NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 are the part of NCERT Solutions for Class 8 Maths. Here you can find the NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5.

Solve the following linear equations.

Ex 2.5 Class 8 Maths Question 1.

Solution:
30x – 12 = 20x + 15
30x – 20x = 15 + 12                  (Transposing 20x to LHS and 12 to RHS)
10x = 27
x = 27/10

Thus, x = 27/10 is the required solution of the given linear equation.

Ex 2.5 Class 8 Maths Question 2.

Solution:
LCM of 2, 4 and 6 = 12
(Multiplying both sides by 12)
6n – 9n + 10n = 252
7n = 252
n = 252 ÷ 7
n = 36

Thus, n = 36 is the required solution of the given linear equation.

Ex 2.5 Class 8 Maths Question 3.

Solution:
-10x + 42 = 17 – 15x
-10x + 15x = 17 – 42         [Transposing 15x to LHS and 42 to RHS]
5x = -25
x = -25 ÷ 5                         [Transposing 5 to RHS]
x = -5

Thus, x = -5 is the required solution of the given linear equation.

Ex 2.5 Class 8 Maths Question 4.

Solution:

(x – 5) × 5 = (x – 3) × 3
5x – 25 = 3x – 9            (Solving the brackets)
5x – 3x = 25 – 9            (Transposing 3x to LHS and 25 to RHS)
2x = 16
x = 16 ÷ 2                      (Transposing 2 to RHS)
x = 8

Thus, x = 8 is the required solution of the given linear equation.

Ex 2.5 Class 8 Maths Question 5.

Solution:
(3t – 2) × 3 – (2t + 3) × 4 = 2 × 4 – 12t
9t – 6 – 8t – 12 = 8 – 12t         (Solving the brackets)
t – 18 = 8 – 12t
t + 12t = 8 + 18                         (Transposing 12t to LHS and 18 to RHS)
13t = 26

t = 26 ÷ 13                                (Transposing 13 to RHS)
t = 2

Hence, t = 2 is the required solution of the given linear equation.

Ex 2.5 Class 8 Maths Question 6.

Solution.
6m – (m – 1) × 3 = 6 – (m – 2) × 2
6m – 3m + 3 = 6 – 2m + 4            (Solving the brackets)
3m + 3 = 10 – 2m
3m + 2m = 10 – 3               (Transposing 2m to LHS and 3 to RHS)
5m = 7
m = 7/5                                (Transposing 5 to RHS)

Hence, m = 7/5 is the required solution of the given linear equation.

Simplify and solve the following linear equations.

Ex 2.5 Class 8 Maths Question 7.

3(t – 3) = 5(2t + 1)

Solution:
Given: 3(t – 3) = 5(2t + 1)
3t – 9 = 10t + 5          (Solving the brackets)
3t – 10t = 9 + 5          (Transposing 10t to LHS and 9 to RHS)
-7t = 14
t = -2                          (Transposing -7 to RHS)
Hence, t = -2 is the required solution of the given linear equation.

Ex 2.5 Class 8 Maths Question 8.

15(y – 4) – 2(y – 9) + 5(y + 6) = 0

Solution:
Given: 15(y – 4) – 2(y – 9) + 5(y + 6) = 0
15y – 60 – 2y + 18 + 5y + 30 = 0           (Solving the brackets)
18y – 12 = 0
18y = 12                           (Transposing 12 to RHS)
y = 2/3
Hence, y = 2/3 is the required solution of the given linear equation.

Ex 2.5 Class 8 Maths Question 9.

3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

Solution:
Given: 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
15z – 21 – 18z + 22 = 32z – 52 – 17             (Solving the bracket)
-3z + 1 = 32z – 69
-3z – 32z = – 69 – 1                (Transposing 32z to LHS and 1 to RHS)
-35z = -70
z = 2
Hence, z = 2 is the required solution of the given linear equation.

Ex 2.5 Class 8 Maths Question 10.

0.25(4f – 3) = 0.05(10f – 9)

Solution:
Given: 0.25(4f – 3) = 0.05(10f – 9)
0.25 × 4f – 3 × 0.25 = 0.05 × 10f – 9 × 0.05    (Solving the brackets)
1.00f – 0.75 = 0.5f – 0.45
f – 0.5f = -0.45 + 0.75           (Transposing 0.5f to LHS and 0.75 to RHS)
0.5f = 0.30
f = 0.6
Hence, f = 0.6 is the required solution of the given linear equation.

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