# HBSE 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.1

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 = $$\frac{-8}{5}$$