NCERT Solutions Class 9th Maths Chapter – 4 Linear Equations in Two Variables Exercise – 4.3

Question 1. Draw the graph of each of the following linear equations in two variables:
(i) x + y = 4

Solution: x + y = 4 ⇒ y = 4 – x
If we have, x = 0, then y = 4 – 0 = 4 (0, 4)
x = 1, then y = 4 – 1 = 3 (1, 3)
x = 2, then y = 4 – 2 = 2 (2, 2)

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

(ii) x – y = 2

Solution: x – y = 2 ⇒ y = x – 2
If we have x = 0, then y = 0 – 2 = -2 (0, -2)
x = 1, then y = 1 – 2 = -1 (1, -1)
x = 2, then y = 2 – 2 = 0 (2, 0)

Thus, the i me is the required graph of x – y = 2

(iii) y = 3x

Solution: y = 3x
If we have x = 0,
then y = 3(0) ⇒ y = 0 (0, 0)
x = 1, then y = 3(1) = 3 (1, 3)
x= -1, then y = 3(-1) = -3 (-1, -3)

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

(iv) 3 = 2x + y

Solution: 3 = 2x + y ⇒ y = 3 – 2x
If we have x = 0, then y = 3 – 2(0) = 3 (0, 3)
x = 1,then y = 3 – 2(1) = 3 – 2 = 1 (1, 1)
x = 2,then y = 3 – 2(2) = 3 – 4 = -1 (2, – 1)

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

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.

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

Solution: The given equation is 3y = ax+7
According to the question, x = 3 and y = 4
Now, substituting the values of x and y in the equation 3y = ax+7,
We get,
3 x 4 = a x 3 + 7
⇒ 12 = 3a + 7
⇒ 3a = 12 – 7
⇒ 3a = 5
⇒ a = 5/3
Thus, the required value of a is 5/3.

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
Fare for 1km = Rs. 8
Remaining distance = (x – 1) km
∴ Fare for (x – 1)km = Rs.5 x(x – 1)
Total taxi fare = Rs. 8 + Rs. 5(x – 1)
According to 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
When x = 0, then y = 5(0) + 3 ⇒ y = 3 (0, 3)
x = -1, then y = 5(-1) + 3 ⇒ y = -2 (-1, -2)
x = -2, then y = 5(-2) + 3 ⇒ y = -7 (-2, -7)

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

Question 5. From the choices given below, choose the equation whose graphs are given ¡n 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 (-1, 1) = -1 + 1 = 0 and (1,-1) = 1 + (-1) = 0]
For Fig.(2), the correct linear equation is y = -x + 2
[As(-1,3) 3 = -1(-1) + 2 = 3 = 3 and (0,2)
⇒ 2 = -(0) + 2 ⇒ 2 = 2]

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

Solution: Constant force is 5 units.
Let the distance travelled = x units and work done = y units.
Work done = Force x Distance
⇒ y = 5 x x ⇒ y = 5x
For drawing the graph, we have y = 5x
When x = 0, then y = 5(0) = 0 (0, 0)
x = 1, then y = 5(1) = 5 (1, 5)
x = -1, then y = 5(-1) = -5 (-1, -5)

(i) 2 units

Solution: Distance travelled =2 units i.e., x = 2
∴ If x = 2, then y = 5(2) = 10
⇒ Work done = 10 units.

(ii) 0 unit

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

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 = Rs. x
and the contribution of Fatima Rs. y
∴ We have x + y = 100 ⇒ y = 100 – x
Now, when x = 0, y = 100 – 0 = 100 (0,100)
x = 50, y = 100 – 50 = 50 (50,50)
x = 100, y = 100 – 100 = 0 (100, 0)
∴ We get the following table:

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

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 = (95 )C + 32
(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.

Solution: We have,
F = (95)C + 32
When C = 0 , F = (95 ) x 0 + 32 = 32 (0, 32)
When C = 15, F = (95 )(-15) + 32= -27 + 32 = 5 (-15, 5)
When C = -10, F = 95 (-10)+32 = -18 + 32 = 14 (-10,14)

(ii) If the temperature Is 30°C, what is the temperature in Fahrenheit?
Solution: From the graph, we have 86°F corresponds to 30°C.

(iii) If the temperature is 95°F, what is the temperature in Celsius?
Solution: From the graph, we have 95°F corresponds 35°C.

(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?
Solution: We have, C = 0
From (1), we get
F = (95)0 + 32 = 32
Also, F = 0
From (1), we get
0 = (95)C + 32 ⇒ −32×59 = C ⇒ C = -17.8

(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find It.
Solution: When F = C (numerically)
From (1), we get
F = 95F + 32 ⇒ F – 95F = 32
⇒ −45F = 32 ⇒ F = -40
∴ Temperature is – 40° both in F and C.