**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:
**

**Ex
2.3 Class 8 Maths Question 10.**

3m = 5m – 8/5**Solution:
**Given: