# NCERT Solutions for class 8 Mathematics/Linear Equations in one Variable

CBSE CLASS 8 MATHEMATICS
LINEAR EQUATIONS IN ONE VARIABLE – Chapter 2

Exercise 2.1

1. Solve x – 2 = 7
2. Solve y + 3 = 10
3. Solve 6 = z + 2
4. Solve 3/7 + x = 17/7
5. Solve 6x = 12
6. Solve t/5 = 10
7. Solve 2x/3 = 18
8. Solve 1.6 = y/1.5
9. Solve 7x – 9 = 16
10. Solve 14y – 8 = 13
11. Solve 17 + 6p = 9
12. Solve x/3 + 1 = 7/15

1. x -2 =7
Add 2 on both sides, we get x – 2 + 2 = 7 + 2
Therefore, x = 9

2. y + 3 = 10
Subtract 3 from both sides, we get y + 3 – 3 = 10 – 3
y = 7

3. 6 = z + 2
Subtract 2 from both sides, we get 6 – 2 = z + 2 – 2
Therefore, 4 = z
z = 4.

4. 3/7 + x = 17/7
Subtract 3/7 from both sides, 3/7 + x – 3/7 = 17/7 – 3/7
Therefore, x = 14/7= 2.

5. 6x = 12
Divide 6 on both sides, we get 6x/6 = 12/6
Therefore, x = 2

6. t/5 = 10
Multiply by 5 on both sides, we get t = 10 x 5
Therefore, t = 50.

7. 2x /3 = 18
Multiply by 3 on both sides, we get 2x = 18 x 3
2x = 54
x = 27.

8. 1.6 = y/1.5
Multiply by1.5 on both sides, we get 1.6 x 1.5 = y.
Therefore, y = 2.40

9. 7x – 9 = 16
Add 9 on both sides, we get 7x – 9 + 9 = 16 + 9
7x = 25
Divide 7 on both sides, we get x = 25/7.

10. 14y – 8 = 13
Add 8 on both sides, we get 14y – 8 + 8 = 13 + 8
14y = 21
Divide 14 on both sides, we get y = 21/14= 3/2 (since 7 is a common factor).

11. 17 + 6p = 9
Subtract 17 on both sides, we get 17 + 6p – 17 = 9 – 17
Therefore, 6p = -8.
Divide 6 on both sides, we get p = -8/6 = -4/3 (since 2 is a common factor).

12. x/3 + 1 = 7/15
Subtract I from both sides, we get x/3 = 7/15 – 1
x / 3 = (7- 15) /15 = -8/15
Multiply 3 on both sides, we get x = -8/5.