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 = \(\frac{-8}{5}\)