## 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

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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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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