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

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 2 Linear Equations in One Variable Ex 2.1 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.1

Solve the following equations.

Question 1.
x – 2 = 7
Solution:
x – 2 = 7
∴ x = 7 + 2 (Transposing – 2 to RHS)
∴ x = 9

Question 2.
y + 3 = 10
Solution:
y + 3 = 10
∴ y = 10 – 3 (Transposing 3 to RHS)
∴ y = 7

Question 3.
6 = z + 2
Solution:
6 = z + 2
∴ z + 2 = 6 (Interchanging both the sides)
∴ z = 6 – 2 (Transposing 2 to RHS)
∴ z = 4.

Question 4.
$$\frac {3}{7}$$ + x = $$\frac {17}{7}$$
Solution:
$$\frac {3}{7}$$ + x = $$\frac {17}{7}$$
∴ x = $$\frac{17}{7}-\frac{3}{7}$$ (Transposing $$\frac {3}{7}$$ to RHS)
∴ x = $$\frac{17-3}{7}$$
∴ x = $$\frac {14}{7}$$
∴ x = 2

Question 5.
6x = 12
Solution:
6x = 12
∴ $$\frac{6 x}{6}=\frac{12}{6}$$ (Dividing both the sides by 6)
∴ x = 2

Question 6.
$$\frac{t}{5}$$ = 10
Solution:
$$\frac{t}{5}$$ = 10
∴ $$\frac{t}{5}$$ × 5 = 10 × 5 (Multiplying both the sides by 5)
∴ t = 50

Question 7.
$$\frac{2 x}{3}$$ = 15
Solution:
$$\frac{2 x}{3}$$ = 15
∴ $$\frac{2 x}{3} \times \frac{3}{2}=18 \times \frac{3}{2}$$ (Multiplying both the sides by $$\frac {3}{2}$$)

Question 8.
1.6 = $$\frac{y}{1.5}$$
Solution:
1.6 = $$\frac{y}{1.5}$$
∴ 1.6 × 1.5 = $$\frac{y}{1.5}$$ × 1.5 (Multiplying both the sides by 1.5)
∴ 2.4 = y (∵ 1.6 × 1.5 = 2.4)
∴ y = 2.4

Question 9.
7x – 9 = 16
Solution:
7x – 9 = 16
∴ 7x = 16 + 9 (Transposing – 9 to RHS)
∴ 7x = 25
∴ $$\frac{7 x}{7}=\frac{25}{7}$$ (Dividing both the sides by 7)
∴ x = $$\frac {25}{7}$$

Question 10.
14y – 8 = 13
Solution:
14y – 8 = 13
∴ 14y = 13 + 8 (Transposing – 8 to RHS)
∴ 14y = 21
∴ $$\frac{14 y}{14}=\frac{21}{14}$$ (Dividing both the sides by 14)
∴ y = $$\frac{7 \times 3}{7 \times 2}$$
∴ y = $$\frac {3}{2}$$

Question 11.
17 + 16p = 9
Solution:
17 + 16p = 9
∴ 6p = 9 – 17 (Transposing 17 to RHS)
∴ 6p = -8
∴ $$\frac{6 p}{6}=\frac{-8}{6}$$ (Dividing both the sides by 6)
∴ p = $$\frac{-4 \times 2}{3 \times 2}$$
∴ p = –$$\frac {4}{3}$$

Question 12.
$$\frac{x}{3}+1=\frac{7}{15}$$
Solution:
$$\frac{x}{3}+1=\frac{7}{15}$$
∴ $$\frac{x}{3}=\frac{7}{15}-1$$ (Transposing 1 to RHS)
∴ $$\frac{7-15}{15}$$ (LCM = 15)
∴ $$\frac{x}{3}=\frac{-8}{15}$$
∴ $$\frac{x}{3} \times 3=\frac{-8}{15} \times 3$$ (Multiplying both the sides by 3)
∴ x = –$$\frac {8}{5}$$