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NCERT Solutions for Linear equation in One Variable Class 8 Maths Chapter 2 CBSE Part 5





In this page we have NCERT book Solutions for Class 8th Maths:Linear equations Chapter 2 for EXERCISE 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 NCERT Solutions for Linear equation in One Variable Class 8 Maths Chapter 2
Answer
NCERT Solutions for Linear equation in One Variable Class 8 Maths Chapter 2
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 NCERT Solutions for Linear equation in One Variable Class 8 Maths Chapter 2
Answer
NCERT Solutions for Linear equation in One Variable Class 8 Maths Chapter 2
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 NCERT Solutions for Linear equation in One Variable Class 8 Maths Chapter 2
Answer
NCERT Solutions for Linear equation in One Variable Class 8 Maths Chapter 2
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 NCERT Solutions for Linear equation in One Variable Class 8 Maths Chapter 2
Answer
NCERT Solutions for Linear equation in One Variable Class 8 Maths Chapter 2
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
Answer
NCERT Solutions for Linear equation in One Variable Class 8 Maths Chapter 2
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 NCERT Solutions for Linear equation in One Variable Class 8 Maths Chapter 2
Answer
NCERT Solutions for Linear equation in One Variable Class 8 Maths Chapter 2
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)
Answer
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
Answer  
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
Answer  
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)
Answer
 0.25(4f − 3) = 0.05(10f − 9)
Multiplying both sides by 20, we obtain
5(4f − 3) = 10f − 9
20− 15 = 10f − 9
20− 10f = − 9 + 15
10f = 6
 f==6/10=3/5

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