Haryana State Board HBSE 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.5 Textbook Exercise Questions and Answers.

## Haryana Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Exercise 2.5

Solve the following linear equations

Question 1.

\(\frac{x}{2}-\frac{1}{5}\) = \(\frac{x}{3}+\frac{1}{4}\)

Solution:

Question 2.

\(\frac{n}{2}\) – \(\frac{3n}{4}\) + \(\frac{5n}{6}\) = 21

Solution:

Question 3.

x + 7 – \(\frac{8x}{3}\) = \(\frac{17}{6}\) – \(\frac{5x}{2}\)

Solution:

Question 4.

\(\frac{x-5}{3}\) = \(\frac{x-3}{5}\)

Solution:

\(\frac{x-5}{3}\) = \(\frac{x-3}{5}\)

or, 5(x – 5) = 3 (x – 3)

or, 5x – 25 = 3x – 9

or, 5x – 3x = -9 + 25

or, 2x = 16

or, x = \(\frac{16}{2}\)

∴ x = 8

Question 5.

\(\frac{3t-2}{4}\) – \(\frac{2t-3}{3}\) = \(\frac{2}{3}\) – t

Solution:

or, 3 (13t – 18) = 24

or, 39t – 54 = 24

or, 39t = 24 + 54 = 78

or, 39t = 78

or, t = 2

Question 6.

m – \(\frac{m-1}{2}\) = 1 – \(\frac{m-2}{3}\)

Solution:

or, 5m – 1 = 6

or, 5m = 6 + 1

or, 5m = 7

∴ m = \(\frac{7}{5}\)

Simplify and solve the following linear equations:

Question 7.

3 (t – 3) = 5 (2t + 1)

Solution:

3 (t – 3) = 5 (2t + 1)

or, 3t – 9 = 10t + 5

or, 3t – 10t = 5 + 9

or, -7t = 14

or, t = \(-\frac{14}{7}\) =-2

t = -2

Question 8.

15(y – 4) – 2(y – 9) + 5(y + 6) = 0

Solution:

15(y – 4) – 2(y – 9) + 5(y + 6) = 0

or, 15y – 60 – 2y + 18 + 5y + 30 = 0

or, 20y – 2y – 60 + 48 = 0

or, 18y – 12 = 0

or, 18y = 12

or, y = \(\frac{12}{18}\) = \(\frac{2}{3}\)

∴ y = \(\frac{2}{3}\)

Question 9.

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

Solution:

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

or, 15z – 21 – 18z + 22 = 32z – 52 – 17

or, 15z – 18z – 32z = -52 – 17 + 21 – 22

or, 15z – 50z = 21 – 91

or, -35z = -70

or, 35z = 70

or, z = 2

∴ z = 2

Question 10.

0.25 (4f – 3) = 0.05 (10f – 9)

Solution:

0.25 (4f – 3) = 0.05 (10f – 9)

or, f – 0.75 = 0.5f – 0.45

or, f – 0.5f = -0.45 + 0.75

or, 0.5f = 0.3

f = \(\frac{0.3}{0.5}\) = \(\frac{3}{5}\) = 0.6

∴ f = 0.6