RD Sharma Solutions for Class 7 Maths Chapter 8 Linear Equations in One Variable
By Ankit Gupta
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Updated on 15 Apr 2025, 12:31 IST
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
| x | LHS | RHS | LHS = RHS? |
| 1 | 1 + 3 = 4 | 12 | No |
| 2 | 2 + 3 = 5 | 12 | No |
| 9 | 9 + 3 = 12 | 12 | Yes |
Therefore, x = 9 is the solution.
(ii) Solve x - 7 = 10
| x | LHS | RHS | LHS = RHS? |
| 17 | 17 - 7 = 10 | 10 | Yes |
Therefore, x = 17 is the solution.
(iii) Solve 4x = 28
| x | LHS | RHS | LHS = RHS? |
| 7 | 4 × 7 = 28 | 28 | Yes |
Therefore, x = 7 is the solution.
(iv) Solve (x/2) + 7 = 11
| x | LHS | RHS | LHS = RHS? |
| 8 | (8/2) + 7 = 11 | 11 | Yes |
Therefore, x = 8 is the solution.
(v) Solve 2x + 4 = 3x
| x | LHS | RHS | LHS = RHS? |
| 4 | 2 × 4 + 4 = 12 | 12 | Yes |
Therefore, x = 4 is the solution.
(vi) Solve (x/4) = 12
| x | LHS | RHS | LHS = RHS? |
| 48 | 48/4 = 12 | 12 | Yes |
Therefore, x = 48 is the solution.
(vii) Solve (15/x) = 3
| x | LHS | RHS | LHS = RHS? |
| 5 | 15/5 = 3 | 3 | Yes |
Therefore, x = 5 is the solution.
(viii) Solve (x/18) = 20
| x | LHS | RHS | LHS = RHS? |
| 360 | 360/18 = 20 | 20 | Yes |
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.
FAQs on RD Sharma Solutions for Class 7 Maths Chapter 8 Linear Equations in One Variable
What is a linear equation in one variable?
A linear equation in one variable is an equation where the highest power of the variable is one. It has the form of ax + b = 0, where a and b are constants, and x is the variable. The goal is to solve for the value of x.
How do I solve linear equations in one variable?
To solve linear equations in one variable, isolate the variable on one side of the equation. This can be done by using basic arithmetic operations like addition, subtraction, multiplication, and division. For example, to solve 2x + 5 = 11, subtract 5 from both sides and then divide by 2 to find the value of x.
What are the types of linear equations that we study in Chapter 8?
In Chapter 8, we study different forms of linear equations such as:
Equations with constants on both sides (e.g., 2x + 5 = 11).
Equations involving fractions or decimals (e.g., (3x/4) = 6).
Simple equations like x + 4 = 10.
What are some common mistakes to avoid while solving linear equations in one variable?
Not simplifying the equation properly: Ensure to simplify both sides of the equation before isolating the variable.
Incorrectly handling negative signs: Pay attention when distributing negative signs across terms, especially when both sides involve negative numbers.
Forgetting to check the solution: Always substitute the value of x back into the original equation to verify the solution.
How can RD Sharma Solutions help me understand linear equations in one variable better?
RD Sharma Solutions provide step-by-step solutions to all the problems in the textbook. These solutions help you understand the logical approach needed to solve each type of equation. The explanations are clear and concise, making it easier to grasp the concept of linear equations and solve related problems.
Are the solutions for all types of questions in Chapter 8 available in RD Sharma?
Yes, RD Sharma Solutions for Class 7 Maths Chapter 8 includes solutions to all types of problems, including practice problems, exercises, and additional questions. The solutions cover equations with constants, fractions, decimals, and other complex forms.
How do I practice solving linear equations effectively?
To practice solving linear equations effectively, start with simple equations and gradually move on to more complex ones. Use RD Sharma Solutions to check your work and understand the steps involved. Repetition and consistent practice will help improve your understanding of solving linear equations in one variable.