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4. Linear Equation In Two Variables Mathematics class 9 in English Medium ncert book solutions Exercise 4.1
4. Linear Equation In Two Variables Exercise 4.1 – Complete NCERT Book Solutions for Class 9 Mathematics (English Medium). Get all chapter explanations, extra questions, solved examples and additional practice questions for 4. Linear Equation In Two Variables Exercise 4.1 to help you master concepts and score higher.
4. Linear Equation In Two Variables Mathematics class 9 in English Medium ncert book solutions Exercise 4.1
NCERT Solutions for Class 9 Mathematics play an important role in helping students understand the concepts of the chapter 4. Linear Equation In Two Variables clearly. This chapter includes the topic Exercise 4.1, which is essential from both academic and examination point of view. The solutions provided here are prepared strictly according to the latest NCERT syllabus and follow the guidelines of CBSE to ensure accuracy and relevance. Each question is explained in a simple and student-friendly manner so that learners can grasp the concepts without confusion. These NCERT Solutions are useful for regular study, homework help, and exam preparation. All textbook questions are solved step by step to improve problem-solving skills and conceptual clarity. Students of Class 9 studying Mathematics can use these solutions to revise important topics, understand difficult questions, and practise effectively before examinations. The chapter 4. Linear Equation In Two Variables is explained in a structured way, making it easier for students to connect the theory with the topic Exercise 4.1. By studying these updated NCERT Solutions for Class 9 Mathematics, students can build a strong foundation, boost their confidence, and score better marks in school and board exams.
4. Linear Equation In Two Variables
Exercise 4.1
Chapter 4. Linear Equation In Two Variables
Exercise 4.1
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 pen = y
Let the cost of notebook= x
Then, According To Question,
x = 2y
⇒ x - 2y = 0
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
Solution:
(i) 2x + 3y = 9.35
Expressing the equation in the form of ax + by + c = 0,
∴ 2x+3y-9.35= 0
On Comparing, We have
Then, a= 2, b= 3, c= -9.3
(ii) x – 5y – 10 = 0
Solution:
(ii) x – 5y – 10 = 0
Expressing the equation in the form of ax + by + c = 0,
∴ x- 5y - 10 = 0
On Comparing, We have
Then, a= 1, b= -5, c= -10
(iii) –2x + 3y = 6
Solution:
(iii) –2x + 3y = 6
Expressing the equation in the form of ax + by + c = 0,
∴ -2x + 3y - 6= 0
On Comparing, We have
Then, a= -2, b= 3, c= -6
(iv) x = 3y
Solution:
(iv) x = 3y
Expressing the equation in the form of ax + by + c = 0,
∴ x - 3y= 0
On Comparing, We have
Then, a= 1, b= -3, c= 0
(v) 2x = –5y
Solution:
(v) 2x = –5y
Expressing the equation in the form of ax + by + c = 0,
∴ 2x + 5y= 0
On Comparing, We have
Then, a= 2, b= 5, c= 0
(vi) 3x + 2 = 0
Solution:
(vi) 3x + 2 = 0
Expressing the equation in the form of ax + by + c = 0,
∴ 3x + 2= 0
On Comparing, We have
Then, a= 3, b= 0, c= 2
(vii) y – 2 = 0
Solution:
(vii) y – 2 = 0
Expressing the equation in the form of ax + by + c = 0,
∴ y-2= 0
On Comparing, We have
Then, a= 0, b= 1, c= -2
(viii) 5 = 2x
Solution:
(viii) 5 = 2x
Expressing the equation in the form of ax + by + c = 0,
∴ 2x - 5= 0
On Comparing, We have
Then, a= 2, b= 0, c= -5
See other sub-topics of this chapter:
1. Exercise 4.1 2. Exercise 4.2 3. Exercise 4.3 4. Exercise 4.4