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 = \(\frac{13}{6}\)