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

• Last Updated : 17 Nov, 2020

### Question 1. Solve for x: x – 2 = 7

Solution:

x – 2 = 7

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x=7+2  (Adding two on both sides of equation)

x=9

### Question 2. Solve for y: y + 3 = 10

Solution:

y + 3 = 10

y = 10 –3  (Subtracting 3 from both sides of equation)

y = 7

### Question 3. Solve for z: 6 = z + 2

Solution:

6 = z + 2

z + 2 = 6 (Rearranging to standard form)

z = 6 – 2 (Subtracting two from both sides)

z = 4

### Question 4. Solve for x: 3/7 + x = 17/7

Solution:

3/7 + x = 17/7

x = 17/7 – 3/7 (Subtracting 3/7 on both sides)

x = 14/7 (Simplifying the equation)

x = 2

### Question 5. Solve for x: 6x = 12

Solution:

6x = 12

x = 12/6 (Dividing both sides by 6)

x = 2

### Question 6. Solve for t: t/5 = 10

Solution:

t/5 = 10

t = 10 × 5 (multiplying both sides by 5)

t = 50

### Question 7. Solve for x: 2x/3 = 18

Solution:

2x/3 = 18

2x = 18 × 3 (Multiplying both sides by 3)

2x = 54

x = 54/2 (Dividing both sides by 2)

x = 27

### Question 8. Solve for y: 1.6 = y/1.5

Solution:

1.6 = y/1.5

y/1.5 = 1.6 (Rearranging to standard form)

y = 1.6 × 1.5 (Multiplying both sides by 1.5)

y = 2.4

### Question 9. Solve for x: 7x – 9 = 16

Solution:

Given that 7x – 9 = 16

7x = 16+9 (Adding 9 on both sides of equation)

7x = 25

x = 25/7 (Dividing both sides by 7)

### Question 10. Solve for y: 14y – 8 = 13

Solution:

14y – 8 = 13

14y = 13+8 (Adding 8 on both sides of equation)

14y = 21

y = 21/14 (Dividing both sides by 14)

y = 3/2 (Simplifying the equation)

### Question 11. Solve for p: 17 + 6p = 9

Solution:

17 + 6p = 9

6p = 9 – 17 (Subtracting 17 from both sides of equation)

6p = -8

p = -8/6 (Dividing both sides by 6)

p = -4/3(Simplifying the equation)

### Question 12. Solve for x: x/3 + 1 = 7/15

Solution:

x/3 + 1 = 7/15

x/3 = 7/15 – 1 (Subtracting both sides by 1)

x/3 = (7 – 15)/15 (Making common denominator)

x/3 = -8/15

x = -8/15 × 3 (Multiplying both sides by 3)

x = -8/5

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