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 = –\(\frac {8}{5}\)