Class 9 Exam  >  Class 9 Notes  >  NCERT Solutions Chapter 4 - Linear Equations in Two Variables (II), Class 9, Maths

NCERT Solutions Chapter 4 - Linear Equations in Two Variables (II), Class 9, Maths PDF Download

Exercise 4.3

 1. Draw the graph of each of the following linear equations in two variables:

(i) x + y = 4                 
 (ii) x - y = 2              
 (iii) y = 3x             
 (iv) 3 = 2x + y


Answer
(i) x + y = 4
Put x = 0 then y = 4
Put x = 4 then y = 0

NCERT Solutions Chapter 4 - Linear Equations in Two Variables (II), Class 9, Maths

NCERT Solutions Chapter 4 - Linear Equations in Two Variables (II), Class 9, Maths

(ii) x - y = 2
Put x = 0 then y = -2
Put x = 2 then y = 0

NCERT Solutions Chapter 4 - Linear Equations in Two Variables (II), Class 9, Maths

NCERT Solutions Chapter 4 - Linear Equations in Two Variables (II), Class 9, Maths

(iii) y = 3x
Put x = 0 then y = 0
Put x = 1 then y = 3

 

NCERT Solutions Chapter 4 - Linear Equations in Two Variables (II), Class 9, Maths

NCERT Solutions Chapter 4 - Linear Equations in Two Variables (II), Class 9, Maths

(iv) 3 = 2x + y
Put x = 0 then y = 3
Put x = 1 then y = 1

NCERT Solutions Chapter 4 - Linear Equations in Two Variables (II), Class 9, Maths

NCERT Solutions Chapter 4 - Linear Equations in Two Variables (II), Class 9, Maths

 

2. Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?

Answer

Here, x = 2 and y =14.
Thus, x + y = 1 
also, y = 7x ⇒ y - 7x = 0
∴ The equations of two lines passing through (2, 14) are
x + y = 1 and y - 7x = 0.
There will be infinite such lines because infinite number of lines can pass through a given point.


3. If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.

Answer

The point (3, 4) lies on the graph of the equation. 
∴ Putting x = 3 and y = 4 in the equation 3y = ax + 7, we get
3×4 = a×3 + 7
⇒ 12 = 3a + 7
⇒ 3a = 12 - 7
⇒ a = 5/3


4. The taxi fare in a city is as follows: For the first kilometre, the fare is Rs 8 and for the subsequent distance it is Rs 5 per km. Taking the distance covered as x km and total fare as Rs y, write a linear equation for this information, and draw its graph.

Answer

Total fare = y
Total distance covered = x
Fair for the subsequent distance after 1st kilometre = Rs 5
Fair for 1st kilometre = Rs 8
A/q
y = 8 + 5(x-1)
⇒ y = 8 + 5x - 5
⇒ y = 5x + 3

NCERT Solutions Chapter 4 - Linear Equations in Two Variables (II), Class 9, Maths

NCERT Solutions Chapter 4 - Linear Equations in Two Variables (II), Class 9, Maths

 

5. From the choices given below, choose the equation whose graphs are given in Fig. 4.6 and Fig. 4.7.
For Fig. 4. 6                           

(i) y = x                                 
 (ii) x + y = 0                          
 (iii) y = 2x                             
 (iv) 2 + 3y = 7x        

For Fig. 4.7
 (i) y = x + 2
 (ii) y = x – 2
 (iii) y = –x + 2
 (iv) x + 2y = 6

NCERT Solutions Chapter 4 - Linear Equations in Two Variables (II), Class 9, Maths   NCERT Solutions Chapter 4 - Linear Equations in Two Variables (II), Class 9, Maths

Answer

In fig. 4.6, Points are (0, 0), (-1, 1) and (1, -1).
∴ Equation (ii) x + y = 0 is correct as it satisfies all the value of the points.

In fig. 4.7, Points are (-1, 3), (0, 2) and (2, 0).
∴ Equation (iii) y = –x + 2 is correct as it satisfies all the value of the points.

6. If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is
 (i) 2 units               

(ii) 0 unit

Answer
 

Let the distance traveled by the body be x and y be the work done by the force.
y ∝ x (Given)
⇒ y = 5x (To equate the proportional, we need a constant. Here, it was given 5)
A/q,
(i) When x = 2 units then y = 10 units
(ii) When x = 0 unit then y = 0 unit

NCERT Solutions Chapter 4 - Linear Equations in Two Variables (II), Class 9, Maths

NCERT Solutions Chapter 4 - Linear Equations in Two Variables (II), Class 9, Maths

 

7. Yamini and Fatima, two students of Class IX of a school, together contributed Rs 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as Rs x and Rs y.) Draw the graph of the same.

Answer

Let the contribution amount by Yamini be x and contribution amount by Fatima be y.
A/q,
x + y = 100
When x = 0 then y = 100
When x = 50 then y = 50
When x = 100 then y = 0

NCERT Solutions Chapter 4 - Linear Equations in Two Variables (II), Class 9, Maths

NCERT Solutions Chapter 4 - Linear Equations in Two Variables (II), Class 9, Maths

 

8. In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius:

F = (9/5)C + 32

(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.

(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?

(iii) If the temperature is 95°F, what is the temperature in Celsius?

(iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?

(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

Answer

(i) F = (9/5)C + 32

When C = 0 then F = 32

also, when C = -10 then F = 14

NCERT Solutions Chapter 4 - Linear Equations in Two Variables (II), Class 9, Maths

NCERT Solutions Chapter 4 - Linear Equations in Two Variables (II), Class 9, Maths

(ii) Putting the value of C = 30 in F = (9/5)C + 32, we get
F = (9/5)×30  + 32
⇒ F = 54 + 32 
⇒ F = 86

(iii) Putting the value of F = 95 in F = (9/5)C + 32, we get
95 = (9/5)C  + 32
⇒ (9/5)C = 95 - 32 
⇒ C = 63 × 5/9
⇒ C = 35

(iv) Putting the value of F = 0 in F = (9/5)C + 32, we get
0 = (9/5)C  + 32
⇒ (9/5)C = -32
⇒ C = -32 × 5/9
⇒ C = -160/9

Putting the value of C = 0 in F = (9/5)C + 32, we get
F = (9/5)× 0  + 32
⇒ F = 32

(v) Here, we have to find when F = C.
Therefore, Putting F = C in F = (9/5)C + 32, we get
F = (9/5)F + 32
⇒ F - 9/5 F = 32
⇒ -4/5 F = 32
⇒ F = -40
Therefore at -40, both Fahrenheit and Celsius numerically the same.

Exercise 4.4

1. Give the geometric representations of y = 3 as an equation
 (i) in one variable
 (ii) in two variables


Answer

(i) in one variable, it is represented as
y = 3

Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important

(ii) in two variables, it is represented as a line parallel to X-axis.

0x  y = 3

Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important

 

2. Give the geometric representations of 2x + 9 = 0 as an equation
 (i) in one variable
 (ii) in two variables


Answer

(i) in one variable, it is represented as
x = -9/2

Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important

(ii) in two variables, it is represented as a line parallel to Y-axis.

2x + 0y + 9 = 0

Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important

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FAQs on NCERT Solutions Chapter 4 - Linear Equations in Two Variables (II), Class 9, Maths

1. What are linear equations in two variables?
Ans. Linear equations in two variables are equations that involve two variables, usually represented as x and y, and their coefficients. These equations are of the form ax + by = c, where a, b, and c are constants. The solutions to these equations are ordered pairs (x, y) that satisfy the equation when substituted into it.
2. How do we solve linear equations in two variables?
Ans. To solve linear equations in two variables, we can use different methods such as the substitution method or the elimination method. In the substitution method, we solve one equation for one variable and substitute the value into the other equation. In the elimination method, we manipulate the equations to eliminate one variable by adding or subtracting the equations. Once we find the value of one variable, we substitute it back into one of the equations to find the value of the other variable.
3. Can we have more than one solution for a linear equation in two variables?
Ans. Yes, a linear equation in two variables can have infinitely many solutions. This happens when the equation represents a line. Since a line extends infinitely in both directions, any point on the line will satisfy the equation. Therefore, there are infinitely many solutions in the form of ordered pairs (x, y) that lie on the line.
4. What is the graphical representation of linear equations in two variables?
Ans. The graphical representation of linear equations in two variables is a straight line on a coordinate plane. Each equation represents a line, and the solution to the system of equations is the point of intersection of these lines. If the lines are parallel, there is no intersection and hence no solution. If the lines are coincident, they overlap, and every point on the line is a solution.
5. How are linear equations in two variables used in real-life situations?
Ans. Linear equations in two variables are used to solve various real-life problems. They can be used to determine the relationship between two variables, such as the cost of an item and the number of items purchased. These equations are also used in fields like physics and engineering to model and analyze linear systems. For example, they can be used to calculate the trajectory of a projectile or the flow of electrical currents.
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