Haryana State Board HBSE 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.1 Textbook Exercise Questions and Answers.
Haryana Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Exercise 2.1
Question 1.
Solve the following equations:
1. x – 2 = 7
2. y + 3 = 10
3. 6 = z + 2
4. \(\frac{3}{7}\) + x = \(\frac{17}{7}\)
5. 6x = 12
7. \(\frac{2x}{3}\) = 18
8. 1.6 = \(\frac{y}{1.5}\)
9. 7x – 9 = 16
10. 14y – 8 = 13
11. 17 + 6P = 9
12. \(\frac{x}{3}\) + 1 = \(\frac{7}{15}\)
Solution:
1. x – 2 = 7
or x – 2 + 2 = 7 + 2 (adding 2 both sides)
∴ x = 9
2. y + 3 = 10
y + 3 – 3 = 10 – 3 (subtract 3 from both sides)
∴ y = 7
3. 6 = x + 2
6 – 2 = z + 2 – 2 (subtract 2 from both sides)
4 = z
∴ z = 4
4. \(\frac{3}{7}\) + x = \(\frac{17}{7}\)
x = \(\frac{17}{7}\) – \(\frac{3}{7}\) = \(\frac{14}{7}\) = 2
∴ x = 2
5. 6x = 12
or \(\frac{6x}{6}\) = \(\frac{12}{6}\) (divide both side by 6)
∴ x = 2
6. \(\frac{t}{5}\) = 10
or \(\frac{t}{5}\) × 5 = 10 × 5 (multiply both sides by 5)
∴ t = 50
7. \(\frac{2x}{3}\) = 18
or 2x = 18 × 3
∴ x = \(\frac{18 \times 3}{2}\) = 9 × 3 = 27
∴ x = 27
8. 1.6 = \(\frac{y}{1.5}\)
or y = 1.6 × 1.5 = 2.40
∴ y = 2.4
9. 7x – 9 = 16
or 7x – 9 + 9 = 16 + 9 (adding 9 both sides)
or 7x = 25
∴ x = \(\frac{25}{7}\)
10. 14y – 8 = 13
14y – 8 + 8 = 13 + 8 (adding 8 both sides)
or 14y = 21
or y = \(\frac{21}{14}\) = \(\frac{3}{2}\)
∴ y = \(\frac{3}{2}\)
11. 17 + 6P = 9
17 + 6P – 17 = 9 – 17 (subtract -17 from both sides)
or 6P = -8
or P = \(\frac{-8}{6}\) = \(\frac{-4}{3}\)
∴ P = \(\frac{-4}{3}\)
12. \(\frac{x}{3}\) + 1 = \(\frac{7}{15}\)
or \(\frac{x}{3}\) + 1 – 1 = \(\frac{7}{15}\) – 1 (subtract 1 from both sides)
or \(\frac{x}{3}\) = \(\frac{7-15}{15}\)
or \(\frac{x}{3}\) = \(\frac{-8}{15}\)
∴ x = \(\frac{-8}{15}\) × 3 = \(\frac{-8}{5}\)