Mathematics is an interesting subject that helps students understand patterns, number relationships, and how to solve everyday problems. One important topic in Class 7 Maths is called Linear Equations in One Variable. The RD Sharma Solutions for Class 7 Chapter 8 explain this concept in a simple and step-by-step way, helping students develop strong problem-solving skills and a clear understanding of algebra. In RD Sharma Solutions for Class 7 Chapter 8 Linear Equations in One Variable, students learn about linear equations, which are useful not just in Maths, but also in daily life situations like balancing expenses or finding unknown values.
Also Check: RD Sharma Solutions for Class 7 Maths
What Are Linear Equations in One Variable?
Linear equations are those where the highest power of the variable is one. This makes the equation easy to solve because it involves simple mathematical steps. A basic example of Linear Equations in One Variable Class 7 Chapter 8 Solutions is an equation like:
ax + b = 0, where:
- x is the variable,
- a and b are constant numbers.
The main goal is to find the value of x that makes the equation true.
Example of Solving Linear Equations
Let’s take an example from the Class 7 Chapter 8 Linear Equations in One Variable Solutions. Suppose we have the equation 2x + 3 = 7. To solve this:
- First, subtract 3 from both sides: 2x = 4
- Then, divide both sides by 2: x = 2
By following these simple steps, we can easily find the unknown value of x.
What You Will Learn in RD Sharma Solutions for Class 7 Chapter 8
The RD Sharma Solutions for Class 7 Chapter 8 Linear Equations in One Variable guide students to solve different types of linear equations. The chapter explains:
- How to solve simple linear equations easily.
- How to handle equations with fractions and decimals.
- How to solve word problems using linear equations.
- The importance of keeping both sides of the equation balanced while solving.
The Linear Equations in One Variable Class 7 Chapter 8 Solutions are written in a simple way so students can understand each step clearly. This makes it easy for students to practice more and build confidence in Maths.
Benefits of Using RD Sharma Solutions for Class 7 Chapter 8
By following the RD Sharma Solutions for Class 7 Chapter 8 Linear Equations in One Variable, students can:
- Understand concepts in a clear and simple way.
- Improve their problem-solving speed and accuracy.
- Be prepared for exams with well-explained step-by-step solutions.
- Build a strong base in algebra which will be helpful in higher classes.
The Class 7 Chapter 8 Linear Equations in One Variable Solutions include worked-out examples, extra questions, and exercises. Regular practice from these solutions helps students do well in exams and solve real-life problems confidently.
You can easily get the RD Sharma Solutions for Class 7 Chapter 8 PDF for free. The PDF includes complete answers, solved examples, and extra practice questions from the RD Sharma Solutions for Class 7 Chapter 8 Linear Equations in One Variable book.
Access to the Class 7 Maths Chapter 8 Linear Equations in One Variable Solutions
Q1. Check by Substitution:
(i) Let’s check if x = 4 solves 3x - 5 = 7.
Putting x = 4,
3 × 4 - 5 = 12 - 5 = 7.
Both sides are 7, so x = 4 is correct.
(ii) Let’s check if x = 3 solves 5 + 3x = 14.
Putting x = 3,
5 + 3 × 3 = 5 + 9 = 14.
Both sides are 14, so x = 3 is correct.
(iii) Let’s check if x = 2 solves 3x - 2 = 8x - 12.
Putting x = 2,
3 × 2 - 2 = 6 - 2 = 4,
8 × 2 - 12 = 16 - 12 = 4.
Both sides are 4, so x = 2 is correct.
(iv) Let’s check if x = 4 solves (3x/2) = 6.
Putting x = 4,
(3 × 4) ÷ 2 = 12 ÷ 2 = 6.
Both sides are 6, so x = 4 is correct.
(v) Let’s check if y = 2 solves y - 3 = 2y - 5.
Putting y = 2,
2 - 3 = -1, 2 × 2 - 5 = 4 - 5 = -1.
Both sides are -1, so y = 2 is correct.
(vi) Let’s check if x = 8 solves (1/2)x + 7 = 11.
Putting x = 8,
(1/2) × 8 + 7 = 4 + 7 = 11.
Both sides are 11, so x = 8 is correct.
Q2. Solve by Guessing (Trial and Error):
(i) x + 3 = 12
When x = 9, 9 + 3 = 12, so x = 9.
(ii) x - 7 = 10
When x = 17, 17 - 7 = 10, so x = 17.
(iii) 4x = 28
When x = 7, 4 × 7 = 28, so x = 7.
(iv) (x/2) + 7 = 11
When x = 8, 8 ÷ 2 + 7 = 4 + 7 = 11, so x = 8.
(v) 2x + 4 = 3x
When x = 4, 2 × 4 + 4 = 12 = 3 × 4, so x = 4.
(vi) x ÷ 4 = 12
When x = 48, 48 ÷ 4 = 12, so x = 48.
(vii) 15 ÷ x = 3
When x = 5, 15 ÷ 5 = 3, so x = 5.
(viii) x ÷ 18 = 20
When x = 360, 360 ÷ 18 = 20, so x = 360.
Q3. Solve x - 3 = 5
Adding 3 on both sides: x = 8.
Check: 8 - 3 = 5, correct.
Q4. Solve x + 9 = 13
Subtracting 9 from both sides: x = 4.
Check: 4 + 9 = 13, correct.
Q5. Solve x - 3/5 = 7/5
Adding 3/5 on both sides: x = 2.
Check: 2 - 3/5 = 7/5, correct.
Q6. Solve 3x = 0
Dividing both sides by 3: x = 0.
Check: 3 × 0 = 0, correct.
Q7. Solve x ÷ 2 = 0
Multiplying both sides by 2: x = 0.
Check: 0 ÷ 2 = 0, correct.
Q8. Solve x - 1/3 = 2/3
Adding 1/3 on both sides: x = 1.
Check: 1 - 1/3 = 2/3, correct.
Q9. Solve x + 1/2 = 7/2
Subtracting 1/2 on both sides: x = 3.
Check: 3 + 1/2 = 7/2, correct.
Q10. Solve 10 - y = 6
Subtracting 10 from both sides: -y = -4,
So, y = 4.
Check: 10 - 4 = 6, correct.
Q11. Solve 7 + 4y = -5
Subtracting 7 from both sides: 4y = -12,
Dividing by 4: y = -3.
Check: 7 + 4×(-3) = -5, correct.
Q12. Solve 4/5 - x = 3/5
Subtract 4/5 from both sides: -x = 3/5 - 4/5 = -1/5,
So, x = 1/5.
Check: 4/5 - 1/5 = 3/5, correct.