NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3

# NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3

## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 are the part of NCERT Solutions for Class 8 Maths. Here you can find the NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3.

Solve the following equations and check your results.

### Ex 2.3 Class 8 Maths Question 1.

3x = 2x + 18

Solution:
Given: 3x = 2x + 18
3x – 2x = 18             (Transposing 2x to LHS)
x = 18
Hence, x = 18 is the required solution.
Check: 3x = 2x + 18
Putting x = 18, we have
LHS = 3 × 18 = 54
RHS = 2 × 18 + 18 = 36 + 18 = 54
LHS = RHS
Hence, verified.

### Ex 2.3 Class 8 Maths Question 2.

5t – 3 = 3t – 5

Solution:
Given: 5t – 3 = 3t – 5
5t – 3t – 3 =  -5                 (Transposing 3t to LHS)
2t =  -5 + 3                         (Transposing -3 to RHS)
2t =  -2
t =  -2 ÷ 2
t =  -1
Hence, t = -1 is the required solution.
Check: 5t – 3 = 3t – 5
Putting t =  -1, we have
LHS = 5t – 3 = 5 × (-1) – 3 =  -5 – 3 =  -8
RHS = 3t – 5 = 3 × (-1) – 5 =  -3 – 5 =  -8
LHS = RHS
Hence, verified.

### Ex 2.3 Class 8 Maths Question 3.

5x + 9 = 5 + 3x

Solution:
Given: 5x + 9 = 5 + 3x
5x – 3x + 9 = 5            (Transposing 3x to LHS)

2x + 9 = 5
2x = 5 – 9                    (Transposing 9 to RHS)
2x =  -4
x =  -4 ÷ 2

x =  -2
Hence, x =  -2 is the required solution.
Check: 5x + 9 = 5 + 3x
Putting x =  -2, we have
LHS = 5 × (-2) + 9 =  -10 + 9 =  -1
RHS = 5 + 3 × (-2) = 5 – 6 =  -1
LHS = RHS
Hence, verified.

### Ex 2.3 Class 8 Maths Question 4.

4z + 3 = 6 + 2z

Solution:
Given: 4z + 3 = 6 + 2z
4z – 2z + 3 = 6                    (Transposing 2z to LHS)
2z + 3 = 6
2z = 6 – 3                            (Transposing 3 to RHS)
2z = 3
z = 3/2
Hence, z = 3/2 is the required solution.
Check: 4z + 3 = 6 + 2z
Putting z = 3/2, we have
LHS = 4z + 3 = 4 × 3/2 + 3 = 6 + 3 = 9
RHS = 6 + 2z = 6 + 2 × 3/2 = 6 + 3 = 9
LHS = RHS
Hence, verified.

### Ex 2.3 Class 8 Maths Question 5.

2x – 1 = 14 – x

Solution:
Given: 2x – 1 = 14 – x
2x + x = 14 + 1                  (Transposing x to LHS and 1 to RHS)
3x = 15
x = 15 ÷ 3 = 5
Hence, x = 5 is the required solution.
Check: 2x – 1 = 14 – x
Putting x = 5, we have
LHS = 2x – 1 = 2 × 5 – 1 = 10 – 1 = 9
RHS = 14 – x = 14 – 5 = 9
LHS = RHS
Hence, verified.

### Ex 2.3 Class 8 Maths Question 6.

8x + 4 = 3(x – 1) + 7

Solution:
Given: 8x + 4 = 3(x – 1) + 7
8x + 4 = 3x – 3 + 7         (Solving the bracket)
8x + 4 = 3x + 4
8x – 3x = 4 – 4                (Transposing 3x to LHS and 4 to RHS)
5x = 0
x = 0 ÷ 5                           (Transposing 5 to RHS)
x = 0
Hence, x = 0 is the required solution.
Check: 8x + 4 = 3(x – 1) + 7
Putting x = 0, we have
LHS = 8 × 0 + 4 = 4

RHS = 3(0 – 1) + 7 = -3 + 7 = 4
LHS = RHS
Hence, verified.

### Ex 2.3 Class 8 Maths Question 7.

x = 4/5 (x + 10)

Solution:
Given: x = 4/5 (x + 10)
5 × x = 4(x + 10)          (Transposing 5 to LHS)
5x = 4x + 40                 (Solving the bracket)
5x – 4x = 40                 (Transposing 4x to LHS)
x = 40
Thus x = 40 is the required solution.
Check: x = 4/5 (x + 10)
Putting x = 40, we have
40 = 4/5 (40 + 10)
40 = 4/5 × 50
40 = 4 × 10
40 = 40
LHS = RHS
Hence, verified.

### Ex 2.3 Class 8 Maths Question 8.

2x/3 + 1 = 7x/15 + 3

Solution:
Given: 2x/3 + 1 = 7x/15 + 3
15(2x/3 + 1) = 15(7x/15 + 3)                          [LCM of 3 and 15 is 15]
2x/3 × 15 + 1 × 15 = 7x/15 × 15 + 3 × 15     [Multiplying both sides by 15]
2x × 5 + 15 = 7x + 45
10x + 15 = 7x + 45
10x – 7x = 45 – 15                     (Transposing 7x to LHS and 15 to RHS)
3x = 30
x = 30 ÷ 3 = 10                           (Transposing 3 to RHS)
Thus, the required solution is x = 10.

### Ex 2.3 Class 8 Maths Question 9.

2y + 5/3 = 26/3 – y

Solution:

3m = 5m – 8/5

Solution:
Given: