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=9

 

Question 2 :

Solve : y+3=10

sol : 

y+3=10

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

y= 10-3 = 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

\( \begin{align*}x&=\dfrac{17}{7}-\dfrac{3}{7}\\&=\dfrac{14}{7}\\&=2\end{align*} \)

 

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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