Class 10 Mathematics Practice Assignments: CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Assignment Set 17
Access comprehensive school assignments for Chapter 03 Pair of Linear Equations in Two Variables using the CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Assignment Set 17. Designed to align with the 2026-27 CBSE academic guidelines, these practice sets help Class 10 Mathematics students reinforce core concepts and improve their problem-solving accuracy.
Download Chapter 03 Pair of Linear Equations in Two Variables Assignment PDF with Solutions
Navigate directly to the solved Mathematics assignments using the digital viewer below. Each practice set includes detailed step-by-step solutions, allowing students to instantly cross-check their work and identify areas requiring further revision.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( x + y = 4 \) ...(i), \( y = 4 - x \).
| x | 2 | 3 | 4 |
| y | 2 | 1 | 0 |
For the equation \( 2 x - 3 y = 3 \) ...(ii), \( y = \frac{2 x}{3} - 1 \).
| x | 0 | 3 | 6 |
| y | -1 | 1 | 3 |
From the graph, the two lines intersect at the point \( (3, 1) \). Hence the solution is \( x = 3,\ y = 1 \).
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 \( x + y = 3 \) ...(i), \( y = 3 - x \).
| x | 0 | 1 | 2 |
| y | 3 | 2 | 1 |
For the equation \( 2 x + 5 y = 12 \) ...(ii), \( y = \frac{12}{5} - \frac{2 x}{5} \).
| x | -4 | 1 | 6 |
| y | 4 | 2 | 0 |
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 \( 4 x + 3 y = 9 \) ...(i), \( y = 3 - \frac{4 x}{3} \).
| x | 0 | 3 | 6 |
| y | 3 | -1 | -5 |
For the equation \( 5 x + 2 y = 13 \) ...(ii), \( y = \frac{13}{2} - \frac{5 x}{2} \).
| x | 1 | 3 | 5 |
| y | 4 | -1 | -6 |
From the graph, the two lines intersect at the point \( (3, -1) \). Hence the solution is \( x = 3,\ y = -1 \).
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 \( 2 x + 3 y = 4 \) ...(i), \( y = \frac{4}{3} - \frac{2 x}{3} \).
| x | -4 | -1 | 2 |
| y | 4 | 2 | 0 |
For the equation \( x - y = -3 \) ...(ii), \( y = x + 3 \).
| x | -2 | -1 | 0 |
| y | 1 | 2 | 3 |
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 \( x + y = 7 \) ...(i), \( y = 7 - x \).
| x | 1 | 2 | 3 |
| y | 6 | 5 | 4 |
For the equation \( 5 x + 2 y = 20 \) ...(ii), \( y = 10 - \frac{5 x}{2} \).
| x | 0 | 2 | 4 |
| y | 10 | 5 | 0 |
From the graph, the two lines intersect at the point \( (2, 5) \). Hence the solution is \( x = 2,\ y = 5 \).
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 \( x + 4 y = 0 \) ...(i), \( y = - \frac{x}{4} \).
| x | -4 | 0 | 4 |
| y | 1 | 0 | -1 |
For the equation \( x + 4 y = 0 \) ...(ii), \( y = - \frac{x}{4} \).
| x | -4 | 0 | 4 |
| y | 1 | 0 | -1 |
The two lines coincide, so the system has infinitely many solutions (it is consistent and dependent).
In simple words: Two straight lines either cross at one point, never meet, or lie on top of each other.
Teacher's Note:
1. Check the ratios of coefficients to confirm what the graph shows.
2. Parallel lines mean no solution.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( x + 2 y = 3 \) ...(i), \( y = \frac{3}{2} - \frac{x}{2} \).
| x | -1 | 1 | 3 |
| y | 2 | 1 | 0 |
For the equation \( 2 x + 4 y = 15 \) ...(ii), \( y = \frac{15}{4} - \frac{x}{2} \).
| x | -1/2 | 0 | 1/2 |
| y | 4 | 15/4 | 7/2 |
The two lines are parallel and never meet, so the system has no solution (it is inconsistent).
In simple words: Two straight lines either cross at one point, never meet, or lie on top of each other.
Teacher's Note:
1. Check the ratios of coefficients to confirm what the graph shows.
2. Parallel lines mean no solution.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 3 x + 2 y = 3 \) ...(i), \( y = \frac{3}{2} - \frac{3 x}{2} \).
| x | -1 | 1 | 3 |
| y | 3 | 0 | -3 |
For the equation \( 6 x + 4 y = 15 \) ...(ii), \( y = \frac{15}{4} - \frac{3 x}{2} \).
| x | -1/2 | 0 | 1/2 |
| y | 9/2 | 15/4 | 3 |
The two lines are parallel and never meet, so the system has no solution (it is inconsistent).
In simple words: Two straight lines either cross at one point, never meet, or lie on top of each other.
Teacher's Note:
1. Check the ratios of coefficients to confirm what the graph shows.
2. Parallel lines mean no solution.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 2 x + 3 y = 5 \) ...(i), \( y = \frac{5}{3} - \frac{2 x}{3} \).
| x | -2 | 1 | 4 |
| y | 3 | 1 | -1 |
For the equation \( 2 x + 3 y = 5 \) ...(ii), \( y = \frac{5}{3} - \frac{2 x}{3} \).
| x | -2 | 1 | 4 |
| y | 3 | 1 | -1 |
The two lines coincide, so the system has infinitely many solutions (it is consistent and dependent).
In simple words: Two straight lines either cross at one point, never meet, or lie on top of each other.
Teacher's Note:
1. Check the ratios of coefficients to confirm what the graph shows.
2. Parallel lines mean no solution.
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 \( x = 6,\ y = 0 \).
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 \( 2 x - 3 y = -7 \) ...(i), \( y = \frac{2 x}{3} + \frac{7}{3} \).
| x | -2 | 1 | 4 |
| y | 1 | 3 | 5 |
For the equation \( 2 x - 3 y = -7 \) ...(ii), \( y = \frac{2 x}{3} + \frac{7}{3} \).
| x | -2 | 1 | 4 |
| y | 1 | 3 | 5 |
The two lines coincide, so the system has infinitely many solutions (it is consistent and dependent).
In simple words: Two straight lines either cross at one point, never meet, or lie on top of each other.
Teacher's Note:
1. Check the ratios of coefficients to confirm what the graph shows.
2. Parallel lines mean no solution.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 8 x + 5 y = 9 \) ...(i), \( y = \frac{9}{5} - \frac{8 x}{5} \).
| x | -7 | -2 | 3 |
| y | 13 | 5 | -3 |
For the equation \( 16 x + 10 y = 27 \) ...(ii), \( y = \frac{27}{10} - \frac{8 x}{5} \).
| x | -1/2 | 0 | 1/2 |
| y | 7/2 | 27/10 | 19/10 |
The two lines are parallel and never meet, so the system has no solution (it is inconsistent).
In simple words: Two straight lines either cross at one point, never meet, or lie on top of each other.
Teacher's Note:
1. Check the ratios of coefficients to confirm what the graph shows.
2. Parallel lines mean no solution.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 5 x - 8 y = -1 \) ...(i), \( y = \frac{5 x}{8} + \frac{1}{8} \).
| x | -5 | 3 | 11 |
| y | -3 | 2 | 7 |
For the equation \( 5 x - 8 y = -1 \) ...(ii), \( y = \frac{5 x}{8} + \frac{1}{8} \).
| x | -5 | 3 | 11 |
| y | -3 | 2 | 7 |
The two lines coincide, so the system has infinitely many solutions (it is consistent and dependent).
In simple words: Two straight lines either cross at one point, never meet, or lie on top of each other.
Teacher's Note:
1. Check the ratios of coefficients to confirm what the graph shows.
2. Parallel lines mean no solution.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 2 x - 3 y = 4 \) ...(i), \( y = \frac{2 x}{3} - \frac{4}{3} \).
| x | -4 | -1 | 2 |
| y | -4 | -2 | 0 |
For the equation \( 3 x - 2 y = 1 \) ...(ii), \( y = \frac{3 x}{2} - \frac{1}{2} \).
| x | -3 | -1 | 1 |
| y | -5 | -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 \( 3 x + 4 y = 6 \) ...(i), \( y = \frac{3}{2} - \frac{3 x}{4} \).
| x | -2 | 2 | 6 |
| y | 3 | 0 | -3 |
For the equation \( 3 x + 4 y = 6 \) ...(ii), \( y = \frac{3}{2} - \frac{3 x}{4} \).
| x | -2 | 2 | 6 |
| y | 3 | 0 | -3 |
The two lines coincide, so the system has infinitely many solutions (it is consistent and dependent).
In simple words: Two straight lines either cross at one point, never meet, or lie on top of each other.
Teacher's Note:
1. Check the ratios of coefficients to confirm what the graph shows.
2. Parallel lines mean no solution.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 2 x - 3 y = 6 \) ...(i), \( y = \frac{2 x}{3} - 2 \).
| x | -3 | 0 | 3 |
| y | -4 | -2 | 0 |
For the equation \( 4 x - 6 y = 9 \) ...(ii), \( y = \frac{2 x}{3} - \frac{3}{2} \).
| x | -1/2 | 0 | 1/2 |
| y | -11/6 | -3/2 | -7/6 |
The two lines are parallel and never meet, so the system has no solution (it is inconsistent).
In simple words: Two straight lines either cross at one point, never meet, or lie on top of each other.
Teacher's Note:
1. Check the ratios of coefficients to confirm what the graph shows.
2. Parallel lines mean no solution.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 3 x - 5 y = -1 \) ...(i), \( y = \frac{3 x}{5} + \frac{1}{5} \).
| x | -7 | -2 | 3 |
| y | -4 | -1 | 2 |
For the equation \( 2 x - y = -3 \) ...(ii), \( y = 2 x + 3 \).
| x | -3 | -2 | -1 |
| y | -3 | -1 | 1 |
From the graph, the two lines intersect at the point \( (-2, -1) \). Hence the solution is \( x = -2,\ y = -1 \).
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 \( x + 3 y = 6 \) ...(i), \( y = 2 - \frac{x}{3} \).
| x | 3 | 6 | 9 |
| y | 1 | 0 | -1 |
For the equation \( 3 x - 5 y = 18 \) ...(ii), \( y = \frac{3 x}{5} - \frac{18}{5} \).
| x | 1 | 6 | 11 |
| y | -3 | 0 | 3 |
From the graph, the two lines intersect at the point \( (6, 0) \). Hence the solution is \( x = 6,\ y = 0 \).
Since x = 6 and y = 0, \( K = 7(6) + 3(0) = 42 \).
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 \( 2 x - y = 1 \) ...(i), \( y = 2 x - 1 \).
| x | 1 | 2 | 3 |
| y | 1 | 3 | 5 |
For the equation \( x + 2 y = 8 \) ...(ii), \( y = 4 - \frac{x}{2} \).
| x | 0 | 2 | 4 |
| y | 4 | 3 | 2 |
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 \( (0, -1) \) 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.
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 \( 2 x - y = 4 \) ...(ii), \( y = 2 x - 4 \).
| x | 2 | 3 | 4 |
| y | 0 | 2 | 4 |
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 \( (0, -4) \).
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.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( x + y = 4 \) ...(i), \( y = 4 - x \).
| x | 0 | 1 | 2 |
| y | 4 | 3 | 2 |
For the equation \( 3 x - 2 y = -3 \) ...(ii), \( y = \frac{3 x}{2} + \frac{3}{2} \).
| x | -1 | 1 | 3 |
| y | 0 | 3 | 6 |
From the graph, the two lines intersect at the point \( (1, 3) \). Hence the solution is \( x = 1,\ y = 3 \).
The shaded region bounded by the two lines and the x-axis is the triangle with vertices \( (-1, 0) \), \( (1, 3) \), \( (4, 0) \).
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 \( 2 x + y = 8 \) ...(i), \( y = 8 - 2 x \).
| x | 3 | 4 | 5 |
| y | 2 | 0 | -2 |
For the equation \( 3 x - 2 y = 12 \) ...(ii), \( y = \frac{3 x}{2} - 6 \).
| x | 2 | 4 | 6 |
| y | -3 | 0 | 3 |
From the graph, the two lines intersect at the point \( (4, 0) \). Hence the solution is \( x = 4,\ y = 0 \).
The lines meet the x-axis at \( (4, 0) \) and \( (4, 0) \).
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 \( x - y = 1 \) ...(i), \( y = x - 1 \).
| x | 2 | 3 | 4 |
| y | 1 | 2 | 3 |
For the equation \( 2 x + y = 8 \) ...(ii), \( y = 8 - 2 x \).
| x | 2 | 3 | 4 |
| y | 4 | 2 | 0 |
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 \( (0, -1) \), \( (3, 2) \), \( (0, 8) \).
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 \( - x + 2 y = 8 \) ...(i), \( y = \frac{x}{2} + 4 \).
| x | -2 | 0 | 2 |
| y | 3 | 4 | 5 |
For \( - x + 5 y = 14 \) ...(ii), \( y = \frac{x}{5} + \frac{14}{5} \).
| x | -4 | 1 | 6 |
| y | 2 | 3 | 4 |
For \( - 2 x + y = 1 \) ...(iii), \( y = 2 x + 1 \).
| x | -1 | 0 | 1 |
| y | -1 | 1 | 3 |
From the graph, the lines meet pairwise at \( (-4, 2) \), \( (2, 5) \), \( (1, 3) \). These are the vertices of the triangle formed.
In simple words: Three lines, taken two at a time, meet in three points, which are the corners of a triangle.
Teacher's Note:
1. Solve each pair of equations to confirm the vertices read from the graph.
2. Plot the three vertices and join them to see the triangle.
Answer:
Draw the graph of each equation by finding three points on it.
For the equation \( 4 x - 3 y = -4 \) ...(i), \( y = \frac{4 x}{3} + \frac{4}{3} \).
| x | -1 | 2 | 5 |
| y | 0 | 4 | 8 |
For the equation \( 4 x + 3 y = 20 \) ...(ii), \( y = \frac{20}{3} - \frac{4 x}{3} \).
| x | -1 | 2 | 5 |
| y | 8 | 4 | 0 |
From the graph, the two lines intersect at the point \( (2, 4) \). Hence the solution is \( x = 2,\ y = 4 \).
The shaded region bounded by the two lines and the x-axis is the triangle with vertices \( (-1, 0) \), \( (2, 4) \), \( (5, 0) \).
Its base along the x-axis is \( 6 \) units and its height is \( 4 \) units, so the area is \( \frac{1}{2} \times 6 \times 4 = 12 \) square units.
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)
Access structured practice assignments for Chapter 03 Pair of Linear Equations in Two Variables designed in alignment with the latest CBSE curriculum for Class 10 Mathematics. These printable sets cover objective and descriptive problem types to support thorough revision.
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Effective Strategy for Class 10 Mathematics Assignments
- Initial Reading: Begin by reading the NCERT book for Class 10 Mathematics to build a baseline understanding.
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- Supplementary Aids: Leverage revision notes and worksheets whenever you encounter difficult topics.
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