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NCERT Solutions

Class 8

Maths

Chapter 2 Linear Equations In One Variable Exercise 2.2

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations In One Variable Ex 2.2

CBSE Maths NCERT Solutions Exercise 2.2 Class 8 - FREE PDF Download

The NCERT Solutions for Class 8 Maths Chapter 2 – Linear Equations in One Variable are prepared by the subject experts to help the students. These NCERT Solutions make the students familiar with various concepts. CBSE Maths Chapter 2 Ex 2.2 Class 8 consists of questions that comprise various topics carrying different marks, so students must be well aware of all concepts to score well in the annual exam. Practising the NCERT solutions repeatedly is an easy way to learn the concepts covered in Class 8 Maths Ex 2.2.

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Access NCERT Solutions for Maths Chapter 2 – Linear Equations in One Variable Ex 2.2 Class 8

Exercise 2.2

1. Find the solution of the linear equation\[\frac{\text{x}}{\text{2}}\text{-}\frac{\text{1}}{\text{5}}\text{=}\frac{\text{x}}{\text{3}}\text{+}\frac{\text{1}}{\text{4}}\]

Ans: We have an equation \[\frac{\text{x}}{\text{2}}\text{-}\frac{\text{1}}{\text{5}}\text{=}\frac{\text{x}}{\text{3}}\text{+}\frac{\text{1}}{\text{4}}\]

The L.C.M of denominators \[\text{2,3,4}\] and \[\text{5}\] is \[\text{60}\]

Multiplying both the sides by \[\text{60}\],

\[\Rightarrow \text{60}\left( \frac{\text{x}}{\text{2}}\text{-}\frac{\text{1}}{\text{5}} \right)\text{=60}\left( \frac{\text{x}}{\text{3}}\text{+}\frac{\text{1}}{\text{4}} \right)\]

\[\Rightarrow \text{60}\left( \frac{\text{5x-2}}{\text{10}} \right)\text{=60}\left( \frac{\text{4x+3}}{\text{12}} \right)\]

\[\Rightarrow \text{6}\left( \text{5x-2} \right)\text{=5}\left( \text{4x+3} \right)\]

\[\Rightarrow \text{30x-12=20x+15}\]

Shifting \[\text{20x}\] to left hand side and \[\text{12}\] on right hand side

\[\Rightarrow \text{30x-20x=15+12}\]

\[\Rightarrow \text{10x=27}\]

\[\Rightarrow \text{x=}\frac{\text{27}}{\text{10}}\]

2. Find the solution of the linear equation\[\frac{\text{n}}{\text{2}}\text{-}\frac{\text{3n}}{\text{4}}\text{+}\frac{\text{5n}}{\text{6}}\text{=21}\]

Ans: We have an equation \[\frac{\text{n}}{\text{2}}\text{-}\frac{\text{3n}}{\text{4}}\text{+}\frac{\text{5n}}{\text{6}}\text{=21}\]

The L.C.M of denominators \[\text{2,4}\] and \[\text{6}\] is \[\text{12}\]

Multiplying both the sides by \[\text{12}\],

\[\Rightarrow \text{12}\left( \frac{\text{n}}{\text{2}}\text{-}\frac{\text{3n}}{\text{4}}\text{+}\frac{\text{5n}}{\text{6}} \right)\text{=12}\left( \text{21} \right)\]

\[\Rightarrow \text{12}\left( \frac{\text{6n-9n+10n}}{\text{12}} \right)\text{=12}\left( \text{21} \right)\]

\[\Rightarrow \text{7n=252}\]

Dividing the equation by \[\text{7}\]

\[\Rightarrow \text{n=36}\]

3. Find the solution of the linear equation\[\text{x+7-}\frac{\text{8x}}{\text{3}}\text{=}\frac{\text{17}}{\text{6}}\text{-}\frac{\text{5x}}{\text{2}}\]

Ans: We have an equation \[\text{x+7-}\frac{\text{8x}}{\text{3}}\text{=}\frac{\text{17}}{\text{6}}\text{-}\frac{\text{5x}}{\text{2}}\]

The L.C.M of denominators \[\text{2,3}\] and \[\text{6}\] is \[\text{6}\]

Multiplying both the sides by \[\text{6}\],

\[\Rightarrow \text{6}\left( \text{x+7-}\frac{\text{8x}}{\text{3}} \right)\text{=6}\left( \frac{\text{17}}{\text{6}}\text{-}\frac{\text{5x}}{\text{2}} \right)\]

\[\Rightarrow \text{6}\left( \frac{\text{6x+42-16x}}{\text{6}} \right)\text{=6}\left( \frac{\text{17-15x}}{\text{6}} \right)\]

\[\Rightarrow \text{6x+42-16x=17-15x}\]

Shifting \[\text{15x}\] to left hand side and \[\text{42}\] to right hand side

\[\Rightarrow \text{-10x+15x=17-42}\]

\[\Rightarrow \text{5x=-25}\]

Dividing both the sides by \[\text{5}\]

\[\Rightarrow \text{x=-5}\]

4. Find the solution of the linear equation\[\frac{\text{x-5}}{\text{3}}\text{=}\frac{\text{x-3}}{\text{5}}\]

Ans: We have an equation \[\frac{\text{x-5}}{\text{3}}\text{=}\frac{\text{x-3}}{\text{5}}\]

The L.C.M of denominators \[\text{3}\] and \[\text{5}\] is \[\text{15}\]

Multiplying both the sides by \[\text{15}\],

\[\Rightarrow \text{15}\left( \frac{\text{x-5}}{\text{3}} \right)\text{=15}\left( \frac{\text{x-3}}{\text{5}} \right)\]

\[\Rightarrow \text{5}\left( \text{x-5} \right)\text{=3}\left( \text{x-3} \right)\]

\[\Rightarrow \text{5x-25=3x-9}\]

Shifting \[\text{3x}\] to left hand side and \[\text{25}\] to right hand side

\[\Rightarrow \text{5x-3x=-9+25}\]

\[\Rightarrow \text{2x=16}\]

Dividing both the sides by \[\text{2}\]

\[\Rightarrow \text{x=8}\]

5. Find the solution of the linear equation\[\frac{\text{3t-2}}{\text{4}}\text{-}\frac{\text{2t+3}}{\text{3}}\text{=}\frac{\text{2}}{\text{3}}\text{-t}\]

Ans: We have an equation \[\frac{\text{3t-2}}{\text{4}}\text{-}\frac{\text{2t+3}}{\text{3}}\text{=}\frac{\text{2}}{\text{3}}\text{-t}\]

The L.C.M of denominators \[\text{3}\] and \[\text{4}\] is \[\text{12}\]

Multiplying both the sides by \[\text{12}\],

\[\Rightarrow \text{12}\left( \frac{\text{3t-2}}{\text{4}}\text{-}\frac{\text{2t+3}}{\text{3}} \right)\text{=12}\left( \frac{\text{2}}{\text{3}}\text{-t} \right)\]

\[\Rightarrow \text{12}\left( \frac{\text{9t-6-8t-12}}{\text{12}} \right)\text{=12}\left( \frac{\text{8-12t}}{\text{12}} \right)\]

\[\Rightarrow \text{t-18=8-12t}\]

Shifting \[\text{12t}\] to left hand side and \[\text{18}\] to right hand side

\[\Rightarrow \text{t+12t=8+18}\]

\[\Rightarrow \text{13t=26}\]

Dividing both the sides by \[\text{13}\]

\[\Rightarrow \text{t=2}\]

6. Find the solution of the linear equation \[\text{m-}\frac{\text{m-1}}{\text{2}}\text{=1-}\frac{\text{m-2}}{\text{3}}\]

Ans: We have an equation \[\text{m-}\frac{\text{m-1}}{\text{2}}\text{=1-}\frac{\text{m-2}}{\text{3}}\]

The L.C.M of denominators \[\text{3}\] and \[\text{2}\] is \[\text{6}\]

Multiplying both the sides by \[\text{6}\],

\[\Rightarrow \text{6}\left( \frac{\text{2m-m+1}}{\text{2}} \right)\text{=6}\left( \frac{\text{3-m+2}}{\text{3}} \right)\]

\[\Rightarrow \text{3}\left( \text{m+1} \right)\text{=3}\left( \text{-m+5} \right)\]

\[\Rightarrow \text{3m+3=10-2m}\]

Shifting \[\text{2m}\] to left hand side and \[\text{3}\] to right hand side

\[\Rightarrow \text{3m+2m=10-3}\]

\[\Rightarrow \text{5m=7}\]

Dividing both the sides by \[\text{5}\]

\[\Rightarrow \text{m=}\frac{\text{7}}{\text{5}}\]

7. Find the solution of the linear equation \[\text{3}\left( \text{t-3} \right)\text{=5}\left( \text{2t+1} \right)\]

Ans: We have an equation \[\text{3}\left( \text{t-3} \right)\text{=5}\left( \text{2t+1} \right)\]

\[\Rightarrow \text{3t-9=10t+5}\]

Shifting \[\text{5}\] to left hand side and \[\text{3t}\] to right hand side

\[\Rightarrow \text{-9-5=10t-3t}\]

\[\Rightarrow \text{-14=7t}\]

Dividing both the sides by \[\text{7}\]

\[\Rightarrow \text{-2=t}\]

8. Find the solution of the linear equation \[\text{15}\left( \text{y-4} \right)\text{-2}\left( \text{y-9} \right)\text{+5}\left( \text{y+6} \right)\text{=0}\]

Ans: We have an equation \[\text{15}\left( \text{y-4} \right)\text{-2}\left( \text{y-9} \right)\text{+5}\left( \text{y+6} \right)\text{=0}\]

\[\Rightarrow \text{15y-60-2y+18+5y+30=0}\]

\[\Rightarrow \text{18y-12=0}\]

Shifting \[\text{12}\] to right hand side

\[\Rightarrow \text{18y=12}\]

Dividing both the sides by \[\text{18}\]

\[\Rightarrow \text{y=}\frac{\text{12}}{\text{18}}\]

\[\Rightarrow \text{y=}\frac{\text{2}}{\text{3}}\]

9. Find the solution of the linear equation \[\text{3}\left( \text{5z-7} \right)\text{-2}\left( \text{9z-11} \right)\text{=4}\left( \text{8z-13} \right)\text{-17}\]

Ans: We have an equation \[\text{3}\left( \text{5z-7} \right)\text{-2}\left( \text{9z-11} \right)\text{=4}\left( \text{8z-13} \right)\text{-17}\]

\[\Rightarrow \text{15z-21-18z+22=32z-52-17}\]

\[\Rightarrow \text{-3z+1=32z-69}\]

Shifting \[\text{69}\] to left hand side and \[\text{3z}\] to right hand side

\[\Rightarrow \text{1+69=32z+3z}\]

\[\Rightarrow \text{70=35z}\]

Dividing both the sides by \[\text{35}\]

\[\Rightarrow \text{z=2}\]

10. Find the solution of the linear equation \[\text{0}\text{.25}\left( \text{4f-3} \right)\text{=0}\text{.05}\left( \text{10f-9} \right)\]

Ans: We have an equation \[\text{0}\text{.25}\left( \text{4f-3} \right)\text{=0}\text{.05}\left( \text{10f-9} \right)\]

\[\Rightarrow \text{f-0}\text{.75=0}\text{.5f-0}\text{.45}\]

Shifting \[\text{0}\text{.5f}\] to left hand side and \[\text{0}\text{.75}\] to right hand side

\[\Rightarrow \text{0}\text{.5f=-0}\text{.45+0}\text{.75}\]

\[\Rightarrow \text{0}\text{.5f=0}\text{.3}\]

Dividing both the sides by \[\text{0}\text{.5}\]

\[\Rightarrow \text{f=0}\text{.6}\]

Conclusion

Class 8 Maths Chapter 2 Exercise 2.2 solutions, "Linear Equations in One Variable," focuses on solving linear equations involving one variable. Exercise 2.2 Class 8 introduces complex problems, requiring careful manipulation. It covers isolating the variable, simplifying the equation, and verifying solutions. Exercises include fractions, integers, and word problems, solidifying understanding of different forms of linear equations. Practice is crucial for developing problem-solving skills and preparing for exams.

Class 8 Maths Chapter 2: Exercises Breakdown

Exercise	Number of Questions
Exercise 2.1	10 Questions with Solutions

CBSE Class 8 Maths Chapter 2 Exercise 2.2 Other Study Materials

S. No	Important Links for Class 8 Ex 2.2 Maths Exercise 2.2: Linear Equations in One Variable
1	Class 8 Linear Equations in One Variable Important Questions
2	Class 8 Linear Equations in One Variable Revision Notes
3	Class 8 Linear Equations in One Variable Important Formulas
4	Class 8 Linear Equations in One Variable NCERT Exemplar Solution
5	Class 8 Linear Equations in One Variable RD Sharma Solutions
6	Class 8 Linear Equations 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.

S. No	NCERT Solutions Class 8 Maths Chapter-wise List
1	Chapter 1 - Rational Numbers Solutions
2	Chapter 2 - Linear Equations in One Variable Solutions
3	Chapter 3 - Understanding Quadrilaterals Solutions
4	Chapter 4 - Data Handling Solutions
5	Chapter 5 - Squares and Square Roots Solutions
6	Chapter 6 - Cubes and Cube Roots Solutions
7	Chapter 7 - Comparing Quantities Solutions
8	Chapter 8 - Algebraic Expressions and Identities Solutions
9	Chapter 9 - Mensuration Solutions
10	Chapter 10 - Exponents and Powers Solutions
11	Chapter 11 - Direct and Inverse Proportions Solutions
12	Chapter 12 - Factorisation Solutions
13	Chapter 13 - Introduction to Graphs Solutions

Important Related Links for CBSE Class 8 Maths

S. No	Other Study Materials for CBSE Class 8 Maths
1	CBSE Class 8 Maths NCERT Textbook
2	CBSE Class 8 Maths Important Questions
3	CBSE Class 8 Maths Sample Papers
4	CBSE Class 8 Maths NCERT Exemplar Solutions
5	CBSE Class 8 Maths RD Sharma Solutions
6	CBSE Class 8 Maths RS Aggarwal Solutions
7	CBSE Class 8 Maths Formulas
8	CBSE Class 8 Maths NCERT Solutions
9	CBSE Class 8 Maths Syllabus
10	CBSE Class 8 Maths Revision Notes

FAQs on NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations In One Variable Ex 2.2

1. What are the stepwise NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.1 as per the latest CBSE 2025–26 syllabus?

The NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.1 follow a stepwise approach to solve linear equations in one variable. Each question is solved using the official CBSE method: first, the equation is simplified using algebraic rules and like terms are grouped, then the variable is isolated. The answer is checked for correctness using substitution into the original equation. All solutions use the format and calculation style given in the updated CBSE 2025–26 NCERT textbook.

2. Where can I find the official answer key for Class 8 Maths Chapter 2 Exercise 2.2 with correct CBSE format explanation?

You can access the official answer key for Class 8 Maths Chapter 2 Exercise 2.2 in NCERT-approved format on Vedantu's NCERT Solutions page. All solutions follow the stepwise explanation as per CBSE guidelines, showing each algebraic step and providing the final answer. The answer key is structured to match NCERT's latest 2025–26 textbook criteria.

3. How do I solve Question 8 of Class 8 Maths Chapter 2 Exercise 2.2 using the NCERT pattern?

To solve Question 8 of Exercise 2.2 in Class 8 Maths Chapter 2, start by carefully translating the word problem into a linear equation as per the NCERT pattern. Next, use algebraic operations to simplify and isolate the variable. Each calculation step should be written clearly, following the standard NCERT answer structure, and the solution should be verified using substitution to confirm its correctness, as expected by CBSE examiners.

4. Are NCERT Solutions for Class 8 Maths Chapter 2 available in Hindi medium for Exercise 2.1 and 2.2?

Yes, NCERT Solutions for Class 8 Maths Chapter 2 are also available in Hindi medium for Exercise 2.1 and 2.2 on Vedantu. All answers are presented in stepwise Hindi language explanation, fully adhering to the CBSE 2025–26 syllabus and NCERT textbook formatting requirements.

5. What is the method for solving Exercise 2.3 in Class 8 Maths Chapter 2 as per the CBSE approved NCERT answer format?

To solve Exercise 2.3 of Class 8 Maths Chapter 2, apply the stepwise CBSE-approved NCERT method: rewrite the equation according to the question, combine like terms, perform algebraic operations to find the value of the variable, and check your answer by substitution. The format used matches the official CBSE examination style outlined in the 2025–26 NCERT Maths textbook.

6. Can I get a free PDF download of NCERT Solutions for Class 8 Maths Chapter 2 for all exercises, including 2.4 and 2.5?

Yes, you can download a free PDF of NCERT Solutions for Class 8 Maths Chapter 2 for all exercises, including 2.4 and 2.5, from Vedantu. The PDF is structured as per the CBSE pattern and follows the latest 2025–26 NCERT syllabus, providing stepwise explanations and correct answers for each question.

7. Are the Vedantu NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.5 based on the updated CBSE 2025–26 syllabus?

Yes, Vedantu NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.5 are fully updated according to the latest CBSE 2025–26 syllabus. All answers are written in the official NCERT format, providing detailed, step-by-step solutions and covering every question as per textbook standards.

8. What type of questions are included in Class 8 Maths Chapter 2 Exercise 2.4 NCERT Solutions?

Class 8 Maths Chapter 2 Exercise 2.4 NCERT Solutions include linear equations involving fractions, simple application problems, and scenarios requiring translating statements into equations. Each solution is presented using a stepwise CBSE method, emphasizing proper variable isolation and answer verification according to the NCERT pattern.

9. Are the answers for Class 8 Maths Chapter 2 Exercises 2.1 to 2.5 verified for accuracy as per NCERT and CBSE guidelines?

Yes, all answers for Class 8 Maths Chapter 2 Exercises 2.1 to 2.5 are carefully verified against the official NCERT answer key and checked for accuracy using the CBSE-approved method. Each step follows the latest 2025–26 textbook, ensuring students get correct and exam-ready solutions.

10. What are the benefits of following the stepwise NCERT Solution approach for Linear Equations in One Variable in Class 8?

Following the stepwise NCERT Solution approach for Linear Equations in One Variable helps students understand each concept clearly, prevents mistakes, and matches the official CBSE examination answer format. This method builds a strong foundation for solving more complex algebraic equations in higher classes and ensures marks are scored according to NCERT marking schemes.

11. How do NCERT Solutions for Class 8 Maths Chapter 2 help in solving word problems efficiently?

NCERT Solutions for Class 8 Maths Chapter 2 guide students to convert word problems into linear equations by identifying variables and relationships as outlined in the question. The stepwise approach helps break down complex statements into manageable algebraic expressions, ensuring a correct and systematic solution as per CBSE standards.

12. What is a common mistake to avoid while solving variables on both sides in Class 8 Maths Chapter 2?

A common mistake is not moving all variables to one side correctly or forgetting to change the sign while shifting terms. To avoid this, always use the standard CBSE algebraic rules—transpose terms carefully and check each step by substituting the found value back into the original equation, as demonstrated in NCERT answers.

13. Can these NCERT Solutions be used for state board exams that follow CBSE/NCERT guidelines?

Yes, these NCERT Solutions for Class 8 Maths Chapter 2 are structured according to the CBSE 2025–26 NCERT textbook, making them suitable for most state boards that adopt the same curriculum or use NCERT-approved mathematics books. They prepare students for any exam following the national standard answer format.

14. Does the latest NCERT Solutions PDF include answers for intext questions and hot questions for Chapter 2?

The latest NCERT Solutions PDF for Class 8 Maths Chapter 2 covers all back exercises and standard intext questions as per CBSE requirements. HOTS (Higher Order Thinking Skills) questions are not part of the standard NCERT answer key but every textbook-prescribed problem is solved stepwise in the correct format.