NCERT Solutions For Class 9 Maths Chapter 4

Ncert Solutions For Class 9 Maths Chapter 4 PDF Free Download

NCERT Solutions For Class 9 Maths Chapter 4 Linear Equation in two variables, are given here to help class 9 students clear all their doubts from this chapter easily. These solutions are prepared by subject experts in a very simple and easily understandable way, according to the latest CBSE syllabus (2018-2019) and guidelines.

It should be noted that linear equations in two variables are one of the most important topics from where questions come in 9th standard examination. So, students need to be thorough with all the questions included in the maths NCERT class 9 book and get acquainted with different types of questions.

NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 1
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 2
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 3
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 4
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 5
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 6
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 7
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 8
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 9
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 10
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 11
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 12
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 13
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 14
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 15
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 16
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 17
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 18
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 19
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 20
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 21
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 22
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 23
NCERT Solutions for class 9 Maths Chapter 4 Linear equations in 2 variables Part 24

Class 9 Maths NCERT Solutions for Linear Equations in Two Variables

The NCERT class 9 solutions for maths chapter 4, Linear Equations in two variables, are provided here so that students can clear their doubts instantly and learn the in-depth concepts of this topic more effectively. Students can download the pdf of these solutions and could practice offline as well. Students are advised to solve sample papers and previous year question papers as well to get an idea of the type of questions asked in class 9 maths exams from chapter Linear Equations in two variables and marks contained by it.

Almost any situation that involves an unknown quantity can be represented in linear equations. We use linear equations to predict profit, calculate income over time, to calculate mileage rates, etc. The topics covered in the chapter are given below for reference:

Section Number Linear Equations in Two Variables Sub-Topics
4.1 Introduction to Linear Equations
4.2 Linear Equations in Two Variables
4.3 Find the solutions to Linear Equations
4.4 Graph of a Linear Equation in Two Variables on on the Cartesian plane
4.5 Equations of Lines Parallel to the x-axis and y-axis
4.6 Summary and Revision Notes

NCERT Solutions For Class 9 Maths Chapter 4 Exercises

 


Q1. Solve the following.

1. A notebook costs twice as much as a pen. Give a linear equation in two variables to represent this statement.

Solution:

Let the pen cost y and the notebook cost x.

A/q (according to question):

Notebook price = pen price = 2y

∴ 2y = x

x – 2y =0.

 

2.Write the given linear equations in the form of ax + by + c = 0. Also, mention the values of  a, b and c in each case:

(i)2x + 5y = 5   (ii)2x-y/2 = 10       (iii)-3x-10 + 3y = 4      (iv)3x + 2 = 0   (v)2x= -6y

 Solution:

(i) 2x + 5y =5

2x + 5y -5 = 0

On comparing it to ax + by + c , We get:

A = 2, b=5 and  c = -5

 

(ii) 2x  – y/2 = 10

2x –y/2 – 10 =0

On comparing it to ax + by +c = 0

a = 2, b= (-y/2) and c = -10

 

(iii) -3x -10 +3y = 4

-3x + 3y +(-10-4) = 0

-3x + 3y +(-14)= 0

On comparing it to ax + by + c =0

a = -3, b = 3 and c = -14.

 

(iv) 3x + 2 =0

3x + 0z + 2 = 0

On comparing it to ax + by +c =0

a=3, b=0 and c =2

 

(v) 2x= -6y

=2x + 6y – 0 = 0

a=0, b = 6 and c =0

 

(vi) 2y – 9 =0

0.x + 2y – 9 = 0

On comparing it to ax + by + c = 0

a=0, b=2 and y = -9.

 

Q2. Fill in the blank with the correct option.

            y  = 4x +2 has ____.

(a)        Infinite solutions.

(b)        Only three solutions.

(c)        A unique solution

Solution:

(a) infinite solution. (Since  y = 4x +2 is  linear equation in two variables)

 

Q3. Give four solutions each for the following equations.

(a)2x+ y = 5     (b) πx + y=7   (c) 2x = 4y

Solution:

(a)2x + y = 5

y= 5 – 2x

let x = 1

∴ y = 5 -2(1)=3

(1,3) is the solution

Or, let x = 0

∴ y = 5-2(0) = 5

(0, 5) is the solution

Or, let x = -1

∴ y = 5-2(-1) = 7

(-1,7) is the solution.

 

Or, let x = 2

∴ y = 5-2(2) = 1

The solution is (2,1)

 

(b) πx + y = 7

y  = 7 – πx

let x = 1

Thus, y = 7 – π(1) = 7 –π

(1,7-π) is the solution.

Let x = 0

Thus, y = 7 –π(0) = 7

(1,7) is the solution.

Let x = -1

Thus, y = 7- π(-1) = 7 +π

(1, 7 +π) is the solution.

Let x =  2

Thus, y = 7 – π(2) = 7-2π

(1, 7 – 2π) is the solution.

 

(c) 4x= 2y

Or, 2x=y

Let x = 1

Thus, y = 2

(1,2) is the solution.

Let x = 0

Thus, y = 0

(0,0) is the solution.

Let x = -1

Thus, y = 2(-1) = -2

(-1,-2) is the solution.

Let x = 2

Thus, y = 2(2) = 4

(2,4) is the solution.

 

Q4. Check which of the following is a solution for the equation 2x + y = 5

(a) (0,3)            (b) (1,0)            (c) (1,1)                        (d) (2,3)            (e) (0,5)

Solution:

(a) (0, 3)

Putting x = 0, y = 3

2(0) + 3 ≠  5.

Therefore, (0,3) is not  a solution for the equation.

 

(b) (1, 0)

                 Putting x = 1, y = 0

                  2(1) + 0 ≠ 5

               Therefore,    (1,0) is not a solution for the equation.

 

c) (1,1)

Putting x=1, y = 0.

2(1) + 0 ≠ 0

Therefore,    (1,1) is not a solution for the equation.

(d) (2, 3)

Putting x = 2, y=3

2(2) + 3 ≠ 5

Therefore, (2,3) is not a solution for the equation.

(e)  (0,5)

Putting x = , y =5

0 + 5 =5

Therefore, (0,5) is a solution for the equation.

 

Q5. Find the value of S,  if x =2 and y= 4 is a solution for  the equation 3x + 4y =11k

Solution:

Given equation ,

3x + 4y =11k

Putting the given values of x and y in the equation.

3(2) + 4(4) = 11k

22=11k

Therefore, k = 2.

 

Q6. For each of the following equations in the two variables:

(a)x -2= y         (b) y+2x =3      (c)4=y + x        (d)x =y/3.

Solution:

(a)x- 2 = y

Let x = 0, then y =-2

Let x= 2, then y = 0

x 0 2
y -2 0

1

(b) y + 2x = 3

Let x = 0, then y = 3

Let x = 1, then y = 1.

x 0 1
Y 3 1

2

(c) 4 = y + x

Or, x + y = 4

Let x = 0, then y = 4

Let x = 4, then y = 0.

x 0 4
y 4 0

3

(d) x = y/3.

Or, 3x = y

Let x = 0 , then  y =0

Let x = 1, then y = 3

x 0 1
y 0 3

4

 

 

Q7. What is the equation of a line passing through (2,10) ? How many other such lines can exist?

Solution:

Any line passing through (2,10) must satisfy the general equation; ax +by =0

For a = 5 and b = -1 the general equation of a line is satisfied.

Thus, the equation of the line will be; 5x –y =0         (Since x = 2, y = 10 )

Infinite such lines exists as (2, 10) is a point and infinite number of lines can pass through a          point.

 

Q8. The point (1,5) lies on the graph of the equation 2y = ax + 6, find the value of a.

Solution:

Given, the point (1,5) lies on the graph of the equation.

Thus, putting the value of x = 1 and y = 5 in the given equation, we get:

2×5 = a x 1 + 6

a = 10 -6 =4

 

Q9. The taxi fare in Bombay follows the following fare scheme; the first kilometer’s fare is Rs.8 and from then on its Rs. 5/km. Assuming the total distance a passenger traveled as x and the fare he paid as Rs.y, write a linear equation for this and draw a graph for it.

Solution:

Given,

Total distance covered = x

Total fare = y

Fair for the first kilometer= 8

Fair after the first 1km = 5

According to the question,

y=8+5(x-1)

y=5x + 3

x 0 -3/5
y 3 0

 5

 

Q10. From the following choices, whose equations is represented in the graph?

(i ) x=y

(ii) x = 2y

(iii)x + y = 0

(iv)2 + 3y = 7x 

6

Solution:

The points in the figure are (0,0),  (-1,1) and (1,-1)

∴  the equation (iii) x +y = 0 has been represented in the graph as it satisfies all the values of the points.

 

Q11. A body moves by some distance which is directly proportional to the force applied on it .Express this in terms of a two variable equation and draw the graph by taking the constant of proportionality as 5. Also,  represent the work done  in the graph when the distance travelled by the body is : (i) 2 units (ii) 0 units

Solution:

Let the distance traveled by the body be x  and the force applied on the body be y.

y ∝ x (given)

y = 5x  (5 is  a constant of proportionality)

According to the question

(i)   when x = 2 units , then y = 10 units

(ii)  when x = 0 units, then y = 0 units.

7

 

 

Q12. Mansi and Sega donated Rs.100 for the victims of a cyclone. Write a linear equation to represent this data. Draw a graph for the same.

Solution:

Let Mansi’s donation be Rs.x and Sega’s donation be Rs.y

According to the question;

x + y = 100

when x  = 0 , y = 100

when x= 50, y = 50

when x = 100, y = 50

X 0 50 100
Y 100 50 0

8

 

 

Q13. The equation which converts Fahrenheit to Celsius is:

                F=(9/5)C + 32

(i) Draw a graph for the above linear equation with Celsius on the x axis.

(ii) If its 30o C, how much is it in Fahrenheit?

(iii) If its 95oF, how much is it in Celsius?

(iv) Find the temperature at which both Fahrenheit and Celsius are numerically the same.

Solution:

(i) F = (9/5)C + 32

When C = 0, F = 32

When C = -10 ,F = 14

C 0 -10
        F       32      14

9

(ii)  When C = 30, F = (9/5)x30 + 32

F =54+32 =86o

(iii)When F = 95, 95 = (9/5)C +32

(9/5)C = 95-32

C =63 x 5/9 = 35.

(iv) We need to find where F = C

Thus, putting F = C in F =(9/5)C + 32 we get

F=(9/5)F + 32

F – 9/5F = 32

-4/5F =32

F= -40o

Therefore, at -40 degrees, both Celsius and Fahrenheit will have the same numerical value.

 

Q14. Write the geometric representation of y = 5 as an equation with

(i)   a single variable

(ii)  two variables.

Solution:

(i) in a single variable it is represented as :

y = 5

(ii)in two variables it is represented as :

y + 0.x = 5

 

Q15. Give the geometric representation of 2x + 4 =0 as an equation with (i)one variable

(ii)two variables

Solution:

(i) with one variable,

x = -2

(ii) with two variables

2x + 0.y + 4 = 0

Almost any situation that involves an unknown quantity can be represented in linear equations. We use linear equations to predict profit, calculate income over time, to calculate mileage rates, etc. In this chapter, students will be introduced to linear equations in two variables. Further, students will learn to find solutions to such equations. Students will learn to represent a linear equation in two variables on the Cartesian plane. Students will also learn about the equation of lines parallel to x-axis and y-axis.

Bottom Line:

Practice these solutions as worksheets and get prepared yourself for class 9 final exam. BYJU’S provides here with online learning materials which students can study online from our website or can download the materials to learn offline. These materials are available for free and are accessible anytime.

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