**Chapter 2 Linear Equations in One Variable Exercise 2.3**

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.

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.

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.

Question 4.

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.

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.

Question 7.

Question 8.

Question 9.

Solution:

Solution:

We have