## MP Board Class 8th Maths Solutions 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:

which is the required solution.

Question 2.

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

Solution:

⇒ n = 6, which is the required solution.

Question 3.

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

Solution:

⇒ x = -5, which is the required solution.

Question 4.

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

Solution:

⇒ x = 8, which is the required solution.

Question 5.

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

Solution:

⇒ t = 2, which is the required solution.

Question 6.

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

Solution:

Simplify and solve the following linear equations.

Question 7.

3(t – 3) = 5(2t + 1)

Solution:

We have, 3(t – 3) = 5(2t + 1)

⇒ 31 – 9 = 101 + 5

⇒ – 9 – 5 = 101 – 31 ⇒ -14 = 71

⇒ 1 = – 2, which is the required solution.

Question 8.

15(y – 4) – 2(y – 9) + 5(y + 6) = 0

Solution:

We have,

15(y – 4) – 2(y – 9) + 5(y + 6) = 0

⇒ 15 y – 60 – 2y + 18 + 5y + 30 = 0

⇒ 18y – 12 = 0 ⇒ 18y = 12

⇒ y = \(\frac{2}{3}\), which is the required solution.

Question 9.

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

Solution:

We have,

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

⇒ 15z – 21 – 18z + 22 = 32z – 52 – 17

⇒ 15z – 18z – 32z = 21 – 22 – 52 – 17 – 70

⇒ -35z = -70 ⇒ z = \(\frac{-70}{-35}\)

⇒ z = 2, which is the required solution.

Question 10.

0.25 (4f- 3) = 0.05(10f – 9)

Solution:

We have, 0.25(4f – 3) = 0.05(10f – 9)

⇒ f – 0.75 = 0.5f – 0.45

⇒ f = 0.6, which is the required solution.