Linear Equations in One Variable Exercise 2.3

 JKBOSE  Solutions for Class 8TH Maths

Linear Equations in One Variable Exercise 2.3

Solve the following equations and check your results.

Ex 2.3 Class 8th Maths

 Q1.3x = 2x + 18

Sol: 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

Q2. 5 t – 3 = 3t – 5

Sol: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

Q3.5x + 9 = 5 + 3x
Sol: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. 

Q4.4z + 3 = 6 + 2z

Sol: 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 = 32
Hence z = 32 is the required solution.
Check: 4z + 3 = 6 + 2z
Putting z = 32, we have
LHS = 4z + 3 = 4 × 32 + 3 = 6 + 3 = 9
RHS = 6 + 2z = 6 + 2 × 32 = 6 + 3 = 9
LHS = RHS.