Class 9 Maths Chapter 4 Exercise 4.1 | Linear Equations in Two Variables | Full NCERT Solution

Class 9 Maths Chapter 4 Exercise 4.1 | Linear Equations in Two Variables | Full NCERT Solution 📚 Class 9 Maths - Chapter 4: Linear Equations in Two Variables 🎯 Exercise 4.1 - Complete NCERT Solutions 📖 In this video, we have explained all the questions from Exercise 4.1 of Chapter 4 (Linear Equations in Two Variables) as per the latest NCERT syllabus. This video is perfect for CBSE Class 9 students looking for a clear and simple explanation of each question. 👨‍🏫 What’s covered in this video: Basics of Linear Equations in Two Variables Step-by-step solutions to all questions Tips to avoid common mistakes Useful for Class 9 exams and school tests 👍 Don’t forget to Like, Share & Subscribe for more helpful content. 📌 Follow our channel for full Chapter 4 solutions and more CBSE Class 9 content. 📅 Updated for 2025 CBSE Syllabus 🔖 Tags class 9 maths chapter 4, linear equations in two variables, class 9 maths exercise 4.1, chapter 4 maths class 9, class 9 ncert maths, cbse class 9 maths, linear equations class 9, exercise 4.1 class 9, class 9 chapter 4 maths explanation, linear equations class 9 in hindi, class 9 maths solutions, ncert maths class 9 chapter 4 =======Follow us social media platforms====== WhatsApp channel link Mohit Maths Guru https://whatsapp.com/channel/0029VaAw... Telegram channel link Mohit Maths Guru https://t.me/mohitmathsguru5 My youtube channel link Mohit Maths Guru    / @mohitmathsguru   Instagram link I'm on Instagram as @mohitmathsguru. Install the app to follow my photos and videos. https://www.instagram.com/mohitmathsg... Ex 4.1 Class 9 Maths Question 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 Rs. x and that of a pen to be Rs.y). Ex 4.1 Class 9 Maths Question 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: (i) 2x + 3y = 9.35 (ii) x−y5−10=0 (iii) – 2x + 3y = 6 (iv) x = 3y (v) 2x = -5y (vi) 3x + 2 = 0 (vii) y – 2 = 0 (viii) 5 = 2x