# NCERT Solutions for Linear equation in One Variable Class 8 Maths Chapter 2 Exercise 2.5

In this page we have NCERT book Solutions for Class 8th Maths:Linear equations Chapter 2 for Exercise 2.5 . Hope you like them and do not forget to like , social share and comment at the end of the page.
Question 1 Solve the linear equation

L.C.M. of the denominators, 2, 3, 4, and 5, is 60.
So Multiplying both sides by 60, we obtain
30− 12 = 20+ 15
Transposing 20x  to R.H.S and 12 to L.H.S, we obtain
30− 20x = 15 + 12
10x = 27
X=27/10=2.7 Question 2 Solve the linear equation

L.C.M. of the denominators, 2, 4, and 6, is 12.
Multiplying both sides by 12, we obtain
6n − 9n + 10n = 252
7n = 252
Dividing by 7 on both the sides
n=36
Question 3- Solve the linear equation

L.C.M. of the denominators, 2, 3, and 6, is 6.
Multiplying both sides by 6, we obtain
6x + 42 − 16= 17 − 15x
Transposing 15x to RHS and 42 to LHS
6x − 16+ 15= 17 − 42
5x = −25
Dividing by 5 on both the sides
x=-5
Question 4- Solve the linear equation

L.C.M. of the denominators, 3 and 5, is 15.
Multiplying both sides by 15, we obtain
5(x − 5) = 3(x − 3)
5x − 25 = 3x − 9
Transposing 3x to RHS and 25 to LHS
5x − 3x = 25 − 9
2x = 16
Dividing both the sides by 2
x=8
Question 5- Solve the linear equation

L.C.M. of the denominators, 3 and 4, is 12.
Multiplying both sides by 12, we obtain
3(3t − 2) − 4(2t + 3) = 8 − 12t
9− 6 − 8t − 12 = 8 − 12t
Transposing 12t to RHS and 6 and 12 to LHS
9− 8t + 12t = 8 + 6 + 12
13t = 26
Dividing by 13 on both the sides
t=2
Question 6  Solve the linear equation

L.C.M. of the denominators, 2 and 3, is 6.
Multiplying both sides by 6, we obtain
6m − 3(m − 1) = 6 − 2(m − 2)
6m − 3m + 3 = 6 − 2m + 4
Transposing 2m to RHS and 3 to LHS
6m − 3m + 2m = 6 + 4 − 3
5m = 7
Dividing by 5 on both the sides
m=7/5

Question 7-   Simplify and solve the linear equation 3(t − 3) = 5(2t + 1)
3(t − 3) = 5(2t + 1)
3t − 9 = 10t + 5
−9 − 5 = 10t − 3t
−14 = 7t
t=-2
Question 8- Simplify and solve the linear equation 15(y − 4) − 2(y − 9) + 5(y + 6) = 0
15(y − 4) − 2(y − 9) + 5(y + 6) = 0
15− 60 − 2y + 18 + 5y + 30 = 0
18− 12 = 0
18= 12 y=12/18=2/3
Question 9- Simplify and solve the linear equation
3(5z − 7) − 2(9z − 11) = 4(8z − 13)−17
3(5z − 7) − 2(9z − 11) = 4(8z − 13)−17
15z − 21 − 18z + 22 = 32z − 52 − 17
−3z + 1 = 32z − 69
−3z − 32z = −69 − 1
−35z = −70
z=2 Question 10- Simplify and solve the linear equation 0.25(4f − 3) = 0.05(10f − 9)