NCERT Solutions Class 9 Chapter 4 Linear Equations in Two Variables Ex 4.1

Exercise 4.1 is the foundation of Chapter 4 in Class 9. It focuses on converting simple statements and relations into linear equations in two variables.

The NCERT Solutions for Linear Equations in Two Variables Class 9 Exercise 4.1 explains every step so you can follow the logic of variable selection, equation formation, and check solutions. 

What does NCERT Maths Class 9 Linear Equations in Two Variables Exercise 4.1 Covers?

You will learn how to write equations from phrases like “sum of two numbers”, “difference is twice”, or “total cost”, and then rearrange the expressions into ax + by + c = 0 form. The exercise emphasizes clear notation and choosing sensible variables for the situation.

Once the equations are formed, the exercise teaches finding particular solution pairs by substituting convenient values for one variable and solving for the other. This numeric practice strengthens reasoning and prepares students for plotting in Exercise 4.2.

Class 9 Chapter 4 Exercise 4.1 Questions and NCERT Solutions

Class 9 Chapter 4 linear equations in two variables Exercise 4.1 questions include short word problems that ask you to form equations and find example pairs that satisfy them. Below are the NCERT Solutions of 4.1 exercise: 

1. 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 ₹ x and that of a pen to be ₹ y)

Solution: Let the cost of a notebook be = ₹ x Let the cost of a pen be = ₹ y According to the question, The cost of a notebook is twice the cost of a pen. i.e., cost of a notebook = 2×cost of a pen x = 2×y x = 2y x-2y = 0 x-2y = 0 is the linear equation in two variables to represent the statement, ‘The cost of a notebook is twice the cost of a pen.’

2. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case.

(ii) x –(y/5)–10 = 0

Solution: The equation x –(y/5)-10 = 0 can be written as, 1x+(-1/5)y +(–10) = 0 Now comparing x+(-1/5)y+(–10) = 0 with ax+by+c = 0 We get, a = 1 b = -(1/5) c = -10

(iii) –2x+3y = 6

Solution: –2x+3y = 6 Re-arranging the equation, we get, –2x+3y–6 = 0 The equation –2x+3y–6 = 0 can be written as, (–2)x+3y+(– 6) = 0 Now, comparing (–2)x+3y+(–6) = 0 with ax+by+c = 0 We get, a = –2 b = 3 c =-6

(iv) x = 3y

Solution: x = 3y Re-arranging the equation, we get, x-3y = 0 The equation x-3y=0 can be written as, 1x+(-3)y+(0)c = 0 Now comparing 1x+(-3)y+(0)c = 0 with ax+by+c = 0 We get a = 1 b = -3 c =0

(v) 2x = –5y

Solution: 2x = –5y Re-arranging the equation, we get, 2x+5y = 0 The equation 2x+5y = 0 can be written as, 2x+5y+0 = 0 Now, comparing 2x+5y+0= 0 with ax+by+c = 0 We get a = 2 b = 5 c = 0

(vi) 3x+2 = 0

Solution: 3x+2 = 0 The equation 3x+2 = 0 can be written as, 3x+0y+2 = 0 Now comparing 3x+0+2= 0 with ax+by+c = 0 We get a = 3 b = 0 c = 2

(vii) y–2 = 0

Solution: y–2 = 0 The equation y–2 = 0 can be written as, 0x+1y+(–2) = 0 Now comparing 0x+1y+(–2) = 0with ax+by+c = 0 We get a = 0 b = 1 c = –2

(viii) 5 = 2x

Solution: 5 = 2x Re-arranging the equation, we get, 2x = 5 i.e., 2x–5 = 0 The equation 2x–5 = 0 can be written as, 2x+0y–5 = 0 Now comparing 2x+0y–5 = 0 with ax+by+c = 0 We get a = 2 b = 0 c = -5

How to Prepare using NCERT Solutions Class 9 Chapter 4 Ex 4.1?

Start by reading each statement slowly and underline the quantities mentioned, which values are unknown and which are known. Choose simple variable names (x and y), translate the words into algebraic expressions, and bring everything to one side to get ax + by + c = 0.

Practice these steps repeatedly and use the following approach to improve:

• Break the sentence into small parts and convert each part into algebraic terms.

• Choose easy values for one variable (like 0, 1, 2) to get sample solution pairs quickly.

• Check your pairs by substituting back into the original equation to avoid mistakes.

• Compare your method to the linear equations in two variables class 9 exercise 4.1 NCERT Solutions to find shorter approaches.

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