RD Sharma Solutions for Class 8 Maths Chapter 9 Linear Equations in One Variable

Mathematics is considered one of the highest-scoring subjects, and with proper preparation, students can achieve full marks in exams. However, preparing for the annual exam can be challenging especially when it comes to solving tricky problems. That’s why having clear, well-structured solutions is so important.

To make your preparation easier, here are comprehensive RD Sharma Class 8 Maths Solutions for Chapter 9 - Linear Equations in One Variable. These solutions are carefully prepared in line with the latest CBSE syllabus and exam pattern, ensuring you're studying exactly what matters. They also include smart tricks and shortcut methods to help you solve problems faster and with better accuracy.

You can download the free PDFs of all exercises for this chapter using the links provided below.

This chapter includes four exercises and focuses on building a strong foundation in solving linear equations. Here's a quick overview of the key concepts covered:

• Understanding linear equations and their basic definitions
• Finding the solution of a linear equation
• Solving equations with variable terms on one side and numbers on the other
• Using the transposition method to solve equations
• Applying the cross-multiplication method
• Real-life applications of linear equations through word problems

Practising these problems regularly not only strengthens your concept clarity but also boosts your confidence for the exam. Make the most of these RD Sharma solutions and score higher with ease.

Download Free RD Sharma Solutions PDF for Class 8 Maths Chapter 9 Linear Equations in One Variable

Get the complete RD Sharma Solutions for Class 8 Maths Chapter 9 – Linear Equations in One Variable in easy-to-understand PDF format. All exercise questions are solved step-by-step by expert teachers to help you revise the full syllabus and score better marks in your exams.

These solutions cover every exercise in Chapter 9 and explain the concepts in a simple way so that you can learn and practice without any confusion. Whether you're preparing for school exams or just want to strengthen your basics, these solutions are a great study tool.

You can download the PDFs for free and use them to revise anytime. These chapter-wise solutions are made to match the latest CBSE guidelines, helping you build confidence and improve your problem-solving skills. If you're looking for clear, reliable answers to RD Sharma Class 8 Maths questions, this is the perfect resource to help you prepare well and perform better.

Access Answers to Maths RD Sharma Solutions for Class 8 Chapter 9 

Q. Solve each of the following equations and also verify your solution:

1. 5x/3 + 2/5 = 1
2. 9 ¼ = y – 1 1/3
3. x/2 + x/8 = 1/8
4.  2x/3 – 3x/8 = 7/12
5. (x + 2) (x + 3) + (x – 3) (x – 2) – 2x(x + 1) = 0

Solutions: 

1. (5x/3) + (2/5) = 1

Solution:

Subtract 2/5 from both sides:

5x/3 = 1 - 2/5 = 3/5

Multiply both sides by 3:

5x = 9/5

Divide both sides by 5:

x = 9/25

x = 9/25

Verification:

(5 × 9/25) ÷ 3 + 2/5 = 45/75 + 2/5 = 3/5 + 2/5 = 1

2. 9¼ = y − 1⅓

Solution:

Convert mixed numbers: 9¼ = 37/4, 1⅓ = 4/3

y = 37/4 + 4/3

LCM of 4 and 3 is 12:

y = 111/12 + 16/12 = 127/12

y = 127/12

Verification:

y − 4/3 = 127/12 − 16/12 = 111/12 = 37/4 = 9¼

3. (x/2) + (x/8) = 1/8

Solution:

Convert to common denominator: (4x + x)/8 = 5x/8 = 1/8

Multiply both sides by 8: 5x = 1 → x = 1/5

x = 1/5

Verification:

(1/5)/2 + (1/5)/8 = 1/10 + 1/40 = 5/40 = 1/8

4. (2x/3) − (3x/8) = 7/12

Solution:

Convert to common denominator: (16x − 9x)/24 = 7x/24

7x/24 = 7/12

Cross-multiply: 84x = 168 → x = 2

x = 2

Verification:

(2×2)/3 − (3×2)/8 = 4/3 − 6/8 = 4/3 − 3/4 = (16 − 9)/12 = 7/12

5. (x + 2)(x + 3) + (x − 3)(x − 2) − 2x(x + 1) = 0

Solution:

(x + 2)(x + 3) = x² + 5x + 6

(x − 3)(x − 2) = x² − 5x + 6

2x(x + 1) = 2x² + 2x

Total: [x² + 5x + 6] + [x² − 5x + 6] − [2x² + 2x]

Simplifies to: 0 − 2x + 12 = 0 → 2x = 12 → x = 6

x = 6

Verification:

(6 + 2)(6 + 3) + (6 − 3)(6 − 2) − 2×6×(6 + 1)

= 8×9 + 3×4 − 2×6×7 = 72 + 12 − 84 = 0

Solve each of the following equations and also check your results in each case:

1. (a-8)/3 = (a-3)/2

Solution:

(a-8)/3 = (a-3)/2

By using cross-multiplication, we get,

(a-8)2 = (a-3)3

2a – 16 = 3a – 9

2a – 3a = -9 + 16

-a = 7

a = -7

Let us verify the given equation now,

(a-8)/3 = (a-3)/2

By substituting the value of ‘a’ we get,

(-7 – 8)/3 = (-7 – 3)/2

-15/3 = -10/2

-5 = -5

2. x – 2x + 2 – 16/3x + 5 = 3 – 7/2x

Solution:

x – 2x + 2 – 16/3x + 5 = 3 – 7/2x

Let us rearrange the equation

x – 2x – 16x/3 + 7x/2 = 3 – 2 – 5

By taking LCM for 2 and 3, which is 6

(6x – 12x – 32x + 21x)/6 = -4

-17x/6 = -4

By cross-multiplying

-17x = -4×6

-17x = -24

x = -24/-17

x = 24/17

Let us verify the given equation now,

x – 2x + 2 – 16/3x + 5 = 3 – 7/2x

By substituting the value of ‘x’, we get,

24/17 – 2(24/17) + 2 – (16/3)(24/17) + 5 = 3 – (7/2)(24/17)

24/17 – 48/17 + 2 – 384/51 + 5 = 3 – 168/34

By taking 51 and 17 as the LCM we get,

(72 – 144 + 102 – 384 + 255)/51 = (102 – 168)/34

-99/51 = -66/34

-33/17 = -33/17

3. (2x+5)/3 = 3x – 10

Solution:

(2x+5)/3 = 3x – 10

Let us simplify,

(2x+5)/3 – 3x = – 10

By taking LCM

(2x + 5 – 9x)/3 = -10

(-7x + 5)/3 = -10

By using cross-multiplication, we get,

-7x + 5 = -30

-7x = -30 – 5

-7x = -35

x = -35/-7

= 5

Let us verify the given equation now,

(2x+5)/3 = 3x – 10

By substituting the value of ‘x’, we get,

(2×5 + 5)/3 = 3(5) – 10

(10+5)/3 = 15-10

15/3 = 5

5 = 5

4. (7y + 2)/5 = (6y – 5)/11

Solution:

(7y + 2)/5 = (6y – 5)/11

By using cross-multiplication, we get,

(7y + 2)11 = (6y – 5)5

77y + 22 = 30y – 25

77y – 30y = -25 – 22

47y = -47

y = -47/47

y = -1

Let us verify the given equation now,

(7y + 2)/5 = (6y – 5)/11

By substituting the value of ‘y’, we get,

(7(-1) + 2)/5 = (6(-1) – 5)/11

(-7 + 2)/5 = (-6 – 5)/11

-5/5 = -11/11

-1 = -1

5. 3/4x + 4x = 7/8 + 6x – 6

Solution:

3/4x + 4x = 7/8 + 6x – 6

Let us rearrange the equation

3/4x + 4x – 6x = 7/8 – 6

By taking 4 and 8 as LCM

(3x + 16x – 24x)/4 = (7 – 48)/8

-5x/4 = -41/8

By cross-multiplying

-5x(8) = -41(4)

-40x = -164

x = -164/-40

= 82/20

= 41/10

Let us verify the given equation now,

3/4x + 4x = 7/8 + 6x – 6

By substituting the value of ‘x’, we get,

(3/4)(41/10) + 4(41/10) = 7/8 + 6(41/10) – 6

123/40 + 164/10 = 7/8 + 246/10 – 6

(123 + 656)/40 = (70 + 1968 – 480)/80

779/40 = 1558/80

779/40 = 779/40

6. (6x+1)/2 + 1 = (7x-3)/3

Solution:

(6x+1)/2 + 1 = (7x-3)/3

(6x + 1 + 2)/2 = (7x – 3)/3

By cross-multiplying

(6x + 3)3 = (7x – 3)2

18x + 9 = 14x – 6

18x – 14x = -6 – 9

4x = -15

x = -15/4

Let us verify the given equation now,

(6x+1)/2 + 1 = (7x-3)/3

By substituting the value of ‘x’, we get,

(6(-15/4) + 1)/2 + 1 = (7(-15/4) – 3)/3

(3(-15/2) + 1)/2 + 1 = (-105/4 -3)/3

(-45/2 + 1)/2 + 1 = (-117/4)/3

(-43/4) + 1 = -117/12

(-43+4)/4 = -39/4

-39/4 = -39/4

7. 7x/2 – 5x/2 = 20x/3 + 10

Solution:

7x/2 – 5x/2 = 20x/3 + 10

Let us rearrange the equation

7x/2 – 5x/2 – 20x/3 = 10

By taking LCM for 2 and 3 is 6

(21x – 15x – 40x)/6 = 10

-34x/6 = 10

By cross-multiplying

-34x = 60

x = 60/-34

= -30/17

Let us verify the given equation now,

7x/2 – 5x/2 = 20x/3 + 10

By substituting the value of ‘x’, we get,

(7-/2)(-30/17) – (5/2)(-30/17) = (20/3)(-30/17) + 10

-210/34 +150/34 = -600/51 + 10

-30/17 = (-600+510)/51

= -90/51

-30/17 = -30/17

Also Read:

• RD Sharma Solutions for Class 8 Maths Chapter 1
• RD Sharma Solutions for Class 8 Maths Chapter 2
• RD Sharma Solutions for Class 8 Maths Chapter 3
• RD Sharma Solutions for Class 8 Maths Chapter 4
• RD Sharma Solutions for Class 8 Maths Chapter 5
• RD Sharma Solutions for Class 8 Maths Chapter 6
• RD Sharma Solutions for Class 8 Maths Chapter 7
• RD Sharma Solutions for Class 8 Maths Chapter 8

Importance of RD Sharma Maths Solutions Class 8 Chapter 9

• This chapter is a student's first deep dive into algebra. RD Sharma Solutions explain concepts like variables, constants, and balance in equations forming a strong base for higher classes.
• Every question is solved with clear steps, which helps students understand how and why each method works, making learning more intuitive.
• With repeated practice of RD Sharma’s questions and solutions, students become quicker and more accurate—resulting in better exam performance and less anxiety.
• The chapter encourages the use of logic through techniques like transposition, cross-multiplication, and elimination, sharpening students’ problem-solving skills.
• Available in PDF format, the solutions can be downloaded anytime, making it convenient for last-minute revision and self-paced learning.