## 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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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

Chapter 2 - Linear Equations in One Variable Exercises in PDF Format | |

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