Class 7 Exam > Class 7 Notes > Mathematics (Maths) Class 7 > RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.3)
Linear Equations in One Variable (Exercise 8.3) 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 8.3 Page No: 8.19
Solve each of the following equations. Also, verify the result in each case.
1. 6x + 5 = 2x + 17
Solution:
Given 6x + 5 = 2x + 17
Transposing 2x to LHS and 5 to RHS, we get
6x – 2x = 17 – 5
4x = 12
Dividing both sides by 4, we get
4x/4 = 12/4
x = 3
Verification:
Substituting x = 3 in the given equation, we get
6 × 3 + 5 = 2 × 3 + 17
18 + 5 = 6 + 17
23 = 23
Therefore LHS = RHS
Hence, verified.
2. 2 (5x – 3) – 3 (2x – 1) = 9
Solution:
Given 2 (5x – 3) – 3 (2x – 1) = 9
Simplifying the brackets, we get
2 × 5x – 2 × 3 – 3 × 2x + 3 × 1 = 9
10x – 6 - 6x + 3 = 9
10x – 6x – 6 + 3 = 9
4x – 3 = 9
Adding 3 to both sides, we get
4x – 3 + 3 = 9 + 3
4x = 12
Dividing both sides by 4, we get
4x/4 = 12/4
Therefore x = 3.
Page 2
Exercise 8.3 Page No: 8.19
Solve each of the following equations. Also, verify the result in each case.
1. 6x + 5 = 2x + 17
Solution:
Given 6x + 5 = 2x + 17
Transposing 2x to LHS and 5 to RHS, we get
6x – 2x = 17 – 5
4x = 12
Dividing both sides by 4, we get
4x/4 = 12/4
x = 3
Verification:
Substituting x = 3 in the given equation, we get
6 × 3 + 5 = 2 × 3 + 17
18 + 5 = 6 + 17
23 = 23
Therefore LHS = RHS
Hence, verified.
2. 2 (5x – 3) – 3 (2x – 1) = 9
Solution:
Given 2 (5x – 3) – 3 (2x – 1) = 9
Simplifying the brackets, we get
2 × 5x – 2 × 3 – 3 × 2x + 3 × 1 = 9
10x – 6 - 6x + 3 = 9
10x – 6x – 6 + 3 = 9
4x – 3 = 9
Adding 3 to both sides, we get
4x – 3 + 3 = 9 + 3
4x = 12
Dividing both sides by 4, we get
4x/4 = 12/4
Therefore x = 3.
Verification:
Substituting x = 3 in LHS, we get
2(5 × 3 – 3) – 3(2 × 3 – 1) = 9
2 × 12 – 3 × 5 = 9
24 – 15 = 9
9 = 9
Thus, LHS = RHS
Hence, verified.
3. (x/2) = (x/3) + 1
Solution:
Given (x/2) = (x/3) + 1
Transposing (x/3) to LHS we get
(x/2) – (x/3) = 1
(3x – 2x)/6 = 1 [LCM of 3 and 2 is 6]
x/6 = 1
Multiplying 6 to both sides we get,
x = 6
Verification:
Substituting x = 6 in given equation we get
(6/2) = (6/3) + 1
3 = 2 + 1
3 = 3
Thus LHS = RHS
Hence, verified.
4. (x/2) + (3/2) = (2x/5) - 1
Solution:
Given (x/2) + (3/2) = (2x/5) - 1
Transposing (2x/5) to LHS and (3/2) to RHS, then we get
(x/2) – (2x/5) = - 1 – (3/2)
(5x -4x)/10 = (-2 – 3)/2 [LCM of 5 and 2 is 10]
x/10 = -5/2
Multiplying both sides by 10 we get,
x/10 × 10 = (-5/2) × 10
Page 3
Exercise 8.3 Page No: 8.19
Solve each of the following equations. Also, verify the result in each case.
1. 6x + 5 = 2x + 17
Solution:
Given 6x + 5 = 2x + 17
Transposing 2x to LHS and 5 to RHS, we get
6x – 2x = 17 – 5
4x = 12
Dividing both sides by 4, we get
4x/4 = 12/4
x = 3
Verification:
Substituting x = 3 in the given equation, we get
6 × 3 + 5 = 2 × 3 + 17
18 + 5 = 6 + 17
23 = 23
Therefore LHS = RHS
Hence, verified.
2. 2 (5x – 3) – 3 (2x – 1) = 9
Solution:
Given 2 (5x – 3) – 3 (2x – 1) = 9
Simplifying the brackets, we get
2 × 5x – 2 × 3 – 3 × 2x + 3 × 1 = 9
10x – 6 - 6x + 3 = 9
10x – 6x – 6 + 3 = 9
4x – 3 = 9
Adding 3 to both sides, we get
4x – 3 + 3 = 9 + 3
4x = 12
Dividing both sides by 4, we get
4x/4 = 12/4
Therefore x = 3.
Verification:
Substituting x = 3 in LHS, we get
2(5 × 3 – 3) – 3(2 × 3 – 1) = 9
2 × 12 – 3 × 5 = 9
24 – 15 = 9
9 = 9
Thus, LHS = RHS
Hence, verified.
3. (x/2) = (x/3) + 1
Solution:
Given (x/2) = (x/3) + 1
Transposing (x/3) to LHS we get
(x/2) – (x/3) = 1
(3x – 2x)/6 = 1 [LCM of 3 and 2 is 6]
x/6 = 1
Multiplying 6 to both sides we get,
x = 6
Verification:
Substituting x = 6 in given equation we get
(6/2) = (6/3) + 1
3 = 2 + 1
3 = 3
Thus LHS = RHS
Hence, verified.
4. (x/2) + (3/2) = (2x/5) - 1
Solution:
Given (x/2) + (3/2) = (2x/5) - 1
Transposing (2x/5) to LHS and (3/2) to RHS, then we get
(x/2) – (2x/5) = - 1 – (3/2)
(5x -4x)/10 = (-2 – 3)/2 [LCM of 5 and 2 is 10]
x/10 = -5/2
Multiplying both sides by 10 we get,
x/10 × 10 = (-5/2) × 10
x = (-50/2)
x = -25
Verification:
Substituting x = -25 in given equation we get
(-25/2) + (3/2) = (-50/5) – 1
(-25 + 3)/2 = -10 – 1
(-22/2) = -11
-11 = -11
Thus LHS = RHS
Hence, verified.
5. (3/4) (x -1) = (x – 3)
Solution:
Given (3/4) (x -1) = (x – 3)
On simplifying the brackets both sides we get,
(3/4) x – (3/4) = (x – 3)
Now transposing (3/4) to RHS and (x – 3) to LHs
(3/4) x – x = (3/4) – 3
(3x – 4x)/4 = (3 – 12)/4
-x/4 = (-9/4)
Multiply both sides by -4 we get
-x/4 × -4 = (-9/4) × -4
x = 9
Verification:
Substituting x = 9 in the given equation:
(3/4) (9 – 1) = (9 -3)
(3/4) (8) = 6
3 × 2 = 6
6 = 6
Thus LHS = RHS
Hence, verified.
6. 3 (x – 3) = 5 (2x + 1)
Solution:
Given 3 (x – 3) = 5 (2x + 1)
Page 4
Exercise 8.3 Page No: 8.19
Solve each of the following equations. Also, verify the result in each case.
1. 6x + 5 = 2x + 17
Solution:
Given 6x + 5 = 2x + 17
Transposing 2x to LHS and 5 to RHS, we get
6x – 2x = 17 – 5
4x = 12
Dividing both sides by 4, we get
4x/4 = 12/4
x = 3
Verification:
Substituting x = 3 in the given equation, we get
6 × 3 + 5 = 2 × 3 + 17
18 + 5 = 6 + 17
23 = 23
Therefore LHS = RHS
Hence, verified.
2. 2 (5x – 3) – 3 (2x – 1) = 9
Solution:
Given 2 (5x – 3) – 3 (2x – 1) = 9
Simplifying the brackets, we get
2 × 5x – 2 × 3 – 3 × 2x + 3 × 1 = 9
10x – 6 - 6x + 3 = 9
10x – 6x – 6 + 3 = 9
4x – 3 = 9
Adding 3 to both sides, we get
4x – 3 + 3 = 9 + 3
4x = 12
Dividing both sides by 4, we get
4x/4 = 12/4
Therefore x = 3.
Verification:
Substituting x = 3 in LHS, we get
2(5 × 3 – 3) – 3(2 × 3 – 1) = 9
2 × 12 – 3 × 5 = 9
24 – 15 = 9
9 = 9
Thus, LHS = RHS
Hence, verified.
3. (x/2) = (x/3) + 1
Solution:
Given (x/2) = (x/3) + 1
Transposing (x/3) to LHS we get
(x/2) – (x/3) = 1
(3x – 2x)/6 = 1 [LCM of 3 and 2 is 6]
x/6 = 1
Multiplying 6 to both sides we get,
x = 6
Verification:
Substituting x = 6 in given equation we get
(6/2) = (6/3) + 1
3 = 2 + 1
3 = 3
Thus LHS = RHS
Hence, verified.
4. (x/2) + (3/2) = (2x/5) - 1
Solution:
Given (x/2) + (3/2) = (2x/5) - 1
Transposing (2x/5) to LHS and (3/2) to RHS, then we get
(x/2) – (2x/5) = - 1 – (3/2)
(5x -4x)/10 = (-2 – 3)/2 [LCM of 5 and 2 is 10]
x/10 = -5/2
Multiplying both sides by 10 we get,
x/10 × 10 = (-5/2) × 10
x = (-50/2)
x = -25
Verification:
Substituting x = -25 in given equation we get
(-25/2) + (3/2) = (-50/5) – 1
(-25 + 3)/2 = -10 – 1
(-22/2) = -11
-11 = -11
Thus LHS = RHS
Hence, verified.
5. (3/4) (x -1) = (x – 3)
Solution:
Given (3/4) (x -1) = (x – 3)
On simplifying the brackets both sides we get,
(3/4) x – (3/4) = (x – 3)
Now transposing (3/4) to RHS and (x – 3) to LHs
(3/4) x – x = (3/4) – 3
(3x – 4x)/4 = (3 – 12)/4
-x/4 = (-9/4)
Multiply both sides by -4 we get
-x/4 × -4 = (-9/4) × -4
x = 9
Verification:
Substituting x = 9 in the given equation:
(3/4) (9 – 1) = (9 -3)
(3/4) (8) = 6
3 × 2 = 6
6 = 6
Thus LHS = RHS
Hence, verified.
6. 3 (x – 3) = 5 (2x + 1)
Solution:
Given 3 (x – 3) = 5 (2x + 1)
On simplifying the brackets we get,
3x – 9 = 10x + 5
Now transposing 10x to LHS and 9 to RHs
3x – 10x = 5 + 9
-7x = 14
Now dividing both sides by -7 we get
-7x/-7 = 14/-7
x = -2
Verification:
Substituting x = -2 in the given equation we get
3 (-2 – 3) = 5 (-4 + 1)
3 (-5) = 5 (-3)
-15 = - 15
Thus LHS = RHS
Hence, verified.
7. 3x – 2 (2x -5) = 2 (x + 3) – 8
Solution:
Given 3x – 2 (2x -5) = 2 (x + 3) – 8
On simplifying the brackets on both sides, we get
3x – 2 × 2x + 2 × 5 = 2 × x + 2 × 3 – 8
3x – 4x + 10 = 2x + 6 – 8
-x + 10 = 2x – 2
Transposing x to RHS and 2 to LHS, we get
10 + 2 = 2x + x
3x = 12
Dividing both sides by 3, we get
3x/3 = 12/3
x = 4
Verification:
Substituting x = 4 on both sides, we get
3(4) – 2{2(4) – 5} = 2(4 + 3) – 8
12 – 2(8 – 5) = 14 – 8
12 – 6 = 6
6 = 6
Thus LHS = RHS
Page 5
Exercise 8.3 Page No: 8.19
Solve each of the following equations. Also, verify the result in each case.
1. 6x + 5 = 2x + 17
Solution:
Given 6x + 5 = 2x + 17
Transposing 2x to LHS and 5 to RHS, we get
6x – 2x = 17 – 5
4x = 12
Dividing both sides by 4, we get
4x/4 = 12/4
x = 3
Verification:
Substituting x = 3 in the given equation, we get
6 × 3 + 5 = 2 × 3 + 17
18 + 5 = 6 + 17
23 = 23
Therefore LHS = RHS
Hence, verified.
2. 2 (5x – 3) – 3 (2x – 1) = 9
Solution:
Given 2 (5x – 3) – 3 (2x – 1) = 9
Simplifying the brackets, we get
2 × 5x – 2 × 3 – 3 × 2x + 3 × 1 = 9
10x – 6 - 6x + 3 = 9
10x – 6x – 6 + 3 = 9
4x – 3 = 9
Adding 3 to both sides, we get
4x – 3 + 3 = 9 + 3
4x = 12
Dividing both sides by 4, we get
4x/4 = 12/4
Therefore x = 3.
Verification:
Substituting x = 3 in LHS, we get
2(5 × 3 – 3) – 3(2 × 3 – 1) = 9
2 × 12 – 3 × 5 = 9
24 – 15 = 9
9 = 9
Thus, LHS = RHS
Hence, verified.
3. (x/2) = (x/3) + 1
Solution:
Given (x/2) = (x/3) + 1
Transposing (x/3) to LHS we get
(x/2) – (x/3) = 1
(3x – 2x)/6 = 1 [LCM of 3 and 2 is 6]
x/6 = 1
Multiplying 6 to both sides we get,
x = 6
Verification:
Substituting x = 6 in given equation we get
(6/2) = (6/3) + 1
3 = 2 + 1
3 = 3
Thus LHS = RHS
Hence, verified.
4. (x/2) + (3/2) = (2x/5) - 1
Solution:
Given (x/2) + (3/2) = (2x/5) - 1
Transposing (2x/5) to LHS and (3/2) to RHS, then we get
(x/2) – (2x/5) = - 1 – (3/2)
(5x -4x)/10 = (-2 – 3)/2 [LCM of 5 and 2 is 10]
x/10 = -5/2
Multiplying both sides by 10 we get,
x/10 × 10 = (-5/2) × 10
x = (-50/2)
x = -25
Verification:
Substituting x = -25 in given equation we get
(-25/2) + (3/2) = (-50/5) – 1
(-25 + 3)/2 = -10 – 1
(-22/2) = -11
-11 = -11
Thus LHS = RHS
Hence, verified.
5. (3/4) (x -1) = (x – 3)
Solution:
Given (3/4) (x -1) = (x – 3)
On simplifying the brackets both sides we get,
(3/4) x – (3/4) = (x – 3)
Now transposing (3/4) to RHS and (x – 3) to LHs
(3/4) x – x = (3/4) – 3
(3x – 4x)/4 = (3 – 12)/4
-x/4 = (-9/4)
Multiply both sides by -4 we get
-x/4 × -4 = (-9/4) × -4
x = 9
Verification:
Substituting x = 9 in the given equation:
(3/4) (9 – 1) = (9 -3)
(3/4) (8) = 6
3 × 2 = 6
6 = 6
Thus LHS = RHS
Hence, verified.
6. 3 (x – 3) = 5 (2x + 1)
Solution:
Given 3 (x – 3) = 5 (2x + 1)
On simplifying the brackets we get,
3x – 9 = 10x + 5
Now transposing 10x to LHS and 9 to RHs
3x – 10x = 5 + 9
-7x = 14
Now dividing both sides by -7 we get
-7x/-7 = 14/-7
x = -2
Verification:
Substituting x = -2 in the given equation we get
3 (-2 – 3) = 5 (-4 + 1)
3 (-5) = 5 (-3)
-15 = - 15
Thus LHS = RHS
Hence, verified.
7. 3x – 2 (2x -5) = 2 (x + 3) – 8
Solution:
Given 3x – 2 (2x -5) = 2 (x + 3) – 8
On simplifying the brackets on both sides, we get
3x – 2 × 2x + 2 × 5 = 2 × x + 2 × 3 – 8
3x – 4x + 10 = 2x + 6 – 8
-x + 10 = 2x – 2
Transposing x to RHS and 2 to LHS, we get
10 + 2 = 2x + x
3x = 12
Dividing both sides by 3, we get
3x/3 = 12/3
x = 4
Verification:
Substituting x = 4 on both sides, we get
3(4) – 2{2(4) – 5} = 2(4 + 3) – 8
12 – 2(8 – 5) = 14 – 8
12 – 6 = 6
6 = 6
Thus LHS = RHS
Hence, verified.
8. x – (x/4) – (1/2) = 3 + (x/4)
Solution:
Given x – (x/4) – (1/2) = 3 + (x/4)
Transposing (x/4) to LHS and (1/2) to RHS
x – (x/4) - (x/4) = 3 + (1/2)
(4x – x – x)/4 = (6 + 1)/2
2x/4 = 7/2
x/2 = 7/2
x = 7
Verification:
Substituting x = 7 in the given equation we get
7 – (7/4) – (1/2) = 3 + (7/4)
(28 – 7 – 2)/4 = (12 + 7)/4
19/4 = 19/4
Thus LHS = RHS
Hence, verified.
9. (6x – 2)/9 + (3x + 5)/18 = (1/3)
Solution:
Given (6x – 2)/9 + (3x + 5)/18 = (1/3)
(6x (2) – 2 (2) + 3x + 5)/18 = (1/3)
(12x – 4 + 3x + 5)/18 = (1/3)
(15x + 1)/ 18 = (1/3)
Multiplying both sides by 18 we get
(15x + 1)/18 × 18 = (1/3) × 18
15x + 1 = 6
Transposing 1 to RHS, we get
= 15x = 6 – 1
= 15x = 5
Dividing both sides by 15, we get
= 15x/15 = 5/15
=x = 1/3
Verification:
Read More
|
76 videos|344 docs|39 tests
|
FAQs on Linear Equations in One Variable (Exercise 8.3) RD Sharma Solutions - Mathematics (Maths) Class 7
| 1. How do you solve linear equations in one variable? |  |
Ans. To solve a linear equation in one variable, you need to isolate the variable on one side of the equation. You can do this by performing various operations such as addition, subtraction, multiplication, and division on both sides of the equation. The goal is to simplify the equation and solve for the unknown variable.
| 2. What is the importance of solving linear equations in one variable? | |
Ans. Solving linear equations in one variable is important as it helps in finding the value of the unknown variable. It is a fundamental concept in mathematics and has applications in various fields such as physics, engineering, finance, and economics. It allows us to solve real-life problems and make predictions based on mathematical models.
| 3. Can linear equations in one variable have more than one solution? | |
Ans. No, linear equations in one variable can have only one solution. This is because a linear equation represents a straight line on a graph, and the intersection of a line with the x-axis represents the solution to the equation. Therefore, there can be either one unique solution or no solution at all.
| 4. How can I check if a given value is a solution to a linear equation in one variable? | |
Ans. To check if a given value is a solution to a linear equation in one variable, substitute the value into the equation and simplify both sides. If the resulting equation is true, then the given value is a solution to the equation. If the resulting equation is false, then the given value is not a solution.
| 5. What are the different methods to solve linear equations in one variable? | |
Ans. There are various methods to solve linear equations in one variable, including:
- Addition and subtraction method: Add or subtract the same number from both sides of the equation to isolate the variable.
- Multiplication and division method: Multiply or divide both sides of the equation by the same number to isolate the variable.
- Cross-multiplication method: This method is used for solving equations involving fractions. It involves multiplying both sides of the equation by the denominators to eliminate the fractions.
- Substitution method: This method involves substituting the value of one variable from one equation into the other equation to solve for the other variable.
- Graphical method: This method involves plotting the equation on a graph and finding the x-coordinate of the point where the line intersects the x-axis, which represents the solution to the equation.
Related Exams
About this Document
|
4.81/5 Rating |
|
Dec 23, 2024 Last updated |
Document Description: RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.3) 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 8.3) have been prepared according to the Class 7 exam syllabus. Information about RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.3) covers topics
like and RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.3) 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 8.3).
Introduction of RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.3) in English is available as part of our Mathematics (Maths) Class 7
for Class 7 & RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.3) 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 8.3) RD Sharma Solutions | Mathematics (Maths) Class 7
Description
Full syllabus notes, lecture & questions for Linear Equations in One Variable (Exercise 8.3) 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 8.3)
In this doc you can find the meaning of RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.3) defined & explained in the simplest way possible. Besides explaining types of
RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.3) theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.3) 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
Linear Equations in One Variable (Exercise 8.3) RD Sharma Solutions | Mathematics (Maths) Class 7
,study material
,MCQs
,Objective type Questions
,Linear Equations in One Variable (Exercise 8.3) RD Sharma Solutions | Mathematics (Maths) Class 7
,Exam
,Summary
,practice quizzes
,Viva Questions
,ppt
,Linear Equations in One Variable (Exercise 8.3) RD Sharma Solutions | Mathematics (Maths) Class 7
,video lectures
,past year papers
,Sample Paper
,Extra Questions
,shortcuts and tricks
,Previous Year Questions with Solutions
,Important questions
,Free
,mock tests for examination
,Semester Notes
;
Additional Information about RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.3) for Class 7 Preparation
RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.3) Free PDF Download
The RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.3) 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 8.3) now and kickstart your journey towards success in the Class 7 exam.
Importance of RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.3)
The importance of RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.3) 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 8.3) Notes
RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.3) 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 8.3).
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 8.3) Notes on EduRev are your ultimate resource for success.
RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.3) Class 7 Questions
The "RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.3) 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 8.3) on the App
Students of Class 7 can study RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.3) 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 8.3),
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 8.3) 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