NCERT Solutions for Class 9 Maths Chapter 4
NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.1, Exercise 4.2, Exercise 4.3 and Exercise 4.4 Linear Equations in Two Variables free PDF to downloads updated for academic session 2020-21 following new NCERT Books. UP Board Solutions for Class 9 Maths Chapter 4 Prashnavali 4.1, Prashnavali 4.2, Prashnavali 4.3 and Prashnavali 4.4 are given below in Hindi Medium. View UP Board Solution and NCERT Solutions for 2020-21 in Video Format Hindi Medium and English Medium or free download in PDF or use it as online all the digital contents. There is not any login or registration condition. Just visit to website and use the contents free of cost.For any inconvenience, contact us for help. We are here to help you without any charge. You can coordinate with us message via social media or text message. Visit to Tiwari Academy Online and Offline Apps page and download Class 9 Maths App for Offline use or Download Kaksha 9 Ganit App for offline use.
NCERT Solutions for Class 9 Maths Chapter 4
|Class: 9||Maths (English and Hindi Medium)|
|Chapter 4:||Linear Equations in Two Variables|
9th Maths Chapter 4 Solutions in English & Hindi Medium
NCERT Solutions for Class 9 Maths Chapter 4 all exercises are given below. NCERT Solutions 2020-21 are updated for the CBSE Exams 2021 based on Latest CBSE Syllabus 2020-2021. NCERT Solutions and UP Board Solutions for Class 9 Maths Chapter 4 as online contents are helpful for all the CBSE, MP Board, Gujrat Board & UP Board (High School) students who are following NCERT Books for their Examination.
9th Maths Exercise 4.1 Solutions
9th Maths Exercise 4.2 Solutions
9th Maths Exercise 4.3 Solutions
9th Maths Exercise 4.4 Solutions
Class 9 Maths Exercise 4.1 & 4.2 Solutions in Video
Class 9 Maths Exercise 4.3 & 4.4 Solutions in Video
When is the solution of a linear equation is not affected?
The solution of a linear equation is not affected when
1.The same number is added to (or subtracted from) both sides of the equation.
2. Both sides of the equation are multiplied (or divided) by the same non-zero number.
What is meant by a solution of the linear equation?
The graph of every linear equation in two variables is a straight line and every point on the graph (straight line) represents a solution of the linear equation.
Important Notes on 9th Maths Chapter 4
- An algebraic equation is a statement of equality of algebraic expression involving one or more variables.
- An equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers (a and b together cannot be zero) and x, y are variables, is called a liner equation in two variables.
- A linear equation in one variable (say x) can be written as linear equation in two variables, by taking coefficient of other variable (say y) as zero.
Important Point in 9th Maths Chapter 4
- Every solution of the linear equation cab be represented by a unique point on the graph of the equation.
x = 0 is the equation of the Y- axis and y = 0 is the equation of the X-axis.
- The graph of x = a represents a straight line parallel to the Y-axis and the graph of y = a represent a straight line parallel to the X-axis.
- An equation of the type y = mx represents a straight line through the origin.
Important Questions on 9th Maths Chapter 4
क्योकि यह एक रैखिक समीकरण है और एक रेखा पर अपरिमित रूप से अनेक बिंदु होते हैं तथा प्रत्येक बिंदु इस रैखिक समीकरण का एक हल होता है।
There are infinite number of lines that can pass through (2, 4) as infinite number of lines passes through a point.
3y = ax + 7.
Putting x = 3 and y = 4,
we have, 3 × 4 = a × 3 + 7
⇒ 12 = 3a + 7
⇒ 12 – 7 = 3a
⇒ a = 5/3
⇒ y = 7 – 2x
x = 0 रखने पर, y = 7 – 2 × 0 = 7,
अतः, (0, 7) समीकरण का एक हल है।
x = 1 रखने पर, y = 7 – 2 × 1 = 5,
अतः, (1, 5) समीकरण का एक हल है।
x = 2 रखने पर, y = 7 – 2 × 2 = 3,
अतः, (2, 3) समीकरण का एक हल है।
x = 3 रखने पर, y = 7 – 2 × 3 = 1,
अतः, (3, 1) समीकरण का एक हल है।
इस प्रकार, (0, 7), (1, 5), (2, 3) और (3, 1) समीकरण 2x + y = 7 के चार हल हैं।
2x + 3y = k में x = 2 और y = 1 रखने पर,
2 × 2 + 3 × 1 = k
⇒ k = 7
Distance traveled = x km and total fare = ₹ y
Total fare = Fare for first km + Fare for remaining distance
Therefore, the equation:
y = 8 + 5×(x – 1)
⇒ y = 5x + 3
the cost of pen =₹ y
According to question,
Cost of notebook = 2 × Cost of Pen
⇒ x = 2y
⇒ x – 2y = 0
⇒ 2x + 3y – 9.35 = 0
Here a = 2, b = 3 and c = – 9.35.