RD Sharma Solutions for Class 7 Maths Exercise 8.3 of Chapter 8 Linear Equation in One Variable are provided in simple PDF format. Learners can easily download the PDF from the given links. The BYJU’S subject experts have solved the questions present in this exercise, which will help students with their exam preparation. We suggest practising RD Sharma Solutions for Class 7 to improve their Maths skills. This exercise discusses the transposition method, which is defined as any term of the equation that can be taken to the other side with its sign changed without affecting equality.
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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:
Substituting x = 1/3 both sides, we get
(6 (1/3) – 2)/9 + (3 (1/3) + 5)/18 = (1/3)
(2 – 2)/9 + (1 + 5)/ 18 = 1/3
(6/18) = (1/3)
(1/3) = (1/3)
Thus LHS = RHS
Hence, verified.
10. m – (m – 1)/2 = 1 – (m – 2)/3
Solution:
Given m – (m – 1)/2 = 1 – (m – 2)/3
(2m – m + 1)/2 = (3 – m + 2)/3
(m + 1)/2 = (5 – m)/3
(m + 1)/2 = (5/3) – (m/3)
(m/2) + (1/2) = (5/3) – (m/3)
Transposing (m/3) to LHS and (1/2) to RHS
(m/2) + (m/3) = (5/3) – (1/2)
(3m + 2m)/6 = (10 – 3)/6
5m/6 = (7/6)
5m = 7
Dividing both sides by 5, we get
5m/5 = 7/5
m = 7/5
Verification:
Substituting m = 7/5 on both sides, we get
(7/5) – (7 – 5)/10 = 1 – (7 – 10)/15
(7/5) – (2/10) = (15 + 3)/15
(14 – 2)/10 = (15 + 3)/15
12/10 = 18/15
(6/5) = (6/5)
Thus LHS = RHS
Hence, verified.
11. (5x – 1)/3 – (2x – 2)/3 = 1
Solution:
Given (5x – 1)/3 – (2x – 2)/3 = 1
(5x – 1 – 2x + 2)/3 = 1
(3x + 1)/3 = 1
Multiplying both sides by 3 we get
(3x + 1)/3 × 3 = 1 × 3
(3x + 1) = 3
Subtracting 1 from both sides we get
3x + 1 – 1 = 3 – 1
3x = 2
Dividing both sides by 3, we get
3x/3 =Â 2/3
x = 2/3
Verification:
Substituting x = 2/3 in LHS, we get
(5 (2/3) – 1)/3 – (2 (2/3) – 2)/3 = 1
(10/3 -1)/3 – (4/3 – 2)/3 = 1
(7/3)/3 – (-2/3)/3 = 1
(7/9) + (2/9) = 1
(9/9) = 1
1 = 1
Thus LHS = RHS
Hence, verified.
12. 0.6x +Â 4/5Â = 0.28x + 1.16
Solution:
Given 0.6x +Â 4/5Â = 0.28x + 1.16
Transposing 0.28x to LHS and 45 to RHS, we get
0.6x – 0.28x = 1.16 – 45
0.32x = 1.16 – 0.8
0.32x = 0.36
Dividing both sides by 0.32, we get
0.32 x 0.32 = 0.360.32
x = 9/8
Verification:
Substituting x = 9/8Â on both sides, we get
0.6(9/8) + 45Â = 0.28(9/8) + 1.16
5.4/8Â + 4/5Â = 2.52/8Â + 1.16
0.675 + 0.8 = 0.315 + 1.16
1.475 = 1.475
Thus LHS = RHS
Hence, verified.
13. 0.5x + (x/3) = 0.25x + 7
Solution:
Given 0.5x + (x/3) = 0.25x + 7
(5/10) x + (x/3) = (25x/100) + 7
(x/2) + (x/3) = (x/4) + 7
Transposing (x/4) to LHS we get
(x/2) + (x/3) – (x/4) = 7
(6x + 4x – 3x)/12 = 7
(7x/12) = 7
Multiplying both sides by 12 we get
(7x/12) × 12 = 7 × 12
7x = 84
Dividing both sides by 7 we get
(7x/7) = (84/7)
x = 12
Verification:
Substituting x = 12 in given equation we get
0.5 (12) + (12/3) = 0.25 (12) + 7
6 + 4 = 3 + 7
10 = 10
Thus LHS = RHS
Hence, verified.
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