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Ncert Solutions for Linear equations Class 8 CBSE Part 5


In this page we have NCERT book Solutions for Class 8th Maths:Linear equations 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 

Answer

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 

Answer

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 

Answer

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 

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)

 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

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 

Answer

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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