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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
Answer
L.C.M. of the denominators, 2, 3, 4, and 5, is 60.
So Multiplying both sides by 60, we obtain
30
x − 12 = 20
x + 15
Transposing 20x to R.H.S and 12 to L.H.S, we obtain
30
x − 20
x = 15 + 12
10
x = 27
X=27/10=2.7
Question 2 Solve the linear equation
Answer
L.C.M. of the denominators, 2, 4, and 6, is 12.
Multiplying both sides by 12, we obtain
6
n − 9
n + 10
n = 252
7
n = 252
Dividing by 7 on both the sides
n=36
Question 3- Solve the linear equation
Answer
L.C.M. of the denominators, 2, 3, and 6, is 6.
Multiplying both sides by 6, we obtain
6
x + 42 − 16
x = 17 − 15
x
Transposing 15x to RHS and 42 to LHS
6
x − 16
x + 15
x = 17 − 42
5
x = −25
Dividing by 5 on both the sides
x=-5
Question 4- Solve the linear equation
Answer
L.C.M. of the denominators, 3 and 5, is 15.
Multiplying both sides by 15, we obtain
5(
x − 5) = 3(
x − 3)
5
x − 25 = 3
x − 9
Transposing 3x to RHS and 25 to LHS
5
x − 3
x = 25 − 9
2
x = 16
Dividing both the sides by 2
x=8
Question 5- Solve the linear equation
Answer
L.C.M. of the denominators, 3 and 4, is 12.
Multiplying both sides by 12, we obtain
3(3
t − 2) − 4(2
t + 3) = 8 − 12
t
9
t − 6 − 8
t − 12 = 8 − 12
t
Transposing 12t to RHS and 6 and 12 to LHS
9
t − 8
t + 12
t = 8 + 6 + 12
13
t = 26
Dividing by 13 on both the sides
t=2
Question 6 Solve the linear equation
Answer
L.C.M. of the denominators, 2 and 3, is 6.
Multiplying both sides by 6, we obtain
6
m − 3(
m − 1) = 6 − 2(
m − 2)
6
m − 3
m + 3 = 6 − 2
m + 4
Transposing 2m to RHS and 3 to LHS
6
m − 3
m + 2
m = 6 + 4 − 3
5
m = 7
Dividing by 5 on both the sides
m=7/5
Question 7- Simplify and solve the linear equation
3(
t − 3) = 5(2
t + 1)
Answer
3(
t − 3) = 5(2
t + 1)
3
t − 9 = 10
t + 5
−9 − 5 = 10
t − 3
t
−14 = 7t
t=-2
Question 8- Simplify and solve the linear equation
15(
y − 4) − 2(
y − 9) + 5(
y + 6) = 0
Answer
15(
y − 4) − 2(
y − 9) + 5(
y + 6) = 0
15
y − 60 − 2
y + 18 + 5
y + 30 = 0
18
y − 12 = 0
18
y = 12
y=12/18=2/3
Question 9- Simplify and solve the linear equation
3(5
z − 7) − 2(9
z − 11) = 4(8
z − 13)−17
Answer
3(5
z − 7) − 2(9
z − 11) = 4(8
z − 13)−17
15
z − 21 − 18
z + 22 = 32
z − 52 − 17
−3
z + 1 = 32
z − 69
−3
z − 32
z = −69 − 1
−35
z = −70
z=2
Question 10- Simplify and solve the linear equation
0.25(4
f − 3) = 0.05(10
f − 9)
Answer
0.25(4
f − 3) = 0.05(10
f − 9)
Multiplying both sides by 20, we obtain
5(4
f − 3) = 10
f − 9
20
f − 15 = 10
f − 9
20
f − 10
f = − 9 + 15
10
f = 6
f==6/10=3/5
Download NCERT Solutions Exercise 2.5 as pdf
Class 8 Maths
Class 8 Science
