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

NCERT Solutions Class 8 Mathematics 
Chapter – 2 (Linear Equations in One Variable) 

The NCERT Solutions in English Language for Class 8 Mathematics Chapter – 2 Linear Equations in One Variable Exercise 2.1 has been provided here to help the students in solving the questions from this exercise. 

Chapter 2: Linear Equations in One Variable

Exercise – 2.1 

Solve the following equations.

1. x – 2 = 7

Solution –
x – 2 = 7
x = 7 + 2
x = 9

2. y + 3 = 10

Solution –
y + 3 = 10
y = 10 –3
y = 7

3. 6 = z + 2

Solution –
6 = z + 2
z + 2 = 6
z = 6 – 2
z = 4

4. \mathbf{\frac{3}{7}} + x = \mathbf{\frac{17}{7}}

Solution –
\frac{3}{7} + x = \frac{17}{7}

x = \frac{17}{7} – \frac{3}{7}

x = \frac{14}{7}
x = 2

5. 6x = 12

Solution –
6x = 12
x = \frac{12}{6}
x = 2

6. \mathbf{\frac{t}{5}}= 10

Solution –
\frac{t}{5} = 10
t = 10 × 5
t = 50

7. \mathbf{\frac{2x}{3}} = 18

Solution –
\frac{2x}{3} = 18
2x = 18 × 3
2x = 54

x = \frac{54}{2}
x = 27

8. 1.6 = \mathbf{\frac{y}{15}}

Solution –
1.6 = \frac{y}{15}

\frac{y}{15} = 1.6
y = 1.6 × 1.5
y = 2.4

9. 7x – 9 = 16

Solution –
7x – 9 = 16
7x = 16+9
7x = 25
x = \frac{25}{7}

10. 14y – 8 = 13

Solution –
14y – 8 = 13
14y = 13+8
14y = 21
y = \frac{21}{14}

y = \frac{3}{2}

11. 17 + 6p = 9

Solution –
17 + 6p = 9
6p = 9 – 17
6p = -8
p = -\frac{8}{6}

p = -\frac{4}{3}

12. \mathbf{\frac{x}{3}} + 1 = \mathbf{\frac{7}{15}}

Solution –

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

\frac{x}{3} = \frac{7}{15} – 1

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

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

x = -\frac{8}{15} × 3

x = -\frac{8}{5}

 

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