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

Linear Equations in One Variable are discussed in Chapter 2 of the Class 8 Mathematics NCERT Textbook. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.1, focus on the topic of Linear Equation in One Variable. Algebraic equations involve variables. The expression on the left-hand side of the equality sign is LHS, whereas on the right-hand side of the equality sign is RHS. For certain values of variables, called the solutions, the expressions on the LHS and RHS should be equal. To make the learning process easy and organised for students, Extramarks provides them with the Class 8 Maths Chapter 2 Exercise 2.1 Solutions so that they can get used to the types of questions that appear in the examinations. These solutions make it easier for them to answer questions correctly in the examinations. Furthermore, Extramarks also provides students with revision notes, important questions, past years’ papers, and much more to assist them in performing well in their examinations.

## 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

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### NCERT Solutions Class 8 Maths Chapter 2 All Exercises

In Chapter 2, Class 8 Mathematics, there are six exercises in the NCERT textbook that involve solving equations that have variables on both sides, etc. A linear equation’s utility lies in its diverse applications, such as solving problems involving numbers, ages, perimeters, and so on. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.1 provided by Extramarks contain stepwise appropriate answers to all the questions of the textbook. In addition to helping students understand the concepts of the chapter, these solutions also help them develop their problem-solving abilities. Students should thoroughly review the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.1 so that they can easily solve any complicated problems related to the concepts of the chapter. These solutions enable students to learn different approaches to solving questions, as well as develop strong fundamentals of the subject. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.1, also help learners to improve their preparation for the subject. As a result, they are able to apply the concepts of the chapter to real-life experiences.

### NCERT Solutions for Class 8

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Q.1 Solve the following equations.

$\begin{array}{l}1.\text{ x}-2=7\\ 2.\text{ y}+3=10\\ 3.\text{ }6=\text{z}+2\end{array}$

$\begin{array}{l}4.\frac{3}{7}+\text{x}=\frac{17}{7}\\ 5.\text{ }6\text{x}=12\\ 6.\frac{\text{t}}{5}=10\end{array}$

$\begin{array}{l}7.\text{\hspace{0.17em}}\frac{2\text{x}}{3}=18\\ 8.\text{\hspace{0.17em}}1.6=\frac{\text{y}}{1.5}\\ 9.\text{\hspace{0.17em}}7\text{x}-9=16\end{array}$

$\begin{array}{l}10.\text{ }14\text{y}-8=13\\ 11.\text{ }17+6\text{p}=9\\ 12.\text{\hspace{0.17em} }\frac{\text{x}}{3}+1=\frac{7}{15}\end{array}$

Ans

$\begin{array}{l}1.\\ \text{​x}-2=7\\ \text{Transposing 2 to R.H.S,we obtain}\\ ⇒\text{x}=7+2\\ ⇒\text{x}=9\\ \\ 2.\text{​}\\ \text{y}+3=10\\ \text{Transposing 3 to R.H.S,we obtain}\\ ⇒\text{y}=10-3\\ ⇒\text{y}=7\\ \\ 3.\\ \text{​}6=\text{z}+2\\ \text{Transposing 2 to L.H.S,we obtain}\\ ⇒6-2=\text{z}\\ ⇒\text{z}=4\\ \\ 4.\text{​}\\ \frac{3}{7}+\text{x}=\frac{17}{7}\\ \text{Transposing}\frac{3}{7}\text{to R.H.S,we obtain}\\ ⇒\text{x}=\frac{17}{7}-\frac{3}{7}\\ ⇒\text{x}=\frac{17-3}{7}\\ ⇒\text{x}=\frac{14}{7}\\ ⇒\text{x}=2\\ \\ 5.\text{​}\\ 6\text{x}=12\\ \text{Dividing both sides by 6,we get}\\ ⇒\frac{6\text{x}}{6}=\frac{12}{6}\\ ⇒\text{x}=2\\ \\ 6.\\ \frac{\text{t}}{5}=10\\ \text{Multiplying both the sides by}\text{5,we get}\\ ⇒\frac{\text{t}}{5}×5=10×5\\ ⇒\text{t}=50\\ \\ 7.\\ \text{\hspace{0.17em}​}\frac{2\text{x}}{3}=18\\ \text{Multiplying both the sides by}\frac{3}{2}\text{,we get}\\ ⇒\frac{2\text{x}}{3}×\frac{3}{2}=18×\frac{3}{2}\\ ⇒\text{x}=27\\ \\ 8.\\ \text{​}1.6=\frac{\text{y}}{1.5}\\ \text{Multiplying both the sides by}1.5\text{,}\text{we get}\\ ⇒1.6×1.5=\frac{\text{y}}{1.5}×1.5\\ ⇒\text{y}=2.4\\ \\ 9.\text{\hspace{0.17em}}\\ 7\text{x}-9=16\\ \text{Transposing 9 to R.H.S,we get}\\ ⇒7\text{x}=16+9\\ ⇒7\text{x}=25\\ \text{Dividing both the sides by 7}\\ ⇒\frac{7\text{x}}{7}=\frac{25}{7}\\ ⇒\text{x}=\frac{25}{7}\\ \\ 10.\text{​}\\ 14\text{y}-8=13\\ \text{Transposing 8 to R.H.S,we get}\\ ⇒14\text{y}=13+8\\ ⇒14\text{y}=21\\ \text{Dividing both the sides by 14}\\ ⇒\frac{14\text{y}}{14}=\frac{21}{14}\\ ⇒\text{y}=\frac{3}{2}\\ \\ 11.\\ \text{​}17+6\text{p}=9\\ \text{Transposing 17 to R.H.S,we get}\\ ⇒6\text{p}=9-17\\ ⇒6\text{p}=-8\\ \text{Dividing both sides by 6}\\ ⇒\frac{6\text{p}}{6}=\frac{-8}{6}\\ ⇒\text{p}=\frac{-4}{3}\\ \\ 12.\text{\hspace{0.17em}}\\ \text{​}\frac{\text{x}}{3}+1=\frac{7}{15}\\ \text{Transposing1 to R.H.S,we get}\\ ⇒\frac{\text{x}}{3}=\frac{7}{15}-1\\ ⇒\frac{\text{x}}{3}=\frac{7-15}{15}\\ ⇒\frac{\text{x}}{3}=\frac{-8}{15}\\ \text{Multiplying both the sides by 3}\\ ⇒\frac{\text{x}}{3}×3=\frac{-8×3}{15}\\ ⇒\text{x}=\frac{-8}{5}\end{array}$

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## 2. Is it essential to practise the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.1 prior to appearing for the examinations?

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The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.1 are not difficult, and practising them will improve students’ performance in the examinations.

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