Thursday, May 28, 2020

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

EXERCISE 2.1



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


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



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



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



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



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



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


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 




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 



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




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




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


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



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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