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

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

$x=-\dfrac{8}{5}$

2 thoughts on “NCERT Solutions for class 8 Maths chapter 2 Linear Equation In One Variable”

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Oya
October 9, 2019 at 6:27 am

Many thanks discussing this short article and rendering it public

EXERCISE 2.1

2 thoughts on “NCERT Solutions for class 8 Maths chapter 2 Linear Equation In One Variable”

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