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 = $$\frac{13}{6}$$