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

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

Vishal kumarUpdated on 21 May 2025, 02:21 PM IST

The term linear equations in two variables offers a fundamental building block of algebra since these expressions draw straight lines on graphs. The fourth chapter of Class 9 Maths teaches students about linear equation formats, along with visualisation methods and variable relationships between such expressions and their graphical representations.

This Story also Contains

  1. NCERT Solutions for Class 9 Maths Chapter 4 – Linear Equations in Two Variables Exercise 4.1
  2. NCERT Solutions Class 9 Maths Chapter 4: Exercise 4.1
  3. Topics Covered in Chapter 4 - Linear Equations in Two Variables (Exercise 4.1):
  4. NCERT Solutions of Class 9 Subject Wise
  5. NCERT Exemplar Solutions of Class 9 Subject Wise
NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.1 - Linear Equations in Two Variables
NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.1 - Linear Equations in Two Variables

Exercise 4.1 of this chapter demonstrates how to convert everyday situations into linear equations through two variables. Our expertly developed NCERT Solutions help students identify mathematical terms and coefficients to master linear equation standard form along with their constant values. Students should utilise NCERT Books to achieve both conceptual understanding and deep mathematical knowledge throughout all subject matter.

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

NCERT Solutions Class 9 Maths Chapter 4: Exercise 4.1

Q1 The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. (Take the cost of a notebook to be Rs x and that of a pen to be Rs y ).

Answer:

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

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

Therefore it means, $x=2y$ or $\Rightarrow x-2y=0$

Q2 (i) Express the following linear equations in the form $ax + by + c = 0$ and indicate the values of a, b and c in each case: $2x + 3y = 9.3\bar{5}$

Answer:

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

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

Q2 (ii) Express the following linear equations in the form $ax + by + c = 0$ and indicate the values of a , b and c in each case: $x - \frac{y}{5} - 10 = 0$

Answer:

Given case: $x - \frac{y}{5} - 10 = 0$ or $\Rightarrow x - \frac{y}{5} - 10 = 0$

Thus , a = 1, $b = \frac{-1}{5}$ and c = -10

Answer:

Given case: $x =3y$ or $\Rightarrow x -3y=0$

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

Q2 (v) Express the following linear equations in the form $ax + by + c = 0$ and indicate the values of a , b and c in each case: $2x = - 5y$

Answer:

Given case: $2x = - 5y$ or $\Rightarrow 2x + 5y=0$

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

Q2 (vi) Express the following linear equations in the form $ax + by + c = 0$ and indicate the values of a , b and c in each case: $3x + 2 = 0$

Answer:

Given case: $3x + 2 = 0$ or $\Rightarrow 3x + 2 = 0$

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

Q2 (vii) Express the following linear equations in the form $ax + by + c = 0$ and indicate the values of a , b and c in each case: $y -2 = 0$

Answer:

Given case: $y -2 = 0$ or $\Rightarrow 0x+y -2 = 0$

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

Q2 (viii) Express the following linear equations in the form $ax + by + c = 0$ and indicate the values of a , b and c in each case: $5 = 2x$

Answer:

Given case: $5 = 2x$ or $\Rightarrow 2x+0y-5=0$

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


Also Read-

Topics Covered in Chapter 4 - Linear Equations in Two Variables (Exercise 4.1):

  • Formulating equations from real-world statements: We can create linear equations by translating real-life situations into mathematical expressions using variables and operations.
  • Form of equation: Standard form of linear equations: ax + by + c = 0
  • Identifying coefficients and constants: In the equation ax + by + c = 0, a and b are coefficients of x and y, while c is the constant term.
  • The standard form requires adjusting the equation arrangement: An equation exists in standard form after moving all variables and constants to one side, alongside zero on the other side.

Also see-

NCERT Exemplar Solutions of Class 9 Subject Wise

Frequently Asked Questions (FAQs)

Q: What is the prerequisite for solving NCERT solutions for Class 9 Maths exercise 4.1?
A:

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

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

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

Q: Give some examples of linear equations with two variables?
A:

5x + 9y - 90 = 0

45x + 90y - 80 = 0

Q: What is the root of the equation as covered in exercise 4.1 Class 9 Maths?
A:

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

Q: How many root linear equations with one variable and linear equations with two variables have?
A:

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

Q: In the NCERT solutions for Class 9 Maths chapter 4 exercise, how many questions are covered?
A:

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

Q: How many solved examples are there before the NCERT solutions for the Class 9 Mathematics chapter 4 exercise?
A:

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

Q: What sorts of questions are included in the NCERT answers for chapter 4 of Maths for Class 9?
A:

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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