NCERT Solutions for Class 8 Maths Chapter 2 - Exercise

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

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

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

Why Choose NCERT Solutions for Class 8 Maths Chapter 2?

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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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Important Topics under NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.1

Chapter 2 of the class 8 maths syllabus is on Linear Equations in One Variable. It is a very important chapter in mathematics that is covered in class 8 and is divided into 5 major sections. The following is a list of the important topics covered under Linear Equations in One Variable. We advise students to go through each one of these topics carefully to make sure they don’t miss out on internalising the concepts of Linear Equations in One Variable.

• Introduction

• Solving equations where linear expressions are on one side and numbers are on the other side of the equation

• Applications of linear equations in one variable

• Solving equations with variables on both the sides

• Reducing the given equations to a simpler form

Important Takeaways from NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable

• Division and multiplication of algebraic expressions with integers as coefficients

• Common errors faced by students

• Factorisation

• Introduction to identities

• Process of solving linear equations in one variable

Free PDF Downloads from NCERT Solutions

You can download the material for Class 8 Math "Linear Equation for One Variable (2)." It's available for free on the website.

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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NCERT Solutions Class 8 Maths Chapter 2 All Exercises