NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.3

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.3

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.3

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.3 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.3.



Ex 4.3 Class 9 Maths Question 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

Solution:
(i) We have x + y = 4 ⇒ y = 4 – x
If we put x = 0, then we get y = 4 – 0 = 4
If we put x = 1, then we get y = 4 – 1 = 3
If we put x = 2, then we get y = 4 – 2 = 2
∴ We have the following table:

x

0

1

2

y

4

3

2

Plot the ordered pairs (0, 4), (1, 3) and (2, 2) on the graph paper.

Joining these points, we get a straight line AB as shown below.

Thus, the line AB is the required graph of x + y = 4.


(ii) We have x – y = 2 ⇒ y = x – 2

If we put x = 0, then we get y = 0 – 2 = -2
If we put x = 1, then we get y = 1 – 2 = -1
If we put x = 2, then we get y = 2 – 2 = 0
∴ We have the following table:

x

0

1

2

y

-2

-1

0

Plot the ordered pairs (0, -2), (1, -1) and (2, 0) on the graph paper.

Joining these points, we get a straight line PQ as shown below.


Thus, the line PQ is the required graph of x – y = 2.


(iii) We have y = 3x
If we put x = 0, then we get y = 3(0) ⇒ y = 0
If we put x = 1, then we get y = 3(1) = 3
If we put x = -1, then we get y = 3(-1) = -3
∴ We have the following table:

x

0

1

-1

y

0

3

-3

Plot the ordered pairs (0, 0), (1, 3) and (-1, -3) on the graph paper.

Joining these points, we get a straight line LM as shown below.


Thus, the line LM is the required graph of y = 3x.

(iv) We have 3 = 2x + y ⇒ y = 3 – 2x
If we put x = 0, then we get y = 3 – 2(0) = 3
If we put x = 1, then we get y = 3 – 2(1) = 3 – 2 = 1
If we put x = 2, then we get y = 3 – 2(2) = 3 – 4 = -1
∴ We have the following table:

x

0

1

2

y

3

1

-1

Plot the ordered pairs (0, 3), (1, 1) and (2, –1) on the graph paper.

Joining these points, we get a straight line CD as shown below.

Thus, the line CD is the required graph of 3 = 2x + y.

 

Ex 4.3 Class 9 Maths Question 2.
Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?

Solution:
(2, 14) means x = 2 and y = 14
Equations which have (2, 14) as the solution are (i) x + y = 16, (ii) 7x – y = 0
There are infinite number of lines which passes through the point (2, 14), because infinite number of lines can be drawn through a point.

Ex 4.3 Class 9 Maths Question 3.
If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.

Solution:
The equation of the given line is 3y = ax + 7
∵ (3, 4) lies on the given line.
∴ It must satisfy the equation 3y = ax + 7
We have the point (3, 4) which means x = 3 and y = 4.
Putting these values in given equation, we get
3 × 4 = a × 3 + 7
⇒ 12 = 3a + 7
⇒ 3a = 12 – 7 = 5

⇒ a = 5/3
Thus, the required value of a is 5/3.

           

Ex 4.3 Class 9 Maths Question 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.

Solution:
Here, total distance covered = x km and total taxi fare = Rs. y
The fare for 1 km = Rs. 8
Remaining distance = (x – 1) km
∴ Fare for (x – 1) km = Rs. 5 × (x – 1)
Total taxi fare = Rs. 8 + Rs. 5(x – 1)
According to the question,
y = 8 + 5(x – 1)

⇒ y = 8 + 5x – 5
⇒ y = 5x + 3,
which is the required linear equation representing the given information.
Graph: We have y = 5x + 3
If we put x = 0, then we get y = 5(0) + 3 ⇒ y = 3
If we put x = -1, then we get y = 5(-1) + 3 ⇒ y = -2
If we put x = -2, then we get y = 5(-2) + 3 ⇒ y = -7
∴ We have the following table:

x

0

-1

-2

y

3

-2

-7

Now, plotting the ordered pairs (0, 3), (-1, -2) and (-2, -7) on a graph paper.

Joining them, we get a straight line PQ as shown below.

Thus, the line PQ is the required graph of the linear equation y = 5x + 3.


Ex 4.3 Class 9 Maths Question 5.
From the choices given below, choose the equation whose graphs are given in Fig. (1) and Fig. (2).
For Fig. (1)
(i) y = x
(ii) x + y = 0
(iii) y = 2x
(iv) 2 + 3y = 7x

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

Solution:
For Fig. (1), the correct linear equation is x + y = 0
[As for (-1, 1) = -1 + 1 = 0 and for (1, -1) = 1 + (-1) = 0]
For Fig. (2), the correct linear equation is y = -x + 2
[As for (-1, 3) ⇒ 3 = -(-1) + 2 = 3 ⇒ 3 = 3 and for (0, 2) ⇒ 2 = -(0) + 2 ⇒ 2 = 2]

Ex 4.3 Class 9 Maths Question 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

Solution:
Constant force is 5 units.
Let the distance travelled = x units and work done = y units.
Work done = Force × Distance
⇒ y = 5 × x

⇒ y = 5x
For drawing the graph, we have y = 5x
If we put x = 0, then we get y = 5(0) = 0
If we put x = 1, then we get y = 5(1) = 5
If we put x = -1, then we get y = 5(-1) = -5

∴ We have the following table:

x

0

1

-1

y

0

5

-5

Plotting the ordered pairs (0, 0), (1, 5) and (-1, -5) on the graph paper.

Joining the points, we get a straight line AB as shown.

From the graph, we get
(i) Distance travelled = 2 units i.e., x = 2
∴ If x = 2, then y = 5(2) = 10
⇒ Work done = 10 units.

(ii) Distance travelled = 0 unit i.e., x = 0
∴ If x = 0 ⇒ y = 5(0) – 0
⇒ Work done = 0 unit.

Ex 4.3 Class 9 Maths Question 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.

Solution:
Let the contribution of Yamini be Rs. x and the contribution of Fatima be Rs. y.
∴ We have x + y = 100 ⇒ y = 100 – x
If we put x = 0, then we get y = 100 – 0 = 100
If we put x = 50, then we get y = 100 – 50 = 50
If we put x = 100, then we get y = 100 – 100 = 0
∴ We have the following table:

x

0

50

100

y

100

50

0

Plotting the ordered pairs (0, 100), (50, 50) and (100, 0) on a graph paper using proper scale.

Joining these points, we get a straight line PQ as shown below.

Thus, the line PQ is the required graph of the linear equation x + y = 100.


Ex 4.3 Class 9 Maths Question 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.

Solution:
(i) We have F = (9/5 )C + 32
If C = 0, then F = (9/5) × 0 + 32 = 32
If C = -15, then F = (9/5)(-15) + 32 = -27 + 32 = 5
If C = -10, then F = (9/5)(-10) + 32 = -18 + 32 = 14
We have the following table:

x

0

-15

-10

y

32

5

14

Plotting the ordered pairs (0, 32), (-15, 5) and (-10, 14) on a graph paper.

Joining these points, we get a straight line AB as shown below.

(ii) From the graph, we have 86°F corresponds to 30°C.
(iii) From the graph, we have 95°F corresponds 35°C.
(iv) We have, C = 0
From (1), we get
F = (9/5)0 + 32 = 32
Also, F = 0
From (1), we get
0 = (9/5)C + 32

⇒ −32 × 5/9 = C

⇒ C = -17.8
(v) If F = C (numerically)
From (1), we get
F = (9/5)F + 32

⇒ F – (9/5)F = 32
⇒ (−4/5)F = 32

⇒ F = −40
∴ Temperature of –40°F is the same as –40°C.


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