# 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 Exercise 2.3**

**Solve the following equations and check your results.**

Ex 2.3 Class 8 Maths Question 1.

3x = 2x + 18

Solution:

We have 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:

We have 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:

We have 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 = -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:

We have 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 = \(\frac { 3 }{ 2 }\)

Hence z = \(\frac { 3 }{ 2 }\) is the required solution.

Check: 4z + 3 = 6 + 2z

Putting z = \(\frac { 3 }{ 2 }\), we have

LHS = 4z + 3 = 4 × \(\frac { 3 }{ 2 }\) + 3 = 6 + 3 = 9

RHS = 6 + 2z = 6 + 2 × \(\frac { 3 }{ 2 }\) = 6 + 3 = 9

LHS = RHS

Hence verified.

Ex 2.3 Class 8 Maths Question 5.

2x – 1 = 14 – x

Solution:

We have 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

LHS we have 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:

We have 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]

or x = 0

Thus x = 0 is the required solution.

Check: 8x + 4 = 3(x – 1) + 7

Putting x = 0, we have

8 × 0 + 4 = 3(0 – 1) + 7

⇒ 0 + 4 = -3 + 7

⇒ 4 = 4

LHS = RHS

Hence verified.

Ex 2.3 Class 8 Maths Question 7.

x = \(\frac { 4 }{ 5 }\) (x + 10)

Solution:

We have x = \(\frac { 4 }{ 5 }\) (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 = \(\frac { 4 }{ 5 }\) (x + 10)

Putting x = 40, we have

40 = \(\frac { 4 }{ 5 }\) (40 + 10)

⇒ 40 = \(\frac { 4 }{ 5 }\) × 50

⇒ 40 = 4 × 10

⇒ 40 = 40

LHS = RHS

Hence verified.

Ex 2.3 Class 8 Maths Question 8.

\(\frac { 2x }{ 3 }\) + 1 = \(\frac { 7x }{ 15 }\) + 3

Solution:

We have \(\frac { 2x }{ 3 }\) + 1 = \(\frac { 7x }{ 15 }\) + 3

15(\(\frac { 2x }{ 3 }\) + 1) = 15(\(\frac { 7x }{ 15 }\) + 3)

LCM of 3 and 15 is 15

\(\frac { 2x }{ 3 }\) × 15 + 1 × 15 = \(\frac { 7x }{ 15 }\) × 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 + \(\frac { 5 }{ 3 }\) = \(\frac { 26 }{ 3 }\) – y

Solution:

Ex 2.3 Class 8 Maths Question 10.

3m = 5m – \(\frac { 8 }{ 5 }\)

Solution:

We have