Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Worksheet

Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Worksheet

Class 10 Maths Worksheets for Chapter 3 (Pair of Linear Equations in Two Variables) are available here online with answers. These worksheets will help students score good marks in their board exams. The questions are made according to the latest CBSE syllabus (2024-2025) and NCERT textbook. Our subject experts have prepared these worksheets following the latest exam pattern. Practicing these questions will help students develop problem-solving skills. To get chapter-wise worksheets, click here. You can also download the PDF for extra practice questions.

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Class 10 Maths Worksheet for Pair of Linear Equations in Two Variables

The CBSE board has released the Class 10 Maths exam datasheet. It’s time for students to review the chapters for the board exam. Students should solve these questions and choose the correct answers. They can check their answers with the provided solutions. Get important questions for Class 10 Maths.

Class 10 Maths Worksheet for Pair of Linear Equations in Two Variables – Download PDF

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Class 10 Maths Worksheet for Pair of Linear Equations in Two Variables

Class 10 Math Chapter 3 Worksheet

Worksheet 1: Basics of Linear Equations

Instructions: Solve the following pairs of linear equations by substitution method.

1. $x + y = 7$

$x – y = 3$

2. $2x + 3y = 12$

$3x – y = 5$

3. $x + 2y = 8$

$2x + y = 7$

Solutions:

1. $x + y = 7$

$x – y = 3$

$2x = 10 \Rightarrow x = 5$

• Substitute
$x = 5$

in

$x + y = 7$

:

$5 + y = 7 \Rightarrow y = 2$

• Solution:
$x = 5, y = 2$

2. $2x + 3y = 12$

$3x – y = 5$

• Solve
$3x – y = 5$

for

$y$

:

$y = 3x – 5$

• Substitute
$y = 3x – 5$

in

$2x + 3y = 12$

:

$2x + 3(3x – 5) = 12 \Rightarrow 11x – 15 = 12 \Rightarrow x = 3$

• Substitute
$x = 3$

in

$y = 3x – 5$

:

$y = 9 – 5 \Rightarrow y = 4$

• Solution:
$x = 3, y = 4$

3. $x + 2y = 8$

$2x + y = 7$

• Solve
$x + 2y = 8$

for

$x$

:

$x = 8 – 2y$

• Substitute
$x = 8 – 2y$

in

$2x + y = 7$

:

$2(8 – 2y) + y = 7 \Rightarrow 16 – 4y + y = 7 \Rightarrow y = 3$

• Substitute
$y = 3$

in

$x = 8 – 2y$

:

$x = 8 – 6 \Rightarrow x = 2$

• Solution:
$x = 2, y = 3$

Worksheet 2: Graphical Method

Instructions: Solve the following pairs of linear equations by graphical method. Draw the graphs on a graph paper.

1. $x + y = 4$

$x – y = 2$

2. $2x + y = 5$

$x – y = 1$

Solutions:

1. $x + y = 4$

$x – y = 2$

• Equation 1:
$x + y = 4$

can be rewritten as

$y = 4 – x$

• Equation 2:
$x – y = 2$

can be rewritten as

$y = x – 2$

• Plotting these equations on graph paper, the lines intersect at
$(3, 1)$

• Solution:
$x = 3, y = 1$

2. $2x + y = 5$

$x – y = 1$

• Equation 1:
$2x + y = 5$

can be rewritten as

$y = 5 – 2x$

• Equation 2:
$x – y = 1$

can be rewritten as

$y = x – 1$

• Plotting these equations on graph paper, the lines intersect at
$(2, 1)$

• Solution:
$x = 2, y = 1$

Worksheet 3: Elimination Method

Instructions: Solve the following pairs of linear equations by elimination method.

1. $3x + 4y = 10$

$2x – y = 3$

2. $x + 3y = 6$

$2x – 3y = 12$

Solutions:

1. $3x + 4y = 10$

$2x – y = 3$

• Multiply second equation by 4:
$8x – 4y = 12$

$3x + 4y + 8x – 4y = 10 + 12 \Rightarrow 11x = 22 \Rightarrow x = 2$

• Substitute
$x = 2$

in

$2x – y = 3$

:

$4 – y = 3 \Rightarrow y = 1$

• Solution:
$x = 2, y = 1$

2. $x + 3y = 6$

$2x – 3y = 12$

$x + 3y + 2x – 3y = 6 + 12 \Rightarrow 3x = 18 \Rightarrow x = 6$

• Substitute
$x = 6$

in

$x + 3y = 6$

:

$6 + 3y = 6 \Rightarrow 3y = 0 \Rightarrow y = 0$

• Solution:
$x = 6, y = 0$

Worksheet 4: Word Problems

Instructions: Solve the following word problems using the concept of linear equations in two variables.

1. The sum of two numbers is 14, and their difference is 2. Find the numbers.
2. The cost of 5 pens and 3 pencils is ₹23. The cost of 2 pens and 2 pencils is ₹10. Find the cost of each pen and pencil.

Solutions:

1. Let the numbers be
$x$

and

$y$

.

• $x + y = 14$

• $x – y = 2$

$2x = 16 \Rightarrow x = 8$

• Substitute
$x = 8$

in

$x + y = 14$

:

$8 + y = 14 \Rightarrow y = 6$

• Solution: The numbers are 8 and 6.
2. Let the cost of one pen be
$x$

and one pencil be

$y$

.

• $5x + 3y = 23$

• $2x + 2y = 10$

• Simplify second equation:
$x + y = 5$

• Solve
$x + y = 5$

for

$y$

:

$y = 5 – x$

• Substitute
$y = 5 – x$

in

$5x + 3y = 23$

:

$5x + 3(5 – x) = 23 \Rightarrow 5x + 15 – 3x = 23 \Rightarrow 2x = 8 \Rightarrow x = 4$

• Substitute
$x = 4$

in

$x + y = 5$

:

$4 + y = 5 \Rightarrow y = 1$

• Solution: The cost of each pen is ₹4 and each pencil is ₹1.

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