Gujarat BoardĀ 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

ā“ x = 9

(ii) y + 3 = 10

Transposing 3 to RHS, we have

y = 10 – 3

ā“ y = 7

(iii) 6 = z + 2

Transposing 2 to LHS, we have

6 – 2 = z

or 4 = z

ā“ 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

ā“ x = 2

(v) 6x = 12

Dividing both sides by 6, we have:

\(\frac { 6x }{ 6 }\) = \(\frac { 12 }{ 6 }\) or x = 2

ā“ x = 2

(vi) \(\frac { t }{ 5 }\) = 10

Multiplying both sides by 5, we have

\(\frac { t }{ 5 }\) Ć5 = 10 Ć5 or t = 50

ā“ t = 50

(vii) \(\frac { 2x }{ 3 }\) = 18

Multiplying both sides by 3, we have

(\(\frac { t }{ 5 }\)) Ć 3 = 18 Ć 3 or 2x = 54

Dividing both sides by 2, we have

\(\frac { 2x }{ 2 }\) = \(\frac { 54 }{ 2 }\)

or x = \(\frac { 54 }{ 2 }\) = 27

ā“ 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 \(\frac { 7x }{ 7 }\) = \(\frac { 25 }{ 7 }\)

ā“ 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 = \(\frac { 21 }{ 14 }\) = \(\frac { 3 }{ 4 }\)

ā“ 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 = \(\frac { -8 }{ 6 }\) = –\(\frac { 4 }{ 3 }\)

ā“ 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

\(\frac { x }{ 3 }\) Ć3 = \(\frac { -8 }{ 15 }\) Ć3 = – \(\frac { -8 }{ 5 }\)

ā“ x = \(\frac { -8 }{ 15 }\)