Class 7 Exam > Class 7 Notes > Mathematics (Maths) Class 7 > RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2)
Linear Equations in One Variable (Exercise 8.2) RD Sharma Solutions | Mathematics (Maths) Class 7 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1 Exercise 8.2 Page No: 8.12 Solve each of the following equations and check your answers: 1. x – 3 = 5 Solution: Given x – 3 = 5 Adding 3 to both sides we get, x – 3 + 3 = 5 + 3 x = 8 Verification: Substituting x = 8 in LHS, we get LHS = x – 3 and RHS = 5 LHS = 8 – 3 = 5 and RHS = 5 LHS = RHS Hence, verified. 2. x + 9 = 13 Solution: Given x + 9 = 13 Subtracting 9 from both sides i.e. LHS and RHS, we get x + 9 – 9 = 13 – 9 x = 4 Verification: Substituting x = 4 on LHS, we get LHS = 4 + 9 = 13 = RHS LHS = RHS Hence, verified. 3. x – (3/5) = (7/5) Solution: Given x – (3/5) = (7/5) Add (3/5) to both sides, we get x – (3/5) + (3/5) = (7/5) + (3/5) x = (7/5) + (3/5) Page 2 Exercise 8.2 Page No: 8.12 Solve each of the following equations and check your answers: 1. x – 3 = 5 Solution: Given x – 3 = 5 Adding 3 to both sides we get, x – 3 + 3 = 5 + 3 x = 8 Verification: Substituting x = 8 in LHS, we get LHS = x – 3 and RHS = 5 LHS = 8 – 3 = 5 and RHS = 5 LHS = RHS Hence, verified. 2. x + 9 = 13 Solution: Given x + 9 = 13 Subtracting 9 from both sides i.e. LHS and RHS, we get x + 9 – 9 = 13 – 9 x = 4 Verification: Substituting x = 4 on LHS, we get LHS = 4 + 9 = 13 = RHS LHS = RHS Hence, verified. 3. x – (3/5) = (7/5) Solution: Given x – (3/5) = (7/5) Add (3/5) to both sides, we get x – (3/5) + (3/5) = (7/5) + (3/5) x = (7/5) + (3/5) x = (10/5) x = 2 Verification: Substitute x = 2 in LHS of given equation, then we get 2 – (3/5) = (7/5) (10 – 3)/5 = (7/5) (7/5) = (7/5) LHS = RHS Hence, verified 4. 3x = 0 Solution: Given 3x = 0 On dividing both sides by 3 we get, (3x/3) = (0/3) x = 0 Verification: Substituting x = 0 in LHS we get 3 (0) = 0 And RHS = 0 Therefore LHS = RHS Hence, verified. 5. (x/2) = 0 Solution: Given x/2 = 0 Multiplying both sides by 2, we get (x/2) × 2 = 0 × 2 x = 0 Verification: Substituting x = 0 in LHS, we get LHS = 0/2 = 0 and RHS = 0 LHS = 0 and RHS = 0 Therefore LHS = RHS Hence, verified. Page 3 Exercise 8.2 Page No: 8.12 Solve each of the following equations and check your answers: 1. x – 3 = 5 Solution: Given x – 3 = 5 Adding 3 to both sides we get, x – 3 + 3 = 5 + 3 x = 8 Verification: Substituting x = 8 in LHS, we get LHS = x – 3 and RHS = 5 LHS = 8 – 3 = 5 and RHS = 5 LHS = RHS Hence, verified. 2. x + 9 = 13 Solution: Given x + 9 = 13 Subtracting 9 from both sides i.e. LHS and RHS, we get x + 9 – 9 = 13 – 9 x = 4 Verification: Substituting x = 4 on LHS, we get LHS = 4 + 9 = 13 = RHS LHS = RHS Hence, verified. 3. x – (3/5) = (7/5) Solution: Given x – (3/5) = (7/5) Add (3/5) to both sides, we get x – (3/5) + (3/5) = (7/5) + (3/5) x = (7/5) + (3/5) x = (10/5) x = 2 Verification: Substitute x = 2 in LHS of given equation, then we get 2 – (3/5) = (7/5) (10 – 3)/5 = (7/5) (7/5) = (7/5) LHS = RHS Hence, verified 4. 3x = 0 Solution: Given 3x = 0 On dividing both sides by 3 we get, (3x/3) = (0/3) x = 0 Verification: Substituting x = 0 in LHS we get 3 (0) = 0 And RHS = 0 Therefore LHS = RHS Hence, verified. 5. (x/2) = 0 Solution: Given x/2 = 0 Multiplying both sides by 2, we get (x/2) × 2 = 0 × 2 x = 0 Verification: Substituting x = 0 in LHS, we get LHS = 0/2 = 0 and RHS = 0 LHS = 0 and RHS = 0 Therefore LHS = RHS Hence, verified. 6. x – (1/3) = (2/3) Solution: Given x – (1/3) = (2/3) Adding (1/3) to both sides, we get x – (1/3) + (1/3) = (2/3) + (1/3) x = (2 + 1)/3 x = (3/3) x =1 Verification: Substituting x = 1 in LHS, we get 1 – (1/3) = (2/3) (3 – 1)/3 = (2/3) (2/3) = (2/3) Therefore LHS = RHS Hence, verified. 7. x + (1/2) = (7/2) Solution: Given x + (1/2) = (7/2) Subtracting (1/2) from both sides, we get x + (1/2) – (1/2) = (7/2) – (1/2) x = (7 – 1)/2 x = (6/2) x = 3 Verification: Substituting x = 3 in LHS we get 3 + (1/2) = (7/2) (6 + 1)/2 = (7/2) (7/2) = (7/2) Therefore LHS = RHS Hence, verified. 8. 10 – y = 6 Solution: Page 4 Exercise 8.2 Page No: 8.12 Solve each of the following equations and check your answers: 1. x – 3 = 5 Solution: Given x – 3 = 5 Adding 3 to both sides we get, x – 3 + 3 = 5 + 3 x = 8 Verification: Substituting x = 8 in LHS, we get LHS = x – 3 and RHS = 5 LHS = 8 – 3 = 5 and RHS = 5 LHS = RHS Hence, verified. 2. x + 9 = 13 Solution: Given x + 9 = 13 Subtracting 9 from both sides i.e. LHS and RHS, we get x + 9 – 9 = 13 – 9 x = 4 Verification: Substituting x = 4 on LHS, we get LHS = 4 + 9 = 13 = RHS LHS = RHS Hence, verified. 3. x – (3/5) = (7/5) Solution: Given x – (3/5) = (7/5) Add (3/5) to both sides, we get x – (3/5) + (3/5) = (7/5) + (3/5) x = (7/5) + (3/5) x = (10/5) x = 2 Verification: Substitute x = 2 in LHS of given equation, then we get 2 – (3/5) = (7/5) (10 – 3)/5 = (7/5) (7/5) = (7/5) LHS = RHS Hence, verified 4. 3x = 0 Solution: Given 3x = 0 On dividing both sides by 3 we get, (3x/3) = (0/3) x = 0 Verification: Substituting x = 0 in LHS we get 3 (0) = 0 And RHS = 0 Therefore LHS = RHS Hence, verified. 5. (x/2) = 0 Solution: Given x/2 = 0 Multiplying both sides by 2, we get (x/2) × 2 = 0 × 2 x = 0 Verification: Substituting x = 0 in LHS, we get LHS = 0/2 = 0 and RHS = 0 LHS = 0 and RHS = 0 Therefore LHS = RHS Hence, verified. 6. x – (1/3) = (2/3) Solution: Given x – (1/3) = (2/3) Adding (1/3) to both sides, we get x – (1/3) + (1/3) = (2/3) + (1/3) x = (2 + 1)/3 x = (3/3) x =1 Verification: Substituting x = 1 in LHS, we get 1 – (1/3) = (2/3) (3 – 1)/3 = (2/3) (2/3) = (2/3) Therefore LHS = RHS Hence, verified. 7. x + (1/2) = (7/2) Solution: Given x + (1/2) = (7/2) Subtracting (1/2) from both sides, we get x + (1/2) – (1/2) = (7/2) – (1/2) x = (7 – 1)/2 x = (6/2) x = 3 Verification: Substituting x = 3 in LHS we get 3 + (1/2) = (7/2) (6 + 1)/2 = (7/2) (7/2) = (7/2) Therefore LHS = RHS Hence, verified. 8. 10 – y = 6 Solution: Given 10 – y = 6 Subtracting 10 from both sides, we get 10 – y – 10 = 6 – 10 -y = -4 Multiplying both sides by -1, we get -y × -1 = - 4 × - 1 y = 4 Verification: Substituting y = 4 in LHS, we get 10 – y = 10 – 4 = 6 and RHS = 6 Therefore LHS = RHS Hence, verified. 9. 7 + 4y = -5 Solution: Given 7 + 4y = -5 Subtracting 7 from both sides, we get 7 + 4y – 7 = -5 -7 4y = -12 Dividing both sides by 4, we get y = -12/ 4 y = -3 Verification: Substituting y = -3 in LHS, we get 7 + 4y = 7 + 4(-3) = 7 – 12 = -5, and RHS = -5 Therefore LHS = RHS Hence, verified. 10. (4/5) – x = (3/5) Solution: Given (4/5) – x = (3/5) Subtracting (4/5) from both sides, we get (4/5) – x – (4/5) = (3/5) – (4/5) - x = (3 -4)/5 - x = (-1/5) Page 5 Exercise 8.2 Page No: 8.12 Solve each of the following equations and check your answers: 1. x – 3 = 5 Solution: Given x – 3 = 5 Adding 3 to both sides we get, x – 3 + 3 = 5 + 3 x = 8 Verification: Substituting x = 8 in LHS, we get LHS = x – 3 and RHS = 5 LHS = 8 – 3 = 5 and RHS = 5 LHS = RHS Hence, verified. 2. x + 9 = 13 Solution: Given x + 9 = 13 Subtracting 9 from both sides i.e. LHS and RHS, we get x + 9 – 9 = 13 – 9 x = 4 Verification: Substituting x = 4 on LHS, we get LHS = 4 + 9 = 13 = RHS LHS = RHS Hence, verified. 3. x – (3/5) = (7/5) Solution: Given x – (3/5) = (7/5) Add (3/5) to both sides, we get x – (3/5) + (3/5) = (7/5) + (3/5) x = (7/5) + (3/5) x = (10/5) x = 2 Verification: Substitute x = 2 in LHS of given equation, then we get 2 – (3/5) = (7/5) (10 – 3)/5 = (7/5) (7/5) = (7/5) LHS = RHS Hence, verified 4. 3x = 0 Solution: Given 3x = 0 On dividing both sides by 3 we get, (3x/3) = (0/3) x = 0 Verification: Substituting x = 0 in LHS we get 3 (0) = 0 And RHS = 0 Therefore LHS = RHS Hence, verified. 5. (x/2) = 0 Solution: Given x/2 = 0 Multiplying both sides by 2, we get (x/2) × 2 = 0 × 2 x = 0 Verification: Substituting x = 0 in LHS, we get LHS = 0/2 = 0 and RHS = 0 LHS = 0 and RHS = 0 Therefore LHS = RHS Hence, verified. 6. x – (1/3) = (2/3) Solution: Given x – (1/3) = (2/3) Adding (1/3) to both sides, we get x – (1/3) + (1/3) = (2/3) + (1/3) x = (2 + 1)/3 x = (3/3) x =1 Verification: Substituting x = 1 in LHS, we get 1 – (1/3) = (2/3) (3 – 1)/3 = (2/3) (2/3) = (2/3) Therefore LHS = RHS Hence, verified. 7. x + (1/2) = (7/2) Solution: Given x + (1/2) = (7/2) Subtracting (1/2) from both sides, we get x + (1/2) – (1/2) = (7/2) – (1/2) x = (7 – 1)/2 x = (6/2) x = 3 Verification: Substituting x = 3 in LHS we get 3 + (1/2) = (7/2) (6 + 1)/2 = (7/2) (7/2) = (7/2) Therefore LHS = RHS Hence, verified. 8. 10 – y = 6 Solution: Given 10 – y = 6 Subtracting 10 from both sides, we get 10 – y – 10 = 6 – 10 -y = -4 Multiplying both sides by -1, we get -y × -1 = - 4 × - 1 y = 4 Verification: Substituting y = 4 in LHS, we get 10 – y = 10 – 4 = 6 and RHS = 6 Therefore LHS = RHS Hence, verified. 9. 7 + 4y = -5 Solution: Given 7 + 4y = -5 Subtracting 7 from both sides, we get 7 + 4y – 7 = -5 -7 4y = -12 Dividing both sides by 4, we get y = -12/ 4 y = -3 Verification: Substituting y = -3 in LHS, we get 7 + 4y = 7 + 4(-3) = 7 – 12 = -5, and RHS = -5 Therefore LHS = RHS Hence, verified. 10. (4/5) – x = (3/5) Solution: Given (4/5) – x = (3/5) Subtracting (4/5) from both sides, we get (4/5) – x – (4/5) = (3/5) – (4/5) - x = (3 -4)/5 - x = (-1/5) x = (1/5) Verification: Substituting x = (1/5) in LHS we get (4/5) – (1/5) = (3/5) (4 -1)/5 = (3/5) (3/5) = (3/5) Therefore LHS =RHS Hence, verified. 11. 2y – (1/2) = (-1/3) Solution: Given 2y – (1/2) = (-1/3) Adding (1/2) from both the sides, we get 2y – (1/2) + (1/2) = (-1/3) + (1/2) 2y = (-1/3) + (1/2) 2y = (-2 + 3)/6 [LCM of 3 and 2 is 6] 2y = (1/6) Now divide both the side by 2, we get y = (1/12) Verification: Substituting y = (1/12) in LHS we get 2 (1/12) – (1/2) = (-1/3) (1/6) – (1/2) = (-1/3) (2 – 6)/12 = (-1/3) [LCM of 6 and 2 is 12] (-4/12) = (-1/3) (-1/3) = (-1/3) Therefore LHS = RHS Hence, verified. 12. 14 = (7x/10) – 8 Solution: Given 14 = (7x/10) – 8 Adding 8 to both sides we get, 14 + 8 = (7x/10) – 8 + 8 22 = (7x/10)Read More
|
76 videos|344 docs|39 tests
|
Related Exams
About this Document
|
4.72/5 Rating |
|
Dec 23, 2024 Last updated |
Document Description: RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) for Class 7 2024 is part of Mathematics (Maths) Class 7 preparation.
The notes and questions for RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) have been prepared according to the Class 7 exam syllabus. Information about RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) covers topics
like and RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) Example, for Class 7 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2).
Introduction of RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) in English is available as part of our Mathematics (Maths) Class 7
for Class 7 & RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) in Hindi for Mathematics (Maths) Class 7 course.
Download more important topics related with notes, lectures and mock test series for Class 7
Exam by signing up for free. Class 7: Linear Equations in One Variable (Exercise 8.2) RD Sharma Solutions | Mathematics (Maths) Class 7
Description
Full syllabus notes, lecture & questions for Linear Equations in One Variable (Exercise 8.2) RD Sharma Solutions | Mathematics (Maths) Class 7 - Class 7 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 7 | Best notes, free PDF download
Information about RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2)
In this doc you can find the meaning of RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) defined & explained in the simplest way possible. Besides explaining types of
RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) tests, examples and also practice Class 7
tests
|
Explore Courses for Class 7 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Viva Questions
,MCQs
,study material
,ppt
,Summary
,Linear Equations in One Variable (Exercise 8.2) RD Sharma Solutions | Mathematics (Maths) Class 7
,Sample Paper
,Objective type Questions
,Semester Notes
,Linear Equations in One Variable (Exercise 8.2) RD Sharma Solutions | Mathematics (Maths) Class 7
,Linear Equations in One Variable (Exercise 8.2) RD Sharma Solutions | Mathematics (Maths) Class 7
,Previous Year Questions with Solutions
,Extra Questions
,Important questions
,Exam
,practice quizzes
,video lectures
,past year papers
,Free
,mock tests for examination
,shortcuts and tricks
;
Additional Information about RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) for Class 7 Preparation
RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) Free PDF Download
The RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) is an invaluable resource that delves deep into the core of the Class 7 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) now and kickstart your journey towards success in the Class 7 exam.
Importance of RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2)
The importance of RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) cannot be overstated, especially for Class 7 aspirants.
This document holds the key to success in the Class 7 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) Notes
RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2).
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) Notes on EduRev are your ultimate resource for success.
RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) Class 7 Questions
The "RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) Class 7 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 7 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) on the App
Students of Class 7 can study RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2),
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) is prepared as per the latest Class 7 syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup