# Exercise 2.1

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

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

Substituting x in equation: 3 x = 2 x + 18 ⟹ 3( ) = 2( ) + 18.

**So, LHS** to RHS.

**2.**5 t − 3 = 3 t − 5

⟹ 2t = − 2 ⟹ t =− 2 2 ⟹ t =

Substituting t in equation: 5 t − 3 = 3 t − 5 ⟹ 5( ) - 3 = 3( ) - 5

⟹ -5-3 ⟹ -3-5 ⟹ =

**So, LHS** to RHS.

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

Substracting 3x in both sides: 5 x − 3 x + 9 = 5 + 3 x − 3 x ⟹ 2 x + 9 = 5

Subtracting 9 from both sides: 2 x + 9 − 9 = 5 − 9 ⟹ 2x =

Dividing both sides by 2 to solve: 2 x 2 = − 4 2 ⟹ x =

Substituting x in the equation: 5 ( ) + 9 = 5 + 3( )

**So, LHS** to RHS.

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

Subtracting 2z from both sides: 4 z − 2 z + 3 = 6 + 2 z − 2 z ⟹ 2 z + 3 = 6

Substracting 3 from both sides: 2 z + 3 − 3 = 6 − 3 ⟹ 2 z =

Dividing both sides by 2: 2 z 2 = 3 2 ⟹ z =

Substituting in equation: 4( ) + 3 = 6 + 2( )

6 + 3 = 6 + 3 ⇒ =

**So, LHS** to RHS.

**5.**2 x − 1 = 14 − x

Adding x to both sides: 2 x + x − 1 = 14 − x + x ⇒ 3x - =

Adding 1 to both sides: 3 x − 1 + 1 = 14 + 1 ⇒ 3x =

Dividing both sides by 3: 3 x 3 = 15 3 ⇒ x =

Substituting x in equation: 2 × - 1 = 14 − 5

⇒ 10 − 1 = 9 ⇒ =

**So, LHS** to RHS.

**6.**8 x + 4 = 3 x − 1 + 7

Distributing the 3 on the right side: 8 x + 4 = 3 x − 3 + 7

Combining like terms on the right side: 8 x + 4 = 3 x + 4

Subtracting 3x from both sides: 8 x − 3 x + 4 = 4 ⇒ 5 x + 4 =

Subtracting 4 from both sides: 5 x + 4 − 4 = 4 − 4 ⇒ 5x =

Dividing both sides by 5: 5 x 5 = 0 5 ⇒ x =

Substituting in equation: 8 × + 4 = 3 ( -1) + 7 ⇒ 4 = 3 − 1 + 7 ⇒ 4 = 3 − 1 + 7 ⇒ 4 = − 3 + 7 ⇒ = .

**So, LHS** to RHS.

**7. x =**4 x + 10 5

Expanding RHS: x = 4 x 5 + 8

Subtracting 4 x 5 from both sides: x − 4 x 5 = 8 ⇒ = 8

Multiplying both sides by 5: x = 5 × 8 ⇒ x =

Substituting x in equation: 40 = 4 5 × 40 + 10

40 = 4 5 × 50 ⇒ =

**So, LHS** to RHS.

**8.**2 x 3 + 1 = 7 x 15 + 3

Subtracting 1 from both sides: 2 x 3 + 1 - 1 = 7 x 15 + 3 - 1 ⇒ 2 x 3 = 7 x 15 + 2.

Subtracting 7 x 15 from both sides: 2 x 3 - 7 x 15 = 2

Finding common denominator: 10 x 15 - 7 x 15 = 2 ⇒ = 2.

Simplify the Equation: 3 x 15 = 2 ⇒ = 2.

Multiplying both sides by 5: x = 2 × 5 ⇒ x = .

Substituting x in equation: 2 × /3 + 1 = 7 × /15 + 3

⇒ 20 3 + 1 = 70 15 + 3

Convert 70 15 to a common denominator: 20 3 + 1 = 14 3 + 3 ⇒ 20 3 + 3 3 = 14 3 + 9 3 ⇒ / = /

**So, LHS** to RHS.

**9.**2 y + 5 3 = 26 3 − y

Simplifying the equation: 2 y + y + 5 3 = 26 3 ⇒ 3 y + 5 3 = 26 3 ⇒ 3y = /

3y = 21 3 ⇒ 3y = 7 ⇒ y =

Substituting y in equation: 2 7 3 + 5 3 = 26 3 -

⇒ 14 3 + 5 3 = 19 3 ⇒ =

**So, LHS** to RHS.

**10.**3 m = 5 m − 8 5

Subtracting 5m from both sides: 3 m − 5 m = 5 m − 5 m − 8 5 ⇒ − 2 m = − 8 5

Dividing both sides by − 2 : − 2 m − 2 = − 8 5 − 2 ⇒ m =

Substituting m in equation: 3 × 4 5 = 5 × 4 5 - 8 5

⇒ 12 5 = 20 5 - 8 5 ⇒ =

**So, LHS** to RHS.