AP State Syllabus AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.1 Textbook Questions and Answers.

## AP State Syllabus 8th Class Maths Solutions 2nd Lesson Linear Equations in One Variable Exercise 2.1

Question

Solve the following Simple Equations:

(i) 6m = 12

(ii) 14p -42

(iii) -5y = 30

(iv) – 2x = – 12

(v) 34x = – 51

(vi) \(\frac{n}{7}\) = -3

(vn) \(\frac{2x}{3}\) = 8

(vui) 3x+1 = 16

(ix) 3p – 7 = 0

(x) 13 – 6n = 7

(xi) 200y – 51 = 49

(xii) 11n + 1 = 1

(xiii) 7x – 9 = 16

(xiv) 8x + \(\frac{5}{2}\) =13

(xv) 4x – \(\frac{5}{3}\) = 9

(xvi) x – \(\frac{4}{3}\) = 3\(\frac{1}{2}\)

Solution:

i) 6m = 12 ⇒ m = \(\frac{12}{6}\) ⇒ m = 2

ii) 14p = – 42p ⇒ P = \(\frac{-42}{14}\)

∴ p = -3

iii) -5y = 30 ⇒ y = \(\frac{30}{-5}\) = -6

∴ y = -6

iv) -2x = -12

⇒ 2x = 12

x = \(\frac{30}{-5}\)

= 6

∴ x = 6

v) 34x = -51

⇒ \(\frac{-3}{2}\) = \(\frac{-3}{2}\)

∴ x = \(\frac{-3}{2}\)

vi) \(\frac{n}{7}\) = -3

⇒ n = -3 x 7 = -21

∴ n = -21

vii) \(\frac{2x}{3}\) = 18 ⇒ 18 x \(\frac{3}{2}\) = 27

∴ x = 27

viii) 3x + 1 = 16

3x = 16 – 1 = 15

3x = 15

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

∴ x = 5

ix) 3p – 7 = 0

⇒ 3p = 7

∴ p = \(\frac{7}{3}\)

x) 13 – 6n = 7 ⇒ -6n = 7 – 13

⇒ -6n = -6 ⇒ n= \(\frac{-6}{-6}\)

∴ n = 1

xi) 200y – 51 = 49

⇒ 200y = 49 + 51

⇒ 200y = 100

⇒ y = \(\frac{100}{200}\)

∴ y = \(\frac{1}{2}\)

xii) 11n + 1 = 1

⇒ 11n = 1 – 1

= 11n = 0

⇒ n = \(\frac{0}{11}\) = 0

∴ n = 0

xiii) 7x – 9 = 16

⇒ 7x = 16 + 9

⇒ 7x = 25

∴ x = \(\frac{25}{7}\)

xiv) 8x + \(\frac{5}{2}\) = 13

xv) 4x – \(\frac{5}{3}\)

xvi) x + \(\frac{4}{3}\) = 3\(\frac{1}{2}\)

⇒ \(x+\frac{4}{3}=\frac{7}{2}\)

⇒\(\frac{7}{2}-\frac{4}{3}=\frac{21-8}{6}\)

∴ x = \(\frac{13}{6}\)