NCERT Solutions for class 8 Maths chapter 2 Linear Equation In One Variable

Exercise 2.1  Exercise 2.2  Exercise 2.3  Exercise 2.4  Exercise 2.5  Exercise 2.6

EXERCISE 2.1

Question 1 :

Solve : x – 2=7

sol : x – 2=7

On transposing 2 to R.H.S , we obtain 

x = 7 + 2

x = 9

 

Question 2 :

Solve : y+3=10

sol : 

y+3=10

On transposing 3 to R.H.S , we obtain

y= 10 – 3

y = 7

 

Question 3 :

Solve : 6 =z+2

sol : 

6 = z+2

On transposing 2 to L.H.S , we obtain

6-2 = z

z=4

Question 4 :

Solve : ​\dfrac{3}{7}+x=\dfrac{17}{7}

sol :

\dfrac{3}{7}+x=\dfrac{17}{7}

On transposing ​\dfrac{3}{7}​ to R.H.S , we obtain

x=\dfrac{17}{7}-\dfrac{3}{7}

=\dfrac{14}{7}

= 2

Question 5 :

Solve : 6x=12

sol :

6x=12

Dividing both side by 6 , we obtain 

\dfrac{6x}{6}=\dfrac{12}{6}

Question 6 :

Solve : ​\dfrac{t}{5}=10

sol :

\dfrac{t}{5}=10

Multiplying both side by 5, we obtain

\dfrac{t}{5}\times5=10\times5

t=50

Question 7 :

Solve : ​\dfrac{2x}{3}=18

sol :

\dfrac{2x}{3}=18

Multiplying both side by ​\dfrac{3}{2}​ , we obtain

\dfrac{2x}{3}\times\dfrac{3}{2}=18\times\dfrac{3}{2}

x=27

Question 8 :

Solve : ​1.6=\dfrac{y}{1.5}

sol : 

1.6=\dfrac{y}{1.5}

Multiplying both sides by 1.5 , we obtain

1.6\times{1.5}=\dfrac{y}{1.5}\times{1.5}

2.4=y

Question 9 :

Solve : 7x-9 = 16

sol : 

7x-9 = 16

On transposing 9 to R.H.S , we obtain 

7x=16+9

7x = 25

Dividing both side by 7 , we obtain 

\dfrac{7x}{7}=\dfrac{25}{7}

x=\dfrac{25}{7}

Question 10 :

Solve : ​14y-8=13

sol : 

14y-8=13

On transposing 8 to R.H.S , we obtain 

\\14y=13+8\\14y=21

Dividing both side by 14 , we obtain 

\dfrac{14y}{14}=\dfrac{21}{14}

y=\dfrac{3}{2}

Question 11 :

Solve : ​17+6p=9

sol :

17+6p=9

On transposing 17 to R.H.S , we obtain

6p=9-17

6p=-8

Dividing both side by 6 , we obtain

\dfrac{6p}{6}=-\dfrac{8}{6}

p=-\dfrac{4}{3}

Question 12 :

Solve : ​\dfrac{x}{3}+1=\dfrac{7}{15}

sol :

\dfrac{x}{3}+1=\dfrac{7}{15}

On transposing 1 to R.H.S , we obtain

\dfrac{x}{3}=\dfrac{7}{15}-1

\dfrac{x}{3}=\dfrac{7-15}{15}

\dfrac{x}{3}=-\dfrac{8}{15}

Multiplying both sides by 3 , we obtain 

\dfrac{x}{3}\times{3}=-\dfrac{8}{15}\times{3}

x=-\dfrac{8}{5}

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