Gujarat Board GSEB Textbook Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.1 Textbook Questions and Answers.
Gujarat Board Textbook Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.1
Question 1.
Solve the following equations.
(i) x- 2 = 7
(ii) y + 3 = 10
(iii) 6 = z + 2
(iv) \frac { 3 }{ 7 } + = \frac { 17 }{ 7 }
(v) 6x = 12
(vi) \frac { t }{ 5 } = 10
(vii) \frac { 2x }{ 3 } = 18
(viii) 1.6 = \frac { y }{ 1.5 }
(ix) 7x – 9 = 16
(x) 14y – 8 = 13
(xi) 17 + 6p = 9
(xii) \frac { x }{ 3 } + 1 = \frac { 7 }{ 15 }
Solutions:
(i) x- 2 = 7
Transposing (-2) to RHS, we have
x = 7 + 2
∴ x = 9
(ii) y + 3 = 10
Transposing 3 to RHS, we have
y = 10 – 3
∴ y = 7
(iii) 6 = z + 2
Transposing 2 to LHS, we have
6 – 2 = z
or 4 = z
∴ z = 4
(iv) \frac { 3 }{ 7 } + = \frac { 17 }{ 7 }
Transposing \frac { 3 }{ 7 } to RHS, we have
x = \frac { 17 }{ 7 } – \frac { 3 }{ 7 } = \frac { 17 – 3 }{ 7 } = \frac { 14 }{ 7 } = 2
∴ x = 2
(v) 6x = 12
Dividing both sides by 6, we have:
\frac { 6x }{ 6 } = \frac { 12 }{ 6 } or x = 2
∴ x = 2
(vi) \frac { t }{ 5 } = 10
Multiplying both sides by 5, we have
\frac { t }{ 5 } ×5 = 10 ×5 or t = 50
∴ t = 50
(vii) \frac { 2x }{ 3 } = 18
Multiplying both sides by 3, we have
(\frac { t }{ 5 }) × 3 = 18 × 3 or 2x = 54
Dividing both sides by 2, we have
\frac { 2x }{ 2 } = \frac { 54 }{ 2 }
or x = \frac { 54 }{ 2 } = 27
∴ x = 27
(viii) 1.6 = \frac { y }{ 1.5 }
Multiplying both sides by 1.5, we have
(ix) 7x – 9 = 16
Transposing (-9) to RHS, we have
7x = 16 + 9 = 25
Dividing both sides by 7, we have \frac { 7x }{ 7 } = \frac { 25 }{ 7 }
∴ x = \frac { 25 }{ 7 }
(x) 14y – 8 = 13
Transposing -8 to RHS, we have
14y = 13 + 8 = 21
Dividing both sides by 14, we have
\frac { 14y }{ 14 } or y = \frac { 21 }{ 14 } = \frac { 3 }{ 4 }
∴ y = \frac { 3 }{ 2 }
(xi) 17 + 6p = 9
Transposing 17 to RHS, we have
6p = 9 – 17 = -8
Dividing both sides by 6, we have
\frac { 6p }{ 6 } = \frac { -8 }{ 6 } or p = \frac { -8 }{ 6 } = –\frac { 4 }{ 3 }
∴ p = \frac { 4 }{ 3 }
(xii) \frac { x }{ 3 } + 1 = \frac { 7 }{ 15 }
Transposing 1 to RHS, we have
\frac { x }{ 3 } = \frac { 7 }{ 15 } – 1
or \frac { x }{ 3 } = \frac { 7 – 15 }{ 15 } = \frac { -8 }{ 15 }
Multiplying both sides by 3, we have
\frac { x }{ 3 } ×3 = \frac { -8 }{ 15 } ×3 = – \frac { -8 }{ 5 }
∴ x = \frac { -8 }{ 15 }