Verify by substitution that the root of 3x - 5 = 7 is x = 4
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RD Sharma SolutionsClass 7Chapter 8 Linear Equations In One VariableExercise 8.1
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RD Sharma Solutions For Class 7 Maths Exercise 8.1 Chapter 8 Linear Equations in One Variable
Get free PDF of RD Sharma Solutions for Class 7 Maths Exercise 8.1 of Chapter 8 Linear Equation in One Variable from the provided links. These PDFs can be easily downloaded from the students. Our experts have uniquely formulated these questions to approach a student’s mentality. RD Sharma Solutions for Class 7 is one of the best study material designed for CBSE students. This exercise includes linear equations. Some of the topics covered in this exercise are listed below:
Equation – A statement of equality which involves one or more literals
Linear equation – An equation in which the highest power of the variables involved is 1
Solution of an linear equation
Solving a linear equation by trial-and-error method
Download the PDF of RD Sharma Solutions For Class 7 Maths Chapter 8 – Linear Equations in One variable Exercise 8.1
rd sharma class 7 maths solution chapter 8 ex 1
rd sharma class 7 maths solution chapter 8 ex 1
rd sharma class 7 maths solution chapter 8 ex 1
rd sharma class 7 maths solution chapter 8 ex 1
rd sharma class 7 maths solution chapter 8 ex 1
rd sharma class 7 maths solution chapter 8 ex 1
Access answers to Maths RD Sharma Solutions For Class 7 Chapter 8 – Linear Equation in One Variable Exercise 8.1
1. Verify by substitution that:
(i) x = 4 is the root of 3x – 5 = 7
(ii) x = 3 is the root of 5 + 3x = 14
(iii) x = 2 is the root of 3x – 2 = 8x – 12
(iv) x = 4 is the root of (3x/2) = 6
(v) y = 2 is the root of y – 3 = 2y – 5
(vi) x = 8 is the root of (1/2)x + 7 = 11
Solution:
(i) Given x = 4 is the root of 3x – 5 = 7
Now, substituting x = 4 in place of ‘x’ in the given equation, we get
= 3(4) – 5 = 7
= 12 – 5 = 7
7 = 7
Since, LHS = RHS
Hence, x = 4 is the root of 3x – 5 = 7.
(ii) Given x = 3 is the root of 5 + 3x = 14.
Now, substituting x = 3 in place of ‘x’ in the given equation, we get
= 5 + 3(3) = 14
= 5 + 9 = 14
14 = 14
Since, LHS = RHS
Hence, x = 3 is the root of 5 + 3x = 14.
(iii) Given x = 2 is the root of 3x – 2 = 8x – 12.
Now, substituting x = 2 in place of ‘x’ in the given equation, we get
= 3(2) – 2 = 8(2) – 12
= 6 – 2 = 16 – 12
4 = 4
Since, LHS = RHS
Hence, x = 2 is the root of 3x – 2 = 8x – 12.
(iv) Given x = 4 is the root of 3x/2 = 6.
Now, substituting x = 4 in place of ‘x’ in the given equation, we get
= (3 × 4)/2 = 6
= (12/2) = 6
6 = 6
Since, LHS = RHS
Hence, x = 4 is the root of (3x/2) = 6.
(v) Given y = 2 is the root of y – 3 = 2y – 5.
Now, substituting y = 2 in place of ‘y’ in the given equation, we get
= 2 – 3 = 2(2) – 5
= -1 = 4 – 5
-1 = -1
Since, LHS = RHS
Hence, y = 2 is the root of y – 3 = 2y – 5.
(vi) Given x = 8 is the root of (1/2)x + 7 = 11.
Now, substituting x = 8 in place of ‘x’ in the given equation, we get
= (1/2)(8) + 7 =11
= 4 + 7 = 11
= 11 = 11
Since, LHS = RHS
Hence, x = 8 is the root of 12x + 7 = 11.
2. Solve each of the following equations by trial – and – error method:
(i) x + 3 =12
(ii) x -7 = 10
(iii) 4x = 28
(iv) (x/2) + 7 = 11
(v) 2x + 4 = 3x
(vi) (x/4) = 12
(vii) (15/x) = 3
(vii) (x/18) = 20
Solution:
(i) Given x + 3 =12
Here LHS = x +3 and RHS = 12
x LHS RHS Is LHS = RHS
1 1 + 3 = 4 12 No
2 2 + 3 = 5 12 No
3 3 + 3 = 6 12 No
4 4 + 3 = 7 12 No
5 5 + 3 = 8 12 No
6 6 + 3 = 9 12 No
7 7 + 3 = 10 12 No
8 8 + 3 = 11 12 No
9 9 + 3 = 12 12 Yes
Therefore, if x = 9, LHS = RHS.
Hence, x = 9 is the solution to this equation.
(ii) Given x -7 = 10
Here LHS = x -7 and RHS = 10
x LHS RHS Is LHS = RHS
9 9 – 7 = 2 10 No
10 10 -7 = 3 10 No
11 11 – 7 = 4 10 No
12 12 – 7 = 5 10 No
13 19 – 7 = 6 10 No
14 14 – 7 = 7 10 No
15 15 – 7 = 8 10 No
16 16 – 7 = 9 10 No
17 17 – 7 = 10 10 Yes
Therefore if x = 17, LHS = RHS
Hence, x = 17 is the solution to this equation.
(iii) Given 4x = 28
Here LHS = 4x and RHS = 28
x LHS RHS Is LHS = RHS
1 4 × 1 = 4 28 No
2 4 × 2 = 8 28 No
3 4 × 3 = 12 28 No
4 4 × 4 = 16 28 No
5 4 × 5 = 20 28 No
6 4 × 6 = 24 28 No
7 4 × 7 = 28 28 Yes
Therefore if x = 7, LHS = RHS
Hence, x = 7 is the solution to this equation.
(iv) Given (x/2) + 7 = 11
Here LHS = (x/2) + 7 and RHS = 11
Since RHS is a natural number, (x/2) must also be a natural number, so we must substitute values of x that are multiples of 2.
x LHS RHS Is LHS = RHS
2 (2/2) + 7 = 1 + 7 =8 11 No
4 (4/2) + 7 = 2 + 7 = 9 11 No
6 (6/2) + 7 = 3 + 7 = 10 11 No
8 (8/2) + 7 = 4 + 7 = 11 11 YesSwipe left
Therefore if x = 8, LHS = RHS
Hence, x = 8 is the solutions to this equation.
(v) Given 2x + 4 = 3x
Here LHS = 2x + 4 and RHS = 3x
Step-by-step explanation: