NCERT Solutions For Class 9 Maths Chapter 4

NCERT Solutions Class 9 Maths Linear Equations in Two Variables

Ncert Solutions For Class 9 Maths Chapter 4 PDF Download

NCERT solutions class 9 maths chapter 4 linear equation in two variables is one of the most common topics from where questions come in 9th standard examination. The NCERT solutions for class 9 maths chapter 4 linear equation in two variables is provided here so that students can learn the concepts of the topic more effectively.

The topics covered in the chapter are given below

Section Number

Topic

4.1

Introduction

4.2

Linear Equations

4.3

Solution of a Linear Equation

4.4

Graph of a Linear Equation in Two Variables

4.5

Equations of Lines Parallel to the x-axis and y-axis

4.6

Summary

The NCERT solutions for class 9 maths chapter 4 is created by expert mathematics teachers according to the latest CBSE syllabus. Check the NCERT solutions for maths chapter 4 class 9 pdf given below.

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, they will learn to find solutions to such equations. Students will learn to represent 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.

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