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

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

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

Solve the following linear equations:

Question 1.

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 = 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 = 0.30/0.50=30/50
⇒ f = 3/5
⇒ f = 0.6