You can Download KSEEB Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Questions and Answers helps you to revise the complete syllabus.

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

Solve the following linear equations:

Question 1.

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

Solution:

Question 2.

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

Solution:

Question 3.

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

Solution:

Question 4.

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

Solution:

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

⇒ 5(x – 5) = 3(x – 3)

⇒ 5x – 25 = 3x – 9

⇒ 5x – 3x = -9 + 25

⇒ 2x = 16

⇒ x = 8

Question 5.

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

Solution:

Question 6.

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

Solution:

⇒ 6m – 3m + 3 + 2m – 4 = 6

⇒ 5m – 1 = 6

⇒ 5m = 6 + 1

⇒ 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)

⇒ 3t – 9 = 10t + 5

⇒ -9 – 5 = 10t – 3t

⇒ -14 = 7t

⇒ 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

⇒ 15y – 60 – 2y + 18 + 5y + 30 = 0

⇒ 15y – 2y + 5y – 60 + 18 + 30 = 0

⇒ 18y – 12 = 0

⇒ 18y = 12

⇒ 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

⇒ 15z – 21 – 18z + 22 = 32z – 52 – 17

⇒ 15z – 18z – 21 + 22 = 32z – 69

⇒ -3z + 1 = 32z – 69

⇒ -3z – 32z = -69 – 1

⇒ -35z = -70

⇒ z = 2

Question 10.

0.25(4f – 3) = 0.05(10f – 9)

Solution:

0.25 (4f – 3) = 0.05 (10f – 9)

⇒ 1f – 0.75 = 0.50f – 0.45

⇒ f – 0.50f = -0.45 + 0.75

⇒ 0.50f = 0.30

⇒ f = \(\frac{0.30}{0.50}=\frac{30}{50}\)

⇒ f = \(\frac{3}{5}\)

⇒ f = 0.6