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

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

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