WorksheetClass 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.

    Click here to download the PDF of additional worksheets for practice on Pair of Linear Equations in Two Variables, Chapter of Class 10 Maths, along with the answer key:

    Real Numbers Worksheet Class 10 | Class 10 Math Polynomials Worksheets

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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=7x + y = 7

        xy=3x – y = 3


      2. 2x+3y=122x + 3y = 12

        3xy=53x – y = 5


      3. x+2y=8x + 2y = 8

        2x+y=72x + y = 7

      Solutions:


      1. x+y=7x + y = 7

        xy=3x – y = 3

        • Add both equations:
          2x=10x=52x = 10 \Rightarrow x = 5

        • Substitute
          x=5x = 5

          in

          x+y=7x + y = 7

          :

          5+y=7y=25 + y = 7 \Rightarrow y = 2

        • Solution:
          x=5,y=2x = 5, y = 2


      2. 2x+3y=122x + 3y = 12

        3xy=53x – y = 5

        • Solve
          3xy=53x – y = 5

          for

          yy

          :

          y=3x5y = 3x – 5

        • Substitute
          y=3x5y = 3x – 5

          in

          2x+3y=122x + 3y = 12

          :

          2x+3(3x5)=1211x15=12x=32x + 3(3x – 5) = 12 \Rightarrow 11x – 15 = 12 \Rightarrow x = 3

        • Substitute
          x=3x = 3

          in

          y=3x5y = 3x – 5

          :

          y=95y=4y = 9 – 5 \Rightarrow y = 4

        • Solution:
          x=3,y=4x = 3, y = 4


      3. x+2y=8x + 2y = 8

        2x+y=72x + y = 7

        • Solve
          x+2y=8x + 2y = 8

          for

          xx

          :

          x=82yx = 8 – 2y

        • Substitute
          x=82yx = 8 – 2y

          in

          2x+y=72x + y = 7

          :

          2(82y)+y=7164y+y=7y=32(8 – 2y) + y = 7 \Rightarrow 16 – 4y + y = 7 \Rightarrow y = 3

        • Substitute
          y=3y = 3

          in

          x=82yx = 8 – 2y

          :

          x=86x=2x = 8 – 6 \Rightarrow x = 2

        • Solution:
          x=2,y=3x = 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=4x + y = 4

        xy=2x – y = 2


      2. 2x+y=52x + y = 5

        xy=1x – y = 1

      Solutions:


      1. x+y=4x + y = 4

        xy=2x – y = 2

        • Equation 1:
          x+y=4x + y = 4

          can be rewritten as

          y=4xy = 4 – x

        • Equation 2:
          xy=2x – y = 2

          can be rewritten as

          y=x2y = x – 2

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

        • Solution:
          x=3,y=1x = 3, y = 1


      2. 2x+y=52x + y = 5

        xy=1x – y = 1

        • Equation 1:
          2x+y=52x + y = 5

          can be rewritten as

          y=52xy = 5 – 2x

        • Equation 2:
          xy=1x – y = 1

          can be rewritten as

          y=x1y = x – 1

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

        • Solution:
          x=2,y=1x = 2, y = 1

      Worksheet 3: Elimination Method

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


      1. 3x+4y=103x + 4y = 10

        2xy=32x – y = 3


      2. x+3y=6x + 3y = 6

        2x3y=122x – 3y = 12

      Solutions:


      1. 3x+4y=103x + 4y = 10

        2xy=32x – y = 3

        • Multiply second equation by 4:
          8x4y=128x – 4y = 12

        • Add both equations:
          3x+4y+8x4y=10+1211x=22x=23x + 4y + 8x – 4y = 10 + 12 \Rightarrow 11x = 22 \Rightarrow x = 2

        • Substitute
          x=2x = 2

          in

          2xy=32x – y = 3

          :

          4y=3y=14 – y = 3 \Rightarrow y = 1

        • Solution:
          x=2,y=1x = 2, y = 1


      2. x+3y=6x + 3y = 6

        2x3y=122x – 3y = 12

        • Add both equations:
          x+3y+2x3y=6+123x=18x=6x + 3y + 2x – 3y = 6 + 12 \Rightarrow 3x = 18 \Rightarrow x = 6

        • Substitute
          x=6x = 6

          in

          x+3y=6x + 3y = 6

          :

          6+3y=63y=0y=06 + 3y = 6 \Rightarrow 3y = 0 \Rightarrow y = 0

        • Solution:
          x=6,y=0x = 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
        xx

        and

        yy

        .


        • x+y=14x + y = 14


        • xy=2x – y = 2

        • Add both equations:
          2x=16x=82x = 16 \Rightarrow x = 8

        • Substitute
          x=8x = 8

          in

          x+y=14x + y = 14

          :

          8+y=14y=68 + y = 14 \Rightarrow y = 6

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

        and one pencil be

        yy

        .


        • 5x+3y=235x + 3y = 23


        • 2x+2y=102x + 2y = 10

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

        • Solve
          x+y=5x + y = 5

          for

          yy

          :

          y=5xy = 5 – x

        • Substitute
          y=5xy = 5 – x

          in

          5x+3y=235x + 3y = 23

          :

          5x+3(5x)=235x+153x=232x=8x=45x + 3(5 – x) = 23 \Rightarrow 5x + 15 – 3x = 23 \Rightarrow 2x = 8 \Rightarrow x = 4

        • Substitute
          x=4x = 4

          in

          x+y=5x + y = 5

          :

          4+y=5y=14 + y = 5 \Rightarrow y = 1

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