Gujarat BoardĀ GSEB Textbook Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3

Question 1.

Solve the following equations and check your results.

(i) 3x = 2x + 18

(ii) 5t – 3 = 3t – 5

(iii) 5x + 9 = 5 + 3x

(iv) 4z + 3 = 6 + 2z

(v) 2x – 1 = 14 – x

(vi) 8x + 4 = 3(x – 1) + 7

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

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

(ix) 2y + \(\frac { 5 }{ 3 }\) = \(\frac { 26 }{ 3 }\) – y

(x) 3m = 5m – \(\frac { 8 }{ 5 }\)

Solution:

(i) 3x = 2x + 18

Transposing 2x from RHS to LHS, we have

3x – 2x = 18 or x = 18

Check: Put x = 18 in LHS and RHS of the equation

LHS=3x

= 3 x 18 = 54

RHS = 2x + 18

= 2 x 18 + 18

= 36 + 18 = 54

Thus, LHS = RHS

(ii) 5t – 3 = 3t – 5

Transposing (-3) to RHS, we have

5t = 3t – 5 + 3 or 5t = 3t – 2

Transposing 31 to LHS, we have

5t – 3t = -2 or 21 = – 2

Diving both sides by 2, we have \(\frac { 2t }{ 2 }\) = \(\frac { -2 }{ 2 }\) or t = -1

Check: Put t = -1 in LHS and RHS of the equation

LHS = 5(-1) -3 = -5 -3 = -8

RHS = 3(-1) -5 = -3 -5 = -8

Thus, LHS = RHS

(iii) 5x + 9 = 5 + 3x

Transposing 9 to RHS, we have

5x = 5 + 3x -9 or 5x = -4 + 3x

Transposing 3x to LHS, we have

5x – 3x = -4 or 2x = -4

Dividing both sides by 2, we have

x = \(\frac { -4 }{ 2 }\) = \(\frac { 1 }{ 2 }\) – 2 or x = -2

Check: Put x = -2 in LHS and RHS of the equation

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

LHS = 5 + 3(-2) = 5-6 = -1

Thus, LHS = RHS

(iv) 4z + 3 = 6 + 2z

Transposing 3 to RHS, we have

4z = 6 – 3+ 2z or 4z = 3 + 2z

Transposing 2z to LHS, we have

4z – 2z = 3 or 2z = 3

Dividing both sides by 2, we have z = \(\frac { 3 }{ 2 }\)

Check: Put z = \(\frac { 3 }{ 2 }\) in LHS and RHS of the equation

LHS= 4Ć\(\frac { 3 }{ 2 }\)+3=6+3=9

RHS=6+2Ć\(\frac { 1 }{ 2 }\)=6+3=9 2

Thus, LHS = RHS

(v) 2x – 1 = 14 – x

Transposing (-1) to RHS, we have

2x = 14-x + 1 or 2x = 15 – x

Transposing (-x) to LHS, we have

2x + x = 15 or 3x = 15

Dividing both sides by 3, we have

x = \(\frac { 15 }{ 3 }\) = 5

ā“ x = 5

Check: Put x = 5 in LHS and RHS of the equation

LHS = 2 Ć 5 – 1 = 10 – 1 = 9

RHS = 14 – 5 = 9

Thus, LHS = RHS

(vi) 8x + 4 = 3(x – 1) + 7

or 8x + 4 = 3x-3 + 7

or 8r + 4 = 3x + 4

Transposing 4 to RHS, we have

8x = 3x + 4 – 4

Transposing 3x to LHS, we have

8x – 3x = 0 or 5x =0

ā“ x = 0

Check: Put x = 0 in LHS and RHS of the equation

LHS = 8X0 + 4 = 0 + 4 = 4

RHS = 3(0 – 1) + 7 = -3 + 7 = 4

Thus, LHS = RHS

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

Transposing \(\frac { 4 }{ 5 }\) x to LHS, we have

x – \(\frac { 4 }{ 5 }\) x = 8 or \(\frac { 5x – 4x }{ 5 }\) = 8 or \(\frac { x }{ 5 }\) = 8

Multiplying both sides by 5, we have

x = 8 x 5 = 40

ā“ x = 40

Check: Put x = 40 in LHS and RHS of the equation

LHS = 40

RHS = \(\frac { 4 }{ 5 }\)(40+10) = \(\frac { 4 }{ 5 }\) Ć 50 = 4 Ć 10 = 40

Thus, LHS = RHS

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

Transposing 1 to RHS, we have

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

Transposing \(\frac { 7x }{ 15 }\) to LHS, we have

\(\frac { 2x }{ 3 }\) – \(\frac { 7x }{ 15 }\) = 2 or \(\frac { 10x – 7x }{ 15 }\) = 2 or or \(\frac { 3x }{ 15 }\) = 2

Multiplying both sides by 15, we have 3x = 2 x 15 = 30

Dividing both sides by 3, we have

\(\frac { 3x }{ 3 }\) = \(\frac { 30 }{ 3 }\)

ā“x = 10

Check: Put x = 10 in LHS and RHS of the equation

Thus, LHS = RHS

(ix) 2y + \(\frac { 5 }{ 3 }\) = \(\frac { 26 }{ 3 }\) – y

(x) 3m = 5m – \(\frac { 8 }{ 5 }\)

Transposing 5m to LHS, we have

3m – 5m = –\(\frac { 8 }{ 5 }\) or -2m = –\(\frac { 8 }{ 5 }\)

Dividing both sides by (-2), we have

Thus, LHS = RHS