# HBSE 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.5

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