These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.1 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.1

Solve the following equations.

Question 1.

x – 2 = 7

Solution:

x – 2 = 7

Transposing (-2) to R.H.S., we get

x = 7 + 2

∴ x = 9

Question 2.

y + 3 = 10

Solution:

y + 3 = 10

Transposing 3 to R.H.S., we get

y = 10 – 3

∴ y = 7

Question 3.

6 = z + 2

Solution:

6 = z + 2

Transposing 2 to L.H.S., we get

6 – 2 = z

4 = z

∴ z = 4

Question 4.

\(\frac{3}{7}+x=\frac{17}{7}\)

Solution:

\(\frac{3}{7}+x=\frac{17}{7}\)

Transposing \(\frac{3}{7}\) to R.H.S., we get

x = \(\frac{17}{7}-\frac{3}{7}=\frac{17-3}{7}=\frac{14}{7}=2\)

∴ x = 2

Question 5.

6x = 12

Solution:

6x = 12

Divided by 6 on both sides, we get

\(\frac{6 x}{6}=\frac{12}{6}\)

∴ x = 2

Question 6.

\(\frac{t}{5}=10\)

Solution:

\(\frac{\mathrm{t}}{5}\) = 10

Multiplying both sides by 5, we get

\(\frac{\mathrm{t}}{5}\) × 5 = 10 × 5

∴ t = 50

Question 7.

\(\frac{2 x}{3}=18\)

Solution:

\(\frac{2 x}{3}=18\)

Multiplying both sides, by 3, we get

\(\frac{2 x}{3}\) × 3 = 18 × 3

2x = 18 × 3

x = \(\frac{18 \times 3}{2}\) = 9 × 3 = 27

∴ x = 27

Question 8.

1.6 = \(\frac{\mathrm{y}}{1.5}\)

Solution:

1.6 = \(\frac{\mathrm{y}}{1.5}\)

Multiplying both sides by 1.5, we get

1.6 × 1.5 = \(\frac{\mathrm{y}}{1.5}\) × 1.5

2.4 = y

∴ y = 2.4

Question 9.

7x – 9 = 16

Solution:

7x – 9 = 16

Transposing (-9) to R.H.S., we get

7x = 16 + 9

Dividing both sides by 7, we get

∴ x = \(\frac{25}{7}\)

Question 10.

14y – 8 = 13

Solution:

14y – 8 = 13

Transposing (-8) to R.H.S., we get

14y = 13 + 8

14y = 21

Dividing both sides by 14 we get,

\(\frac{14 y}{14}\) = \(\frac{21}{14}\)

∴ y = \(\frac{3}{2}\)

Question 11.

17 + 6p = 9

Solution:

17 + 6P = 9

Transposing 17 to RHS, we get

6P = 9 – 17

6P = -8

Dividing both sides by 6, we have

\(\frac{6 P}{6}=\frac{-8}{6}\)

∴ P = \(\frac{-4}{3}\)

Question 12.

\(\frac{x}{3}+1=\frac{7}{15}\)

Solution:

\(\frac{x}{3}+1=\frac{7}{15}\)

Transposing 1 to R.H.S, we get

\(\frac{x}{3}=\frac{7}{15}-1\)

\(\frac{x}{3}=\frac{7-15}{15}\)

\(\frac{x}{3}=\frac{-8}{15}\)

Multiplying both sides by 3, we have

\(\frac{x}{3} \times 3=\frac{-8}{15} \times 3\)

∴ x = \(\frac{-8}{5}\)