x - 3 y = 1 \) ...(i), \( y = \frac{2 x}{3} - \frac{1}{3} \).
| x | -4 | -1 | 2 |
| y | -3 | -1 | 1 |
For the equation \( 3 x - 4 y = 1 \) ...(ii), \( y = \frac{3 x}{4} - \frac{1}{4} \).
| x | -5 | -1 | 3 |
| y | -4 | -1 | 2 |
From the graph, the two lines intersect at the point \( (-1, -1) \). Hence the solution is \( x = -1,\ y = -1 \).
Put \( x = 3,\ y = 2 \) in the first equation \( 2 x - 3 y = 1 \): LHS is not equal to RHS, so the point does not lie on it.
Put \( x = 3,\ y = 2 \) in the second equation \( 3 x - 4 y = 1 \): LHS = RHS, so the point lies on it.
Hence (3, 2) lies on the line \( 3x - 4y = 1 \).
In simple words: Plot both lines on the same axes; the point where they cross satisfies both equations.
Teacher's Note:
1. Plot at least three points for each line so that an error in one point is easy to spot.
2. Read the intersection carefully and verify it in both equations.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 3 x - 5 y = 19 \) ...(i), \( y = \frac{3 x}{5} - \frac{19}{5} \).
| x | -7 | -2 | 3 |
| y | -8 | -5 | -2 |
For the equation \( - 7 x + 3 y = -1 \) ...(ii), \( y = \frac{7 x}{3} - \frac{1}{3} \).
| x | -5 | -2 | 1 |
| y | -12 | -5 | 2 |
From the graph, the two lines intersect at the point \( (-2, -5) \). Hence the solution is \( x = -2,\ y = -5 \).
Put \( x = 4,\ y = 9 \) in the first equation \( 3 x - 5 y = 19 \): LHS is not equal to RHS, so the point does not lie on it.
Put \( x = 4,\ y = 9 \) in the second equation \( - 7 x + 3 y + 1 = 0 \): LHS = RHS, so the point lies on it.
Hence (4, 9) lies on the line \( 3y - 7x + 1 = 0 \).
In simple words: Plot both lines on the same axes; the point where they cross satisfies both equations.
Teacher's Note:
1. Plot at least three points for each line so that an error in one point is easy to spot.
2. Read the intersection carefully and verify it in both equations.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 2 x + y = 10 \) ...(i), \( y = 10 - 2 x \).
| x | 2 | 3 | 4 |
| y | 6 | 4 | 2 |
For the equation \( 4 x - y = 8 \) ...(ii), \( y = 4 x - 8 \).
| x | 2 | 3 | 4 |
| y | 0 | 4 | 8 |
From the graph, the two lines intersect at the point \( (3, 4) \). Hence the solution is \( x = 3,\ y = 4 \).
Put \( x = 1,\ y = -4 \) in the first equation \( 2 x + y = 10 \): LHS is not equal to RHS, so the point does not lie on it.
Put \( x = 1,\ y = -4 \) in the second equation \( 4 x - y = 8 \): LHS = RHS, so the point lies on it.
Hence (1, -4) lies on the line \( 4x - y = 8 \).
In simple words: Plot both lines on the same axes; the point where they cross satisfies both equations.
Teacher's Note:
1. Plot at least three points for each line so that an error in one point is easy to spot.
2. Read the intersection carefully and verify it in both equations.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 4 x - 5 y = 20 \) ...(i), \( y = \frac{4 x}{5} - 4 \).
| x | 0 | 5 | 10 |
| y | -4 | 0 | 4 |
For the equation \( 3 x + 5 y = 15 \) ...(ii), \( y = 3 - \frac{3 x}{5} \).
| x | 0 | 5 | 10 |
| y | 3 | 0 | -3 |
From the graph, the two lines intersect at the point \( (5, 0) \). Hence the solution is \( x = 5,\ y = 0 \).
The region bounded by the two lines and the y-axis is the triangle with vertices \( (0, -4) \), \( (5, 0) \), \( (0, 3) \).
In simple words: Plot both lines on the same axes; the point where they cross satisfies both equations.
Teacher's Note:
1. Plot at least three points for each line so that an error in one point is easy to spot.
2. Read the intersection carefully and verify it in both equations.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 5 x - 6 y = -30 \) ...(i), \( y = \frac{5 x}{6} + 5 \).
| x | -6 | 0 | 6 |
| y | 0 | 5 | 10 |
For the equation \( 5 x + 4 y = 20 \) ...(ii), \( y = 5 - \frac{5 x}{4} \).
| x | -4 | 0 | 4 |
| y | 10 | 5 | 0 |
From the graph, the two lines intersect at the point \( (0, 5) \). Hence the solution is \( x = 0,\ y = 5 \).
The shaded region bounded by the two lines and the x-axis is the triangle with vertices \( (-6, 0) \), \( (0, 5) \), \( (4, 0) \).
In simple words: Plot both lines on the same axes; the point where they cross satisfies both equations.
Teacher's Note:
1. Plot at least three points for each line so that an error in one point is easy to spot.
2. Read the intersection carefully and verify it in both equations.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 2 x + y = -6 \) ...(i), \( y = - 2 x - 6 \).
| x | -1 | 0 | 1 |
| y | -4 | -6 | -8 |
For the equation \( 3 x - 2 y = 12 \) ...(ii), \( y = \frac{3 x}{2} - 6 \).
| x | -2 | 0 | 2 |
| y | -9 | -6 | -3 |
From the graph, the two lines intersect at the point \( (0, -6) \). Hence the solution is \( x = 0,\ y = -6 \).
The shaded region bounded by the two lines and the x-axis is the triangle with vertices \( (-3, 0) \), \( (0, -6) \), \( (4, 0) \).
In simple words: Plot both lines on the same axes; the point where they cross satisfies both equations.
Teacher's Note:
1. Plot at least three points for each line so that an error in one point is easy to spot.
2. Read the intersection carefully and verify it in both equations.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 2 x + 3 y = 4 \) ...(i), \( y = \frac{4}{3} - \frac{2 x}{3} \).
| x | -4 | -1 | 2 |
| y | 4 | 2 | 0 |
For the equation \( 3 x - y = -5 \) ...(ii), \( y = 3 x + 5 \).
| x | -2 | -1 | 0 |
| y | -1 | 2 | 5 |
From the graph, the two lines intersect at the point \( (-1, 2) \). Hence the solution is \( x = -1,\ y = 2 \).
The shaded region bounded by the two lines and the x-axis is the triangle with vertices \( (- \frac{5}{3}, 0) \), \( (-1, 2) \), \( (2, 0) \).
In simple words: Plot both lines on the same axes; the point where they cross satisfies both equations.
Teacher's Note:
1. Plot at least three points for each line so that an error in one point is easy to spot.
2. Read the intersection carefully and verify it in both equations.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 3 x + y = 1 \) ...(i), \( y = 1 - 3 x \).
| x | -1 | 0 | 1 |
| y | 4 | 1 | -2 |
For the equation \( 2 x - 3 y = -8 \) ...(ii), \( y = \frac{2 x}{3} + \frac{8}{3} \).
| x | -4 | -1 | 2 |
| y | 0 | 2 | 4 |
From the graph, the two lines intersect at the point \( (- \frac{5}{11}, \frac{26}{11}) \). Hence the solution is \( x = - \frac{5}{11},\ y = \frac{26}{11} \).
The shaded region bounded by the two lines and the x-axis is the triangle with vertices \( (-4, 0) \), \( (- \frac{5}{11}, \frac{26}{11}) \), \( (\frac{1}{3}, 0) \).
In simple words: Plot both lines on the same axes; the point where they cross satisfies both equations.
Teacher's Note:
1. Plot at least three points for each line so that an error in one point is easy to spot.
2. Read the intersection carefully and verify it in both equations.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( x + 2 y = 5 \) ...(i), \( y = \frac{5}{2} - \frac{x}{2} \).
| x | -1 | 1 | 3 |
| y | 3 | 2 | 1 |
For the equation \( 2 x - 3 y = -4 \) ...(ii), \( y = \frac{2 x}{3} + \frac{4}{3} \).
| x | -2 | 1 | 4 |
| y | 0 | 2 | 4 |
From the graph, the two lines intersect at the point \( (1, 2) \). Hence the solution is \( x = 1,\ y = 2 \).
The lines meet the x-axis at \( (5, 0) \) and \( (-2, 0) \).
In simple words: Plot both lines on the same axes; the point where they cross satisfies both equations.
Teacher's Note:
1. Plot at least three points for each line so that an error in one point is easy to spot.
2. Read the intersection carefully and verify it in both equations.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 2 x - y = 4 \) ...(i), \( y = 2 x - 4 \).
| x | 2 | 3 | 4 |
| y | 0 | 2 | 4 |
For the equation \( - x + 3 y = 3 \) ...(ii), \( y = \frac{x}{3} + 1 \).
| x | 0 | 3 | 6 |
| y | 1 | 2 | 3 |
From the graph, the two lines intersect at the point \( (3, 2) \). Hence the solution is \( x = 3,\ y = 2 \).
The lines meet the y-axis at \( (0, -4) \) and \( (0, 1) \).
In simple words: Plot both lines on the same axes; the point where they cross satisfies both equations.
Teacher's Note:
1. Plot at least three points for each line so that an error in one point is easy to spot.
2. Read the intersection carefully and verify it in both equations.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 3 x - y = 3 \) ...(i), \( y = 3 x - 3 \).
| x | 1 | 2 | 3 |
| y | 0 | 3 | 6 |
For the equation \( x - 2 y = -4 \) ...(ii), \( y = \frac{x}{2} + 2 \).
| x | 0 | 2 | 4 |
| y | 2 | 3 | 4 |
From the graph, the two lines intersect at the point \( (2, 3) \). Hence the solution is \( x = 2,\ y = 3 \).
The shaded region bounded by the two lines and the x-axis is the triangle with vertices \( (-4, 0) \), \( (2, 3) \), \( (1, 0) \).
In simple words: Plot both lines on the same axes; the point where they cross satisfies both equations.
Teacher's Note:
1. Plot at least three points for each line so that an error in one point is easy to spot.
2. Read the intersection carefully and verify it in both equations.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 4 x - y = 8 \) ...(i), \( y = 4 x - 8 \).
| x | 2 | 3 | 4 |
| y | 0 | 4 | 8 |
For the equation \( 2 x - 3 y = -6 \) ...(ii), \( y = \frac{2 x}{3} + 2 \).
| x | 0 | 3 | 6 |
| y | 2 | 4 | 6 |
From the graph, the two lines intersect at the point \( (3, 4) \). Hence the solution is \( x = 3,\ y = 4 \).
The shaded region bounded by the two lines and the x-axis is the triangle with vertices \( (-3, 0) \), \( (3, 4) \), \( (2, 0) \).
In simple words: Plot both lines on the same axes; the point where they cross satisfies both equations.
Teacher's Note:
1. Plot at least three points for each line so that an error in one point is easy to spot.
2. Read the intersection carefully and verify it in both equations.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 3 x - 4 y = -6 \) ...(i), \( y = \frac{3 x}{4} + \frac{3}{2} \).
| x | -2 | 2 | 6 |
| y | 0 | 3 | 6 |
For the equation \( 3 x + y = 9 \) ...(ii), \( y = 9 - 3 x \).
| x | 1 | 2 | 3 |
| y | 6 | 3 | 0 |
From the graph, the two lines intersect at the point \( (2, 3) \). Hence the solution is \( x = 2,\ y = 3 \).
The shaded region bounded by the two lines and the x-axis is the triangle with vertices \( (-2, 0) \), \( (2, 3) \), \( (3, 0) \).
In simple words: Plot both lines on the same axes; the point where they cross satisfies both equations.
Teacher's Note:
1. Plot at least three points for each line so that an error in one point is easy to spot.
2. Read the intersection carefully and verify it in both equations.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 4 x - 3 y = 6 \) ...(i), \( y = \frac{4 x}{3} - 2 \).
| x | 0 | 3 | 6 |
| y | -2 | 2 | 6 |
For the equation \( x + 3 y = 9 \) ...(ii), \( y = 3 - \frac{x}{3} \).
| x | 0 | 3 | 6 |
| y | 3 | 2 | 1 |
From the graph, the two lines intersect at the point \( (3, 2) \). Hence the solution is \( x = 3,\ y = 2 \).
The region bounded by the two lines and the y-axis is the triangle with vertices \( (0, -2) \), \( (3, 2) \), \( (0, 3) \).
In simple words: Plot both lines on the same axes; the point where they cross satisfies both equations.
Teacher's Note:
1. Plot at least three points for each line so that an error in one point is easy to spot.
2. Read the intersection carefully and verify it in both equations.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 3 x - 2 y = 1 \) ...(i), \( y = \frac{3 x}{2} - \frac{1}{2} \).
| x | 1 | 3 | 5 |
| y | 1 | 4 | 7 |
For the equation \( 2 x - 3 y = -6 \) ...(ii), \( y = \frac{2 x}{3} + 2 \).
| x | 0 | 3 | 6 |
| y | 2 | 4 | 6 |
From the graph, the two lines intersect at the point \( (3, 4) \). Hence the solution is \( x = 3,\ y = 4 \).
The shaded region bounded by the two lines and the x-axis is the triangle with vertices \( (-3, 0) \), \( (3, 4) \), \( (\frac{1}{3}, 0) \).
In simple words: Plot both lines on the same axes; the point where they cross satisfies both equations.
Teacher's Note:
1. Plot at least three points for each line so that an error in one point is easy to spot.
2. Read the intersection carefully and verify it in both equations.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 3 x + 2 y = 12 \) ...(i), \( y = 6 - \frac{3 x}{2} \).
| x | 0 | 2 | 4 |
| y | 6 | 3 | 0 |
For the equation \( 5 x - 2 y = 4 \) ...(ii), \( y = \frac{5 x}{2} - 2 \).
| x | 0 | 2 | 4 |
| y | -2 | 3 | 8 |
From the graph, the two lines intersect at the point \( (2, 3) \). Hence the solution is \( x = 2,\ y = 3 \).
The lines meet the y-axis at \( (0, 6) \) and \( (0, -2) \).
In simple words: Plot both lines on the same axes; the point where they cross satisfies both equations.
Teacher's Note:
1. Plot at least three points for each line so that an error in one point is easy to spot.
2. Read the intersection carefully and verify it in both equations.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 2 x + 3 y = 8 \) ...(i), \( y = \frac{8}{3} - \frac{2 x}{3} \).
| x | -2 | 1 | 4 |
| y | 4 | 2 | 0 |
For the equation \( x + 4 y = 9 \) ...(ii), \( y = \frac{9}{4} - \frac{x}{4} \).
| x | -3 | 1 | 5 |
| y | 3 | 2 | 1 |
From the graph, the two lines intersect at the point \( (1, 2) \). Hence the solution is \( x = 1,\ y = 2 \).
In simple words: Plot both lines on the same axes; the point where they cross satisfies both equations.
Teacher's Note:
1. Plot at least three points for each line so that an error in one point is easy to spot.
2. Read the intersection carefully and verify it in both equations.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 2 x + 3 y = 12 \) ...(i), \( y = 4 - \frac{2 x}{3} \).
| x | 0 | 3 | 6 |
| y | 4 | 2 | 0 |
For the equation \( - x + 2 y = 1 \) ...(ii), \( y = \frac{x}{2} + \frac{1}{2} \).
| x | 1 | 3 | 5 |
| y | 1 | 2 | 3 |
From the graph, the two lines intersect at the point \( (3, 2) \). Hence the solution is \( x = 3,\ y = 2 \).
In simple words: Plot both lines on the same axes; the point where they cross satisfies both equations.
Teacher's Note:
1. Plot at least three points for each line so that an error in one point is easy to spot.
2. Read the intersection carefully and verify it in both equations.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 3 x + y = 5 \) ...(i), \( y = 5 - 3 x \).
| x | 1 | 2 | 3 |
| y | 2 | -1 | -4 |
For the equation \( 2 x - y = 5 \) ...(ii), \( y = 2 x - 5 \).
| x | 1 | 2 | 3 |
| y | -3 | -1 | 1 |
From the graph, the two lines intersect at the point \( (2, -1) \). Hence the solution is \( x = 2,\ y = -1 \).
The lines meet the y-axis at \( (0, 5) \) and \( (0, -5) \).
In simple words: Plot both lines on the same axes; the point where they cross satisfies both equations.
Teacher's Note:
1. Plot at least three points for each line so that an error in one point is easy to spot.
2. Read the intersection carefully and verify it in both equations.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( x + 3 y = 6 \) ...(i), \( y = 2 - \frac{x}{3} \).
| x | 3 | 6 | 9 |
| y | 1 | 0 | -1 |
For the equation \( 2 x - 3 y = 12 \) ...(ii), \( y = \frac{2 x}{3} - 4 \).
| x | 3 | 6 | 9 |
| y | -2 | 0 | 2 |
From the graph, the two lines intersect at the point \( (6, 0) \). Hence the solution is \( x = 6,\ y = 0 \).
The lines meet the y-axis at \( (0, 2) \) and \( (0, -4) \).
In simple words: Plot both lines on the same axes; the point where they cross satisfies both equations.
Teacher's Note:
1. Plot at least three points for each line so that an error in one point is easy to spot.
2. Read the intersection carefully and verify it in both equations.
Free study material for Mathematics
Download Practice Assignments: Class 10 Mathematics Chapter 03 Pair of Linear Equations in Two Variables
Revision Assignment: Chapter 03 Pair of Linear Equations in Two Variables (CBSE)
Review targeted chapter assignments for Class 10 Mathematics Chapter 03 Pair of Linear Equations in Two Variables. Built according to official CBSE guidelines, these downloadable problem sets help students build accuracy and prepare effectively for school tests.
Key Advantages of Solving Chapter 03 Pair of Linear Equations in Two Variables Assignments
- Syllabus Compliance: Sets reflect current CBSE evaluation criteria and official marking frameworks.
- Multi-Format Practice: Includes varied problem types designed to deepen comprehension across all sub-topics.
- Time Management: Routine practice optimizes pacing to finish school examinations comfortably within schedule.
Effective Strategy for Class 10 Mathematics Assignments
- Textbook Review: Always study the core NCERT book for Class 10 Mathematics prior to beginning the assignment.
- Independent Attempt: Solve Chapter 03 Pair of Linear Equations in Two Variables questions on your own initially before cross-checking with expert solutions.
- Error Tracking: Record challenging concepts in a dedicated notebook and practice online MCQ tests for revision.
FAQs
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Yes, our teachers have given solutions for all questions in the Class 10 Mathematics Chapter 03 Pair of Linear Equations in Two Variables assignments. This will help you to understand step-by-step methodology to get full marks in school tests and exams.
Yes. These assignments are designed as per the latest CBSE syllabus for 2026. We have included huge variety of question formats such as MCQs, Case-study based questions and important diagram-based problems found in Chapter 03 Pair of Linear Equations in Two Variables.
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