NCERT Solutions for Exercise 4.1 Class 9 Maths Chapter 4 - Linear Equations in Two Variables

# NCERT Solutions for Exercise 4.1 Class 9 Maths Chapter 4 - Linear Equations in Two Variables

Edited By Vishal kumar | Updated on Oct 05, 2023 12:02 PM IST

## NCERT Solutions for Class 9 Maths Chapter 4: Linear Equations in Two Variables Exercise 4.1- Download Free PDF

NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.1 introduces us to a linear equation with two variables. As this is the first time a linear equation with two variables is introduced in NCERT solutions for Class 9 Maths chapter 4 exercise 4.1 so chapter starts with the prerequisite of linear equation with one variable that we have learnt in an earlier class. Class 9 Maths chapter 4 exercise 4.1 is based on a very basic concept of representation of any equation in the form of a general linear equation with two variables.

NCERT Solutions for Class 9 Maths exercise 4.1 is a base builder exercise for the further exercises in the chapter linear equation with two variables and all the chapters that will be studied in further classes. In exercise 4.1 Class 9 Maths, we have given a rough idea about general equations and it’s representation.

9th class maths exercise 4.1 answers are authored by subject experts at Careers360, and presented in a comprehensible and detailed manner. Each class 9 maths chapter 4 exercise 4.1 comprises a total of two questions with multiple parts. PDF versions of these solutions are readily accessible, allowing students to download and utilize them offline at no cost. Along with Class 9 Maths chapter 4 exercise 4.1, the following exercises are also present.

## Access Linear Equations In Two Variables Class 9 Maths Chapter 4 Exercise: 4.1

Let the cost of a notebook be Rs x and that of a pen be Rs y .

According to the given condition: The cost of a notebook is twice the cost of a pen.

Thus, $x=2y$

$\Rightarrow x-2y=0$

Given : $2x + 3y = 9.3\bar{5}$

$\Rightarrow 2x + 3y - 9.3\bar{5}=0$

Here , a=2, b=3 and c = $- 9.3\bar{5}$

Given:

$x - \frac{y}{5} - 10 = 0$

$\Rightarrow x - \frac{y}{5} - 10 = 0$

Here ,

a=1,

$b=\frac{-1}{5}$

c = -10

Given :

$-2x + 3y = 6$

$\Rightarrow -2x + 3y - 6=0$

Here , a= -2, b=3 and c = -6

Given : $x =3y$

$\Rightarrow x -3y=0$

Here , a= 1, b= -3 and c =0

Given : $2x = - 5y$

$\Rightarrow 2x + 5y=0$

Here , a=2, b= 5 and c =0

Given : $3x + 2 = 0$

$\Rightarrow 3x + 2 = 0$

Here , a= 3, b=0 and c =2

Given : $y -2 = 0$

$\Rightarrow 0.x+y -2 = 0$

Here , a=0, b= 1 and c = -2

Given : $5 = 2x$

$\Rightarrow 2x+0.y-5=0$

Here , a=2, b= 0 and c = -5

## More About NCERT Solution for Class 9 Maths Chapter 4 Exercise 4.1

NCERT syllabus Class 9 Maths chapter 4 exercise 1 is broadly based on the basis of linear equations with two variables. Linear Equations with Two Variables are also used in real-life applications as mentioned before in exercise 4.1 Class 9 Maths in the NCERT book. Two variable equations show us the dependence of two variables on each other for example run scored by player x and run scored by player y a linear equation will show how both of them are related and this is how we can use them in real life. We can also use a linear equation in our daily life by applying it to the things we buy from stationery such as pen and pencil as covered in NCERT solutions for Class 9 Maths exercise 4.1 question 1. NCERT solutions for Class 9 Maths exercise 4.1 covers all types of basic questions that can be formed on linear equations with two variables.

## Benefits of NCERT Solutions for Class 9 Maths Exercise 4.1

• Class 9 Mathematics chapter 4 exercise 4.1 covers a broad range of concepts that will be needed to go to Class 9 Maths chapter 4 exercise 4.2.

• NCERT Class 9 Maths chapter 4 exercise 4.1, will be helpful in Class 10 Maths chapter 3 Linear Equation with Two Variable

• NCERT Class 9 Maths chapter 4 exercise 4.1, will be helpful in Class 11 Maths chapter 6- Linear Inequalities

• Exercise 4.1 Class 9 Maths, is also a base builder concept for many topics of joint entrance exam (JEE Main).

## Key Features of Exercise 4.1 Class 9 Maths

1. Expert-Crafted Solutions: The 9th class maths exercise 4.1 answers have been meticulously crafted by subject experts from Careers360. They are presented in a language that is easy to understand and in great detail.

2. Question Structure: This exercise 4.1 class 9 maths comprises a total of two questions, each containing multiple parts.

3. PDF Availability: PDF versions of the class 9 maths ex 4.1 solution are readily available for students to download. This allows them to access and utilize the solutions offline, and it comes at no cost.

Also see-

## Subject Wise NCERT Exemplar Solutions

1. What is the prerequisite for solving NCERT solutions for Class 9 Maths exercise 4.1?

Knowledge about linear equations with one variable is required to solve a linear equation with two variable

2. What is the general form of linear equation with one variable according to NCERT solutions for Class 9 Maths exercise 4.1?

The general form of linear equation with one variable is ax+by+c = 0

3. Give some examples of linear equations with two variables?

5x + 9y - 90 = 0

45x + 90y - 80 = 0

4. What is the root of the equation as covered in exercise 4.1 Class 9 Maths?

The value which satisfies the equation by putting value in the variable is called the root of equations

5. How many root linear equations with one variable and linear equations with two variables have?

linear equations with one variable have one root while linear equations with two variables have two roots

6. In the NCERT solutions for Class 9 Maths chapter 4 exercise, how many questions are covered?

In Class 9 Maths chapter exercise 4.1, there are two problems. Question 2 includes eight subparts.

7. How many solved examples are there before the NCERT solutions for the Class 9 Mathematics chapter 4 exercise?

Before NCERT answers for Class 9 Maths chapter 4 exercise, there are two solved examples

8. What sorts of questions are included in the NCERT answers for chapter 4 of Maths for Class 9?

Question one is about applying linear equation in real life and question two is about expressing the component of the linear equation which matches to variables of the general equation

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