NCERT Solutions for Exercise 4.2 Class 9 Maths Chapter 4 - Linear Equations in Two Variables

# NCERT Solutions for Exercise 4.2 Class 9 Maths Chapter 4 - Linear Equations in Two Variables

Edited By Vishal kumar | Updated on Oct 05, 2023 03:11 PM IST

## NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables: Exercise 4.2- Download Free PDF

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables: Exercise 4.2- The notion of linear equations in two variables is covered in NCERT Solutions for Class 9 Maths exercise 4.2 as well as how to solve linear equations. Linear equations in two variables are systems of equations with a single solution, no solutions, or an infinite number of solutions of the kind ax+by+c=0, where a, b, c are real numbers. We must first determine the value of variables in order to find the set of solutions to the linear equations with two variables. The algebraic representations of linear equations in two variables were the topic of Class 9 Maths chapter 4 exercise 4.2. The substitution approach is the simplest way to solve a linear equation.

The substitution method is defined as substituting the value of one variable in another equation to obtain the value of another variable. We must first determine the value of variables in order to find the set of solutions to the linear equations with two variables. To use the substitution approach to solve a system of linear equations, we must first determine the value of one variable.

9th class maths exercise 4.2 answers are crafted by subject experts at Careers360 in straightforward language. Students can also download the PDF versions of these class 9 maths chapter 4 exercise 4.2 solutions and make use of them. The following activities are included along with Class 9 Maths chapter 4 exercise 4.2.

## Linear Equations In Two Variables Class 9 Maths Chapter 4 Exercise: 4.2

### (iii) infinitely many solutions

Given : $y = 3x + 5$

This equation is of a line and a line has infinite points on it and each point is a solution Thus, (iii) infinitely many solutions is the correct option.

(i) Given : $2x + y = 7$

Putting x=0, we have , $y=7-2\times 0=7$ means $(0,7)$ is a solution.

Putting x=1, we have , $y=7-2\times 1=5$ means $(1,5)$ is a solution.

Putting x=2, we have , $y=7-2\times 2=3$ means $(2,3)$ is a solution.

Putting x=3, we have , $y=7-2\times 3=1$ means $(3,1)$ is a solution.

The four solutions are : $(0,7),(1,5),(2,3),(3,1)$ .

(ii) Given : $\pi x +y = 9$

Putting x=0, we have , $y=9-\pi \times 0=9$ means $(0,9)$ is a solution.

Putting x=1, we have , $y=9-\pi \times 1=9-\pi$ means $(1,9-\pi )$ is a solution.

Putting x=2, we have , $y=9-\pi \times 2=9-2\pi$ means $(2,9-2\pi )$ is a solution.

Putting x=3, we have , $y=9-\pi \times 3=9-3\pi$ means $(3,9-3\pi )$ is a solution.

The four solutions are : $(0,9),(1,9-\pi ),(2,9-2\pi ),(3,9-3\pi )$ .

(iii) Given : $x = 4y$

Putting x=0, we have , $y=\frac{0}{4}=0$ means $(0,0)$ is a solution.

Putting x=1, we have , $y=\frac{1}{4}$ means $(1,\frac{1}{4})$ is a solution.

Putting x=2, we have , $y=\frac{2}{4}=\frac{1}{2}$ means $(2,\frac{1}{2})$ is a solution.

Putting x=3, we have , $y=\frac{3}{4}$ means $(3,\frac{3}{4})$ is a solution.

The four solutions are : $(0,0)$ , $(1,\frac{1}{4})$ , $(2,\frac{1}{2})$ and $(3,\frac{3}{4})$ .

(i) Given : $x - 2y = 4$

Putting $(0,2)$ ,

we have , $x - 2y = 0-2(2)=-4\neq 4$

Therefore, $(0,2)$ is not a solution of $x - 2y = 4$ .

Given : $x - 2y = 4$

Putting (2,0),

we have , $x - 2y = 2-2(0)=2\neq 4$

Therefore, (2,0) is not a solution of $x - 2y = 4$ .

Given : $x - 2y = 4$

Putting (4,0),

we have , $x - 2y = 4-2(0)=4=4$

Therefore, (4,0) is a solution of $x - 2y = 4$ .

Given : $x - 2y = 4$

Putting $(\sqrt2 , 4\sqrt2)$ ,

we have , $x - 2y =\sqrt{2}-2(4\sqrt{2})=\sqrt{2}-8\sqrt{2}=-7\sqrt{2}\neq 4$

Therefore, $(\sqrt2 , 4\sqrt2)$ is not a solution of $x - 2y = 4$ .

Given : $x - 2y = 4$

Putting (1,1) ,

we have , $x - 2y = 1-2(1)=-1\neq 4$

Therefore, (1,1) is not a solution of $x - 2y = 4$ .

Given : $2x + 3y = k$

Putting (2,1),

we have , $k=2x + 3y = 2(2)+3(1)=4+3=7$

Therefore, k=7 for $2x + 3y = k$ putting x=2 and y=1.

## More About NCERT Solutions for Class 9 Maths Exercise 4.2:

We must solve the linear equation in exercise 4.2 Class 9 Maths. We start by changing different values for x and y. We acquire the value of y by substituting x=0 for x in the given linear equation. The linear equations' solutions are then represented in the form(x, y) . NCERT bookr Class 9 Maths chapter 4 exercise 4.2 comprises four questions, two of which are multiple-choice questions, one of which is the main question with three equations to solve, and the last one is a short answer question. The NCERT solutions for Class 9 Maths exercise 4.2 includes questions to identify whether the given roots are a linear equation's solution or not.

## Benefits of NCERT Solutions for Class 9 Maths Exercise 4.2

• NCERT solutions for Class 9 Maths exercise 4.2 helps students to get practice in solving a variety of maths problems, and also helps in gaining efficiency and accuracy by solving questions.

• In NCER syllabus exercise 4.2 Class 9 Maths, The substitution method is a simple strategy to solve equations that removes an equation from the problem to make it easier to solve.

• By solving the NCERT solution for Class 9 Maths chapter 4 exercise 4.2 exercises, we can get good scores in Mathematics as they understand the subject better and help to understand the concepts more clearly.

## Key Features of 9th Class Maths Exercise 4.2 Answers

1. Expertly Crafted Solutions: The solution for class 9 maths chapter 4 exercise 4.2 have been prepared by subject matter experts. They ensure accuracy and clarity in explanations.

2. Comprehensive Coverage: Exercise 4.2 class 9 maths covers various topics related to quadratic equations and their solutions. It includes problems involving quadratic equations in one variable.

3. Step-by-Step Format: Class 9 maths ex 4.2 Solutions are presented in a step-by-step format, making it easy for students to follow the solution process and understand the concepts.

4. Conceptual Clarity: The primary objective of ex 4.2 class 9 is to help students develop a clear understanding of quadratic equations and their solutions.

5. Variety of Problems: This Exercise 4.2 class 9 maths offers a range of problems with different levels of complexity, allowing students to practice and enhance their problem-solving skills.

6. PDF Availability: Solutions are often available in PDF format, allowing students to download and access them offline for convenient studying.

7. Free of Charge: These resources are typically provided free of charge, making them accessible to all students.

Also see-

## Subject Wise NCERT Exemplar Solutions

1. What is the general form of a linear equation in two variables, according to NCERT solutions for Class 9 Maths chapter 4 exercise 4.2?

ax+by+c=0, is the general form of the linear equation in two variables where a, b,and c are real numbers.

2. Check whether the equation xy-5=8 a linear equation in two variables?

Because of the term xy which is of degree 2 , xy-5=8 is not a linear equation in two variables.

3. Check whether the equation 7x+y=8 a linear equation in two variables ?

7x+y=8 is a linear equation in two variables since the degree of the given expression 7x+y=8 is 1

4. Linear equations in one variable have a ________ solution.

Linear equations in one variable have a unique solution.

5. In exercise 4.2 Class 9 Maths , What is the two-variable equation?

The two-variable equation is nothing but an equation that has two different variables and also two different solutions.

6. What are the coefficients of x and y the equation 9x-y = -12?

The coefficient of x is 9 and the coefficient of y is -1.

7. What is the constant of the equation 2x-4y=-3?

The constant of the equation 2x-4y=-3 is -3

8. According to NCERT solutions for Class 9 Maths chapter 4 exercise 4.2, What is the coefficient of y in the equation 2x-4y=-3?

The coefficient of y in the equation 2x-4y=-3 is -4

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