NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable (EX 2.1) Exercise 2.1

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable (EX 2.1) Exercise 2.1

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Access NCERT solutions for Maths Chapter 2 – Linear Equations in One Variable part-1

Access NCERT solutions for Maths Chapter 2 – Linear Equations in One Variable

1. Find the solution of \[\text{x-2=7}\].

Ans: We have an equation \[\text{x-2=7}\], to solve the following equation we will shift \[\text{2}\] to right hand side, we get

\[\text{x=7+2}\]

\[\text{x=9}\]


2. Find the solution of \[\text{y+3=10}\].

Ans: We have an equation \[\text{y+3=10}\], to solve the following equation we will shift \[\text{3}\] to right hand side, we get

\[\text{y=10-3}\]

\[\text{y=7}\]


3. Find the solution of \[\text{6=z+2}\].

Ans: We have an equation \[\text{6=z+2}\], to solve the following equation we will shift \[\text{2}\] to left hand side, we get

\[\text{6-2=z}\]

\[\text{z=4}\]


4. Find the solution of \[\frac{\text{3}}{\text{7}}\text{+x=}\frac{\text{17}}{\text{7}}\].

Ans: We have an equation \[\frac{\text{3}}{\text{7}}\text{+x=}\frac{\text{17}}{\text{7}}\], to solve the following equation we will shift \[\frac{\text{3}}{\text{7}}\] to right hand side, we get

\[\text{x=}\frac{\text{17}}{\text{7}}\text{-}\frac{\text{3}}{\text{7}}\]

\[\text{x=}\frac{\text{14}}{\text{7}}\]

\[\text{x=2}\]


5. Find the solution of \[\text{6x=12}\].

Ans: We have an equation \[\text{6x=12}\], to solve the following equation, we will shift \[6\] to right hand side and divide \[12\] by \[6\], we get

\[\text{x=}\frac{\text{12}}{6}\]

\[\text{x=2}\]


6. Find the solution of \[\frac{\text{t}}{\text{5}}\text{=10}\].

Ans: We have an equation \[\frac{\text{t}}{\text{5}}\text{=10}\], to solve the following equation, we will shift \[5\] to right hand side and multiply \[10\] by \[5\], we get

\[\text{t=10}\times \text{5}\]

\[\text{t=50}\]


7. Find the solution of \[\frac{\text{2x}}{\text{3}}\text{=18}\].

Ans: We have an equation \[\frac{\text{2x}}{\text{3}}\text{=18}\], to solve the following equation, we will multiply both sides of equation by \[\frac{\text{2}}{\text{3}}\] , we get

\[\text{x=}\frac{\text{18 }\!\!\times\!\!\text{ 3}}{\text{2}}\]

\[\text{x=27}\]


8. Find the solution of \[\text{1}\text{.6=}\frac{\text{y}}{\text{1}\text{.5}}\].

Ans: We have an equation \[\text{1}\text{.6=}\frac{\text{y}}{\text{1}\text{.5}}\], to solve the following equation, we will multiply both sides of equation by \[1.5\] , we get

\[\text{1}\text{.6 }\!\!\times\!\!\text{ 1}\text{.5=y}\]

\[\text{y=2}\text{.4}\]


9. Find the solution of \[\text{7x-9=16}\].

Ans: We have an equation \[\text{7x-9=16}\], to solve the following equation we will shift \[\text{9}\] to right hand side, we get

\[\text{7x=16+9}\]

\[\text{7x=25}\]

Now, we will divide both sides by \[\text{7}\], we get

\[\text{x=}\frac{\text{25}}{\text{7}}\]


10. Find the solution of \[\text{14y-8=13}\].

Ans: We have an equation \[\text{14y-8=13}\], to solve the following equation we will shift \[\text{8}\] to right hand side, we get

\[\text{14y=13+8}\]

\[\text{14y=21}\]

Now, we will divide both sides by \[\text{14}\], we get

\[\text{x=}\frac{\text{21}}{\text{14}}\]

\[\text{x=}\frac{\text{3}}{\text{2}}\]


11. Find the solution of \[\text{17+6p=9}\].

Ans: We have an equation \[\text{17+6p=9}\], to solve the following equation we will shift \[\text{17}\] to right hand side, we get

\[\text{6p=9-17}\]

\[\text{6p=-8}\]

Now, we will divide both sides by \[\text{6}\], we get

\[\text{p=-}\frac{\text{8}}{\text{6}}\]

\[\text{p=-}\frac{\text{4}}{\text{3}}\]


12. Find the solution of \[\frac{\text{x}}{\text{3}}\text{+1=}\frac{\text{7}}{\text{15}}\].

Ans: We have an equation \[\frac{\text{x}}{\text{3}}\text{+1=}\frac{\text{7}}{\text{15}}\], to solve the following equation we will shift \[\text{1}\] to right hand side, we get

\[\frac{\text{x}}{\text{3}}\text{=}\frac{\text{7}}{\text{15}}\text{-1}\]

\[\frac{\text{x}}{\text{3}}\text{=-}\frac{\text{8}}{\text{15}}\]

Now, we will multiply both sides by \[\text{3}\], we get

\[\text{x=-}\frac{\text{8 }\!\!\times\!\!\text{ 3}}{\text{15}}\]

\[\text{x=-}\frac{\text{8}}{\text{5}}\]


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Maths is a difficult subject for so many students. On the other hand, it can be effortless for some. It depends on every individual's efficiency of understanding. To make it simple for you, NCERT Solutions have a practical list of subject experts to simplify the Class 8 Maths Chapter 2. You can obtain the free PDF downloads from our Website. 

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Importance of Linear Equations in One Variable

Learning about Linear Equations helps you modularized the real phenomena that make one variable changing constantly with the other variable involved. 


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