RD Sharma Solutions for Class 7 Maths Chapter 8 Linear Equations in One Variable

Mathematics is a subject that helps us understand patterns, relationships, and solve problems. One of the fundamental topics in Class 7 Maths is Linear Equations in One Variable. Chapter 8 of RD Sharma introduces students to the concept of linear equations, which is crucial for building problem-solving skills and understanding algebra. Linear equations are not only used in mathematics but also in real-life situations, such as balancing budgets or determining unknown values.

In a linear equation, the highest power of the variable is one. These equations are simple to understand and solve, as they involve finding the value of a variable that makes the equation true. A typical linear equation in one variable looks like this: ax + b = 0, where x is the variable, and a and b are constants. The goal is to solve for x, the unknown value.

For example, consider the equation 2x + 3 = 7. To solve it, we need to isolate the variable x on one side of the equation. First, we subtract 3 from both sides to get 2x = 4, and then divide both sides by 2, giving x = 2. This simple process of isolating the variable is the basis of solving linear equations.

RD Sharma Solutions for Class 7 Chapter 8 helps students understand and solve these types of equations step by step. The solutions are designed in a way that encourages students to practice regularly and improve their understanding. The chapter covers various methods for solving linear equations, including handling equations with fractions or decimals, and equations with variables on both sides.

In this chapter, students will learn how to:

• Solve simple linear equations.
• Solve equations that involve fractions or decimals.
• Solve word problems involving linear equations.
• Understand the importance of balancing both sides of the equation to find the correct value of the variable.

RD Sharma’s step-by-step solutions offer detailed explanations for each type of problem, ensuring that students not only learn how to solve equations but also grasp the reasoning behind each solution. With consistent practice, students can gain confidence in solving linear equations, which will be helpful in future mathematical concepts and everyday problem-solving. This chapter lays a strong foundation for more advanced topics in algebra and mathematics.

Download RD Sharma Solutions for Class 7 Maths Chapter 8 Linear Equations in One Variable PDF Here

RD Sharma Class 7 Chapter 8 PDF includes detailed solutions, examples, and extra questions to help you master real numbers and other topics. Click here to download the RD Sharma Class 7 Chapter 8 PDF.

Access Answers to RD Sharma Solutions for Class 7 Maths Chapter 8 Linear Equations in One Variable

In this chapter, students will learn about decimals and how to perform basic operations with them. The solutions provided here are detailed and easy to follow, helping students understand each concept thoroughly.

Q1. Verify by substitution:

(i) x = 4 is the solution to the equation 3x - 5 = 7

Substitute x = 4 into the equation:

3(4) - 5 = 7

12 - 5 = 7

7 = 7

Since both sides are equal, x = 4 is the solution.

(ii) x = 3 is the solution to the equation 5 + 3x = 14

Substitute x = 3 into the equation:

5 + 3(3) = 14

5 + 9 = 14

14 = 14

Since both sides are equal, x = 3 is the solution.

(iii) x = 2 is the solution to the equation 3x - 2 = 8x - 12

Substitute x = 2 into the equation:

3(2) - 2 = 8(2) - 12

6 - 2 = 16 - 12

4 = 4

Since both sides are equal, x = 2 is the solution.

(iv) x = 4 is the solution to the equation (3x/2) = 6

Substitute x = 4 into the equation:

(3 × 4) / 2 = 6

12 / 2 = 6

6 = 6

Since both sides are equal, x = 4 is the solution.

(v) y = 2 is the solution to the equation y - 3 = 2y - 5

Substitute y = 2 into the equation:

2 - 3 = 2(2) - 5

-1 = 4 - 5

-1 = -1

Since both sides are equal, y = 2 is the solution.

(vi) x = 8 is the solution to the equation (1/2)x + 7 = 11

Substitute x = 8 into the equation:

(1/2)(8) + 7 = 11

4 + 7 = 11

11 = 11

Since both sides are equal, x = 8 is the solution.

Q2. Solve by trial and error:

(i) Solve x + 3 = 12

Therefore, x = 9 is the solution.

(ii) Solve x - 7 = 10

Therefore, x = 17 is the solution.

(iii) Solve 4x = 28

Therefore, x = 7 is the solution.

(iv) Solve (x/2) + 7 = 11

Therefore, x = 8 is the solution.

(v) Solve 2x + 4 = 3x

Therefore, x = 4 is the solution.

(vi) Solve (x/4) = 12

Therefore, x = 48 is the solution.

(vii) Solve (15/x) = 3

Therefore, x = 5 is the solution.

(viii) Solve (x/18) = 20

Therefore, x = 360 is the solution.

Q3. Solve x - 3 = 5

Adding 3 to both sides:

x = 8

Verification: Substituting x = 8 into the equation, both sides are equal.

Q4. Solve x + 9 = 13

Subtract 9 from both sides:

x = 4

Verification: Substituting x = 4 into the equation, both sides are equal.

Q5. Solve x - (3/5) = (7/5)

Add (3/5) to both sides:

x = 2

Verification: Substituting x = 2 into the equation, both sides are equal.

Q6. Solve 3x = 0

Dividing both sides by 3:

x = 0

Verification: Substituting x = 0 into the equation, both sides are equal.

Q7. Solve (x/2) = 0

Multiplying both sides by 2:

x = 0

Verification: Substituting x = 0 into the equation, both sides are equal.

Q8. Solve x - (1/3) = (2/3)

Add (1/3) to both sides:

x = 1

Verification: Substituting x = 1 into the equation, both sides are equal.

Q9. Solve x + (1/2) = (7/2)

Subtract (1/2) from both sides:

x = 3

Verification: Substituting x = 3 into the equation, both sides are equal.

Q10. Solve 10 - y = 6

Subtract 10 from both sides:

y = 4

Verification: Substituting y = 4 into the equation, both sides are equal.

Q11. Solve 7 + 4y = -5

Subtract 7 from both sides:

y = -3

Verification: Substituting y = -3 into the equation, both sides are equal.

Q12. Solve (4/5) - x = (3/5)

Subtract (4/5) from both sides:

x = (1/5)

Verification: Substituting x = (1/5) into the equation, both sides are equal.