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NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.1 - Linear Equations in One Variable

Last updated date: 14th Sep 2024
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NCERT Solutions for Class 8 Chapter 2 Maths Linear Equations in One Variable Exercise 2.1 - FREE PDF Download

In Class 8 Maths Chapter 2, "Linear Equations in One Variable," ex 2.1 class 8 focuses on solving linear equations that contain variables on both sides. It's crucial to understand how to isolate the variable to find the solution accurately. Pay attention to the balancing method, which ensures both sides of the equation remain equal as you simplify.

Table of Content
1. NCERT Solutions for Class 8 Chapter 2 Maths Exercise 2.1 - FREE PDF Download
2. Glance of NCERT Solutions for Maths Chapter 2 - Exercise 2.1 Class 8 | Vedantu
3. Access PDF for Class 8 Maths NCERT Chapter 2 Linear Equations in One Variable Exercise 2.1 Solutions
4. Class 8 Maths Chapter 2: Exercises Breakdown
5. CBSE Class 8 Maths Chapter 2 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 8 Maths
FAQs

This chapter lays the foundation for solving more complex algebraic equations. Mastering these basic techniques is important as they are essential for higher-level mathematics. Focus on practicing the given problems to build a strong understanding of the concepts.

Glance of NCERT Solutions for Maths Chapter 2 Linear Equations in One Variable - Exercise 2.1 Class 8 | Vedantu

• This chapter focuses on solving linear equations in one variable, a foundational concept in algebra.

• This chapter includes practical applications of linear equations, helping students understand how to form and solve equations in everyday scenarios.

• The general form of a linear equation: ax+b=c

• This chapter emphasizes the importance of ex 2.1 class 8, which involves solving equations with variables on both sides.

• This chapter helps students build critical problem-solving skills, essential for higher mathematics.

• Maths Class 8  Linear Equations in One Variable article contains chapter notes, important questions, exemplar solutions, exercises and video links for Chapter 2 -  Linear Equations in One Variable , which you can download as PDFs.

• Exercise 2.1 class 8 NCERT solutions has over all 10 Questions & Solutions.

Access PDF for Class 8 Maths NCERT Chapter 2 Linear Equations in One Variable Exercise 2.1 Solutions

1. Solve and Verify the Equation $\text{3x=2x+18}$.

Ans: We have an equation $\text{3x=2x+18}$

To solve the equation, we will shift $\text{2x}$ to left hand side

$\text{3x-2x=18}$

$\text{x=18}$

Now to verify the result, solve L.H.S and R.H.S

L.H.S

$\text{3x=3 }\!\!\times\!\!\text{ 18}$

$\text{54}$

R.H.S

$\text{2x+18=2 }\!\!\times\!\!\text{ 18+18}$

$\text{54}$

L.H.S$\text{=}$R.H.S

Hence Proved

2. Solve and Verify the Equation $\text{5t-3=3t-5}$.

Ans: We have an equation $\text{5t-3=3t-5}$

To solve the equation, we will shift $\text{3t}$ to left hand side and $\text{3}$ to the right hand side, now,

$\text{5t-3t=-5+3}$

$\text{2t=-2}$

Dividing the equation by $\text{2}$

$\text{t=-1}$

Now to verify the result, solve L.H.S and R.H.S

L.H.S

$\text{5t-3=}\left( \text{5 }\!\!\times\!\!\text{ -1} \right)\text{-3}$

$\text{-5-3=-8}$

R.H.S

$\text{3t-5=}\left( \text{3 }\!\!\times\!\!\text{ -1} \right)\text{-5}$

$\text{-3-5=-8}$

L.H.S$\text{=}$R.H.S

Hence Proved

3. Solve and Verify the Equation $\text{5x+9=5+3x}$.

Ans: We have an equation $\text{5x+9=5+3x}$

To solve the equation, we will shift $\text{3x}$ to left hand side and $\text{9}$ to the right hand side, now,

$\text{5x-3x=5-9}$

$\text{2x=-4}$

Dividing the equation by $\text{2}$

$\text{x=-2}$

Now to verify the result, solve L.H.S and R.H.S

L.H.S

$\text{5x+9=}\left( \text{5 }\!\!\times\!\!\text{ -2} \right)\text{+9}$

$\text{-10+9=-1}$

R.H.S

$\text{5+3x=5+}\left( \text{3 }\!\!\times\!\!\text{ -2} \right)$

$\text{5-6=-1}$

L.H.S$\text{=}$R.H.S

Hence Proved

4. Solve and Verify the Equation $\text{4z+3=6+2z}$.

Ans: We have an equation $\text{4z+3=6+2z}$

To solve the equation, we will shift $\text{2z}$ to left hand side and $\text{3}$ to right hand side, now,

$\text{4z-2z=6-3}$

$\text{2z=3}$

Dividing the equation by $\text{2}$

$\text{z=}\frac{\text{3}}{\text{2}}$

Now to verify the result, solve L.H.S and R.H.S

L.H.S

$\text{4z+3=}\left( \text{4 }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{2}} \right)\text{+3}$

$\text{6+3=9}$

R.H.S

$\text{6+2z=6+}\left( \text{2 }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{2}} \right)$

$\text{6+3=9}$

L.H.S$\text{=}$R.H.S

Hence Proved

5. Solve and Verify the Equation $\text{2x-1=14-x}$.

Ans: We have an equation $\text{2x-1=14-x}$

To solve the equation, we will shift $\text{x}$ to left hand side and $\text{1}$ to the right hand side, now,

$\text{2x+x=14+1}$

$\text{3x=15}$

Dividing the equation by $\text{3}$

$\text{x=5}$

Now to verify the result, solve L.H.S and R.H.S

L.H.S

$\text{2x-1=}\left( \text{2 }\!\!\times\!\!\text{ 5} \right)\text{-1}$

$\text{10-1=9}$

R.H.S

$\text{14-x=14-5}$

$\text{14-5=9}$

L.H.S$\text{=}$R.H.S

Hence Proved

6. Solve and Verify the Equation $\text{8x+4=3}\left( \text{x-1} \right)\text{+7}$.

Ans: We have an equation $\text{8x+4=3}\left( \text{x-1} \right)\text{+7}$

$\text{8x+4=3x-3+7}$

$\text{8x+4=3x+4}$

To solve the equation, we will shift $\text{3x}$ to left hand side and $4$ to the right hand side, now,

$\text{8x-3x=4-4}$

$\text{5x=0}$

$\text{x=0}$

Now to verify the result, solve L.H.S and R.H.S

L.H.S

$\text{8x+4=}\left( \text{8 }\!\!\times\!\!\text{ 0} \right)\text{+4}$

$\text{0+4=4}$

R.H.S

$\text{3}\left( \text{x-1} \right)\text{+7=3}\left( \text{0-1} \right)\text{+7}$

$\text{-3+7=4}$

L.H.S$\text{=}$R.H.S

Hence Proved

7. Solve and Verify the Equation $\text{x=}\frac{\text{4}}{\text{5}}\left( \text{x+10} \right)$.

Ans: We have an equation $\text{x=}\frac{\text{4}}{\text{5}}\left( \text{x+10} \right)$

$\text{x=}\frac{\text{4x+40}}{\text{5}}$

Multiplying the equation by $\text{5}$

$\text{5x=4x+40}$

To solve the equation, we will shift $\text{4x}$ to left hand side, now,

$\text{5x-4x=40}$

$\text{x=40}$

Now to verify the result, solve L.H.S and R.H.S

L.H.S

$\text{x=40}$

R.H.S

$\frac{\text{4}}{\text{5}}\left( \text{x+10} \right)\text{=}\frac{\text{4}}{\text{5}}\left( \text{40+10} \right)$

$\frac{\text{4}}{\text{5}}\left( \text{40+10} \right)\text{=}\frac{\text{4 }\!\!\times\!\!\text{ 50}}{\text{5}}$

$\frac{\text{4 }\!\!\times\!\!\text{ 50}}{\text{5}}\text{=40}$

L.H.S$\text{=}$R.H.S

Hence Proved

8. Solve and Verify the Equation $\frac{\text{2x}}{\text{3}}\text{+1=}\frac{\text{7x}}{\text{15}}\text{+3}$.

Ans: We have an equation $\frac{\text{2x}}{\text{3}}\text{+1=}\frac{\text{7x}}{\text{15}}\text{+3}$

To solve the equation, we will shift $\frac{\text{7x}}{\text{15}}$ to left hand side and $\text{1}$ to right hand side, now,

$\frac{\text{2x}}{\text{3}}\text{-}\frac{\text{7x}}{\text{15}}\text{=3-1}$

$\frac{\text{10x-7x}}{\text{15}}\text{=2}$

Multiplying $15$ to the equation, we get

$\text{3x=30}$

$\text{x=10}$

Now to verify the result, solve L.H.S and R.H.S

L.H.S

$\frac{\text{2x}}{\text{3}}\text{+1=}\frac{\left( \text{2 }\!\!\times\!\!\text{ 10} \right)\text{+3}}{\text{3}}$

$\frac{\left( \text{2 }\!\!\times\!\!\text{ 10} \right)\text{+3}}{\text{3}}\text{=}\frac{\text{23}}{\text{3}}$

R.H.S

$\frac{\text{7x}}{\text{15}}\text{+3=}\frac{\left( \text{7 }\!\!\times\!\!\text{ 10} \right)\text{+45}}{\text{15}}$

$\frac{\left( \text{7 }\!\!\times\!\!\text{ 10} \right)\text{+45}}{\text{15}}\text{=}\frac{\text{115}}{\text{15}}$

$\frac{\text{115}}{\text{15}}\text{=}\frac{\text{23}}{\text{3}}$

L.H.S$\text{=}$R.H.S

Hence Proved

9. Solve and Verify the Equation $\text{2y+}\frac{\text{5}}{\text{3}}\text{+}\frac{\text{26}}{\text{3}}\text{-y}$.

Ans: We have an equation $\text{2y+}\frac{\text{5}}{\text{3}}\text{=}\frac{\text{26}}{\text{3}}\text{-y}$

To solve the equation, we will shift $y$ to left hand side and $\frac{\text{5}}{\text{3}}$ to right hand side, now,

$\text{2y+y=}\frac{\text{26}}{\text{3}}\text{-}\frac{\text{5}}{\text{3}}$

$\text{3y=}\frac{\text{21}}{\text{3}}$

$\text{3y=7}$

$\text{y=}\frac{\text{7}}{\text{3}}$

Now to verify the result, solve L.H.S and R.H.S

L.H.S

$\text{2y+}\frac{\text{5}}{\text{3}}\text{=2 }\!\!\times\!\!\text{ }\frac{\text{7}}{\text{3}}\text{+}\frac{\text{5}}{\text{3}}$

$\text{2 }\!\!\times\!\!\text{ }\frac{\text{7}}{\text{3}}\text{+}\frac{\text{5}}{\text{3}}\text{=}\frac{\text{14+5}}{\text{3}}$

$\frac{\text{14+5}}{\text{3}}\text{=}\frac{\text{19}}{\text{3}}$

R.H.S

$\frac{\text{26}}{\text{3}}\text{-y=}\frac{\text{26}}{\text{3}}\text{-}\frac{\text{7}}{\text{3}}$

$\frac{\text{26}}{\text{3}}\text{-}\frac{\text{7}}{\text{3}}\text{=}\frac{\text{19}}{\text{3}}$

L.H.S$\text{=}$R.H.S

Hence Proved

10. Solve and Verify the Equation $\text{3m=5m-}\frac{\text{8}}{\text{5}}$.

Ans: We have an equation $\text{3m=5m-}\frac{\text{8}}{\text{5}}$

To solve the equation, we will shift $\text{5m}$ to left hand side, now,

$\text{3m-5m=-}\frac{\text{8}}{\text{5}}$

$\text{-2m=-}\frac{\text{8}}{\text{5}}$

Dividing the equation by $\text{-2}$, we get,

$\text{m=}\frac{4}{\text{5}}$

Now to verify the result, solve L.H.S and R.H.S

L.H.S

$\text{3m=3 }\!\!\times\!\!\text{ }\frac{\text{4}}{\text{5}}$

$\frac{\text{3 }\!\!\times\!\!\text{ 4}}{\text{5}}\text{=}\frac{\text{12}}{\text{5}}$

R.H.S

$\text{5m-}\frac{\text{8}}{\text{5}}\text{=}\frac{\text{5 }\!\!\times\!\!\text{ 4}}{\text{5}}\text{-}\frac{\text{8}}{\text{5}}$

$\frac{\text{5 }\!\!\times\!\!\text{ 4}}{\text{5}}\text{-}\frac{\text{8}}{\text{5}}=\frac{12}{\text{5}}$

L.H.S$\text{=}$R.H.S

Hence Proved

Conclusion

Ex 2.1 Class 8  Maths Chapter 2 is crucial for mastering linear equations in one variable. It focuses on building a strong foundation by teaching students how to solve equations through simple steps. This exercise emphasizes understanding the process of isolating the variable and verifying solutions. Students should concentrate on practicing various types of problems to enhance their problem-solving skills. It's important to pay attention to the methods used and ensure accuracy in each step. Consistent practice will help in gaining confidence and proficiency in handling linear equations.

Class 8 Maths Chapter 2: Exercises Breakdown

 Chapter 2 - Linear Equations in One Variable Exercises in PDF Format Exercise 2.2 10 Questions with Solutions

CBSE Class 8 Maths Chapter 2 Other Study Materials

 S. No Important Links for Class 8 Maths Chapter 2 Linear Equations in One Variable 1 Class 8 Linear Equations in One Variable Important Questions 2 Class 8 Linear Equations in One Variable Notes 3 Class 8 Linear Equations in One Variable Formula 4 Class 8 Linear Equations in One Variable RD Sharma Solutions 5 Class 8 Linear Equations in One Variable RS Aggarwal Solutions

Chapter-Specific NCERT Solutions for Class 8 Maths

Given below are the chapter-wise NCERT Solutions for Class 8 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.1 - Linear Equations in One Variable

1. How many questions are there in Exercise 2.1 of Class 8 Maths?

There are 10 questions in Exercise 2.1 of Class 8 Maths. Students can get well versed in all the concepts by using NCERT Solutions prepared by expert teachers of Vedantu. Visit the Vedantu website to download the solutions for the problems from all chapters for free of cost. They strictly follow the CBSE curriculum so students need not worry about questions being out of the syllabus. These solutions are written in simple language to help students understand the concept better.

2. Which example is important for class 8 Maths Exercise 2.1?

Exercise 2.1 has only two examples, so practicing on both examples will be a wise choice . To make the practice and revision session more interesting, visit the Vedantu website to find the NCERT Solutions created by the professionals and download the PDF for free of cost. With the help of these solutions, students will secure a perfect score in the exams. Students can simply download the solutions and save them on their computers. They can refer to these solutions for their doubts and queries.

3. Where can I get NCERT Solutions for Class 8 Maths Chapter 2?

Vedantu offers the best NCERT Solutions for the students of Class 8. The NCERT Solutions are provided chapter-wise, so it is convenient for the students to get well versed in each chapter. Visit the page NCERT Solutions for Class 8 Maths Chapter 2 to download the PDF of Chapter 2 Maths Class 8 for free of cost. These solutions are based on the latest guidelines and exam patterns. So these solutions are 100% accurate and reliable.

4.  Do I have to solve all the problems in NCERT Solutions?

Yes, you need to solve all problems because it helps you understand the concepts well and is a great tool for revision. Since experts create these solutions, it will help the students get well versed in every chapter. Visit the page NCERT Solutions for Class 8 Maths Chapter 2 to download the PDF of NCERT Solutions prepared by Vedantu.

5.  How can I understand Chapter 2 of Maths Class 8?

To understand Chapter 2 Maths Class 8 you need to understand the concepts and their application. Vedantu provides online live classes and also NCERT Solutions to clear all the doubts students struggle with. To download the free PDF of NCERT Solutions created by the experts, visit the Vedantu website or use the Vedantu app on your phone and download the solutions for free of cost.

6. What exactly is a linear equation with one variable in Class 8 Exercise 2.1?

In Class 8 Exercise 2.1, a linear equation in one variable is one that only has one variable, which is typically denoted by the symbol x. It can be written as ax+b=0, with a and b constants. The goal is to determine the value of x that makes the equation true. This type of equation is known as "linear" because it represents a straight line when graphed.

7. In Class 8 Maths Ex 2.1, what should my main focus be?

Pay close attention to the steps involved in isolating the variable in Class 8 Maths Ex 2.1 . To get the constants on one side of the equation and the variable x on the other, the equation must be rearranged. Recheck your answers by inserting them back into the original equation to see if both sides are equal. Practice solving various problem kinds.

8. Why is isolating the variable important in class 8 maths ch 2 ex 2.1 ?

In class 8 maths ch 2 ex 2.1 isolating the variable is crucial because it simplifies the equation, making it easier to solve. By getting the variable on one side of the equation and the constants on the other, you can directly find the value of the variable. This method helps to clearly see the solution and ensures accuracy in solving the equation.

9. How do I deal with negative coefficients in class 8 ex 2.1 ?

In class 8 ex 2.1 when dealing with negative coefficients, treat them like any other number. Be mindful of the signs when you add, subtract, multiply, or divide both sides of the equation. For example, if you need to move a term with a negative coefficient, you might add or subtract it from both sides to maintain the balance of the equation.

10. What if the variable appears on both sides of the equation in class 8 maths chapter 2.1?

In class 8 maths chapter 2.1 the variable appears on both sides of the equation, use the transposition method. This involves moving all terms containing the variable to one side and all constant terms to the other. For example, if you have 2x+3=x+7, you would subtract x from both sides to get x+3=7, then isolate x by subtracting 3 from both sides to get x=4.