NCERT Solutions for class 8 Maths chapter 2 Linear Equation In One Variable Exercise 2.1

EXERCISE 2.1

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Introduction - Linear Equations in One Variable - Chapter 2 - NCERT Class 8th Maths

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Q 1 - Ex 2.1 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2

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Question 1 :
Solve : x - 2=7
sol : x - 2=7
On transposing 2 to R.H.S , we obtain 
x = 7 + 2
x = 9

Q 2 - Ex 2.1 -Class 8 Linear Equations in One Variable - NCERT Maths

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Question 2 :
Solve : y+3=10
sol : 
y+3=10
On transposing 3 to R.H.S , we obtain
y= 10 - 3
y = 7

Q 3 - Ex 2.1 - Class 8th - Linear Equations in One Variable - NCERT Maths

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Question 3 :
Solve : 6 =z+2
sol : 
6 = z+2
On transposing 2 to L.H.S , we obtain
6-2 = z
z=4

Q 4 - Ex 2.1 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2

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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

Q 5 - Ex 2.1 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2

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Question 5 :
Solve : 6x=12
sol :
6x=12
Dividing both side by 6 , we obtain 
​$\dfrac{6x}{6}=\dfrac{12}{6}$​

Q 6 - Ex 2.1 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2

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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

Q 7 - Ex 2.1 - Class 8 Linear Equations in One Variable - NCERT Maths 

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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

Q 8 - Ex 2.1 - Class 8 Linear Equations in One Variable - NCERT Maths 

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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

Q 9 - Ex 2.1 - Class 8 Linear Equations in One Variable - NCERT Maths

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

Q 10 - Ex 2.1 - Class 8 Linear Equations in One Variable - NCERT Maths

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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 
​$\dfrac{14y}{14}=\dfrac{21}{14}$​
​$y=\dfrac{3}{2}$​

Q 11 - Ex 2.1 - Class 8 Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2

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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
​$\dfrac{6p}{6}=-\dfrac{8}{6}$​
​$p=-\dfrac{4}{3}$

Q 12 - Ex 2.1 - Class 8 Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2

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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 
​$\dfrac{x}{3}\times{3}=-\dfrac{8}{15}\times{3}$​
​$x=-\dfrac{8}{5}$​