# 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 NCERT Solutions for Class 8 Maths Chapter 2 provided by Vedantu offers clarifications for the students who can go far but lack the concentration. These solutions help the students to score excellent marks in their exams. If you are a student of Class 8, then this NCERT Solutions for Class 8 Maths Chapter 2 PDF file is beneficial to obtain. All the experts at Vedantu have enough skills and followed CBSE guidelines to create the NCERT Solutions of Class 8, which are available on the site for download. Subjects like Science, Maths, English will become easy to study if you have access to NCERT Solution for Class 8 Science, Maths solutions and solutions of other subjects.

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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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