# Linear Equations in One Variable : Exercise 2.3 (Mathematics NCERT Class 8th)

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

**Q.1 3x = 2x + 18
**

**Given, 3x = 2x + 18**

*Sol.*Transposing 2x from RHS to LHS, we get,

3x â€“ 2x = 18

x = 18

Verification:

LHS = 3x = 3 x 18 = 54

RHS = 2x + 18 = 2 x 18 + 18 = 36 + 18 = 54

Here, LHS = RHS

Thus, the result obtained is correct.

**Q.2 5t â€“ 3 = 3t â€“ 5
**

**Given, 5t â€“ 3 = 3t â€“ 5**

*Sol.*Transposing 3t from RHS to LHS and â€“ 3 from LHS to RHS, we get,

5t â€“ 3t = -5 â€“ (- 3)

2t = - 2

Dividing both the sides by 2, we get,

t = -1

Verification:

LHS = 5t - 3 = 5 x (-1) â€“ 3 = - 8

RHS = 3t â€“ 5 = 3 x (-1) â€“ 5 = -3 - 18 = -8

Here, LHS = RHS

Thus, the result obtained is correct.

**Q.3 5x + 9 = 5 + 3x
**

**Given, 5x + 9 = 5 + 3x**

*Sol.*Transposing 3x from RHS to LHS and 9 from LHS to RHS

5x â€“ 3x = 5 â€“ 9

2x = -4

Dividing both the sides by 2, we get,

x = -2

Verification:

LHS = 5x + 9 = 5 x (-2) + 9 = - 1

RHS = 5 + 3x = 5 + 3 x (-2) = 5 - 6 = -1

Here, LHS = RHS

Thus, the result obtained is correct.

**Q.4 4z + 3 = 6 + 2z
**

**Given, 4z + 3 = 6 + 2z**

*Sol.*Transposing 2z from RHS to LHS and 3 from LHS to RHS, we get,

4z â€“ 2z = 6 â€“ 3

2z = 3

Dividing by 2 on both the sides, we get,

Verification:

LHS = 4z + 3 = 4 x + 3 = 6 + 3 = 9

RHS = 6 + 2z = 6 + 2 x = 6 + 3 = 9

Here, LHS = RHS

Thus, the result obtained is correct.

**Q.5 2x â€“ 1 = 14 â€“ x
**

**Given, 2x â€“ 1 = 14 â€“ x**

*Sol.*Transposing x from RHs to LHS and 1 from LHS to RHS, we get,

2x + x = 14 + 1

3x = 15

Dividing by 3 on both the sides, we get,

x = 5

Verification:

LHS = 2x - 1 = 2 x (5) - 1 = 10 - 1= 9

RHS = 14 - x = 14 - 5 = 9

Here, LHS = RHS

Thus, the result obtained is correct.

**Q.6 8x + 4 = 3 (x â€“ 1) + 7
**

**Given, 8x + 4 = 3 (x â€“ 1) + 7**

*Sol.*8x + 4 = 3x â€“ 3 + 7

Transposing 3x from RHS to LHS and 4 from LHS to RHS, we get,

8x â€“ 3x = - 3 + 7 â€“ 4

5x = - 7 + 7

x = 0

Verification:

LHS = 8x + 4 = 8 x (0) + 4 = 4

RHS = 3(x â€“ 1) + 7 = 3(0 â€“ 1) + 7 = - 3 + 7 = 4

Here, LHS = RHS

Thus, the result obtained is correct.

**Q.7 **

** Sol.** Given,

Multiplying by 5 on both the sides, we get,

5x = 4(x + 10)

5x = 4x + 40

Transposing 4x from RHS to LHS, we get,

5x â€“ 4x = 40

x = 40

Verification:

LHS = x = 40

RHS == = = 40

Here, LHS = RHS

Thus, the result obtained is correct.

**Q.8 **

** Sol.** Given,

Transposing from RHS to LHS and 1 from LHS to RHS, we get,

Multiplying by 5 on both the sides, we get,

*x*= 10

Verification:

LHS = = =

RHS = = = =

=

Here, LHS = RHS

Thus, the result obtained is correct.

**Q.9 **

** Sol.** Given,

Dividing by 3 on both the sides, we get,

Verification:

LHS = = = =

RHS = = =

Here, LHS = RHS

Thus, the result obtained is correct.

**Q.10 **

** Sol.** Given,

Transposing 5

*m*from RHS to LHS, we get,

Dividing by -2 on both the sides, we get,

Verification:

LHS =

RHS =

Here, LHS = RHS

Thus, the result obtained is correct.