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

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