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

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.3 Textbook Exercise Questions and Answers.

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

Question 1.
Draw the graph of each of the following linear equations in two variables :
(i) x + y = 4
Answer:
We have : x + y = 4 ⇒ y = 4 – x
When x = 0, y = 4 – 0 = 4
When x = 2,y = 4 – 2 = 2
When x = 4, y = 4 – 4 = 0
Thus, we have the following table :


Plotting the points (0, 4), (2, 2) and (4, 0) on the graph paper and drawing a line joining them, we obtain the graph of the given equation as shown.

(ii) x – y = 2
Answer:
We have : x – y = 2 ⇒ y = x- 2
When x = 0, y = 0- 2 = -2
When x = 2, y = 2- 2 = 0
When x = 4, y = 4 – 2 = 2
Thus, we have the following table :


Plotting the points (0, – 2), (2, 0) and (4, 2) on the graph paper and drawing a line joining them, we obtain the graph of the given equation as shown.

(iii) y = 3x
Answer:
We have : y = 3x
When x = 0, y – 3(0) = 0
When x = 1, y = 3(1) = 3
When x = – 1, y = 3(- 1) = – 3
Thus, we have the following table :


Plotting the points (0, 0), (1, 3) and (-1, -3) on the graph paper and drawing a line joining them, we obtain the graph of the given equation as shown.

(iv) 3 = 2x + y
Answer:
We have : 3 = 2x + y
⇒ y = 3 – 2x
When x = 0, y = 3 – 2(0) = 3-0 = 3
When x = 3, y = 3 – 2(3) = 3 – 6 = – 3
When* = – l,y = 3 – 2(-1) = 3 + 2 = 5
Thus, we have the following table :


Plotting the points (0, 3), (3, – 3) and (- 1, 5) on the graph paper and drawing a line joining them, we obtain the graph of the given equation as shown.

Question 2.
Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
Answer:
Here (2, 14) is a solution of a linear equation we are looking for. One example of such a linear equation is 7x – y = 0. Note that x + y = 16, 2x + y = 18 and 7x + y = 28 are also satisfied by the coordinates of the point (2, 14). So, any line passing through the point (2, 14) is an example of a linear equation for which (2, 14) is a solution. Thus, there are infinite number of fines through (2, 14).

Question 3.
If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
Answer:
Since (3, 4) fie on the graph corresponding to 3y = ax + 7, therefore (3, 4) satisfies the given equation.
That is, 3(4) = a(3) + 7 ⇒ 12 – 7 = 3a
⇒ 3a = 5 ⇒ a = \(\frac{5}{3}\)
Hence, for a = g, (3, 4) lies on the graph of equation 3y = ax + 7.

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

The linear equation for the above information is given by
y = 8 × 1 + 5(x – 1)
or y = 8 + 5x – 5 ⇒ y = 5x + 3
⇒ 5x – y + 3 = 0

When x = 0, y = 5 × 0 + 3 = 0 + 3 = 3
When x = -1, y = 5 × (-1) + 3 = – 5 + 3 = -2
When x = -2, y = 5 × (- 2) + 3 = – 10 + 3 = – 7
Thus, we have the following table :

Plotting the points (0, 3), (-1,-2) and (- 2, – 7) on the graph paper and drawing a line joining them, we obtain the required graph.

Question 5.
From the choices given below, choose the equation whose graphs are given in Fig. (a) and Fig. (b) :

For Fig. (a) :
(i) y = x
(ii) x + y = 0
(iii) y = 2x
(iv) 2 + 3y = 7x

For Fig. (b) :
(i) y = x + 2
(ii) y = x – 2
(iii) y = -x + 2
(iv) x + 2y = 6
Answer:
For Fig. (a), the correct equation from the choices given is clearly x + y = 0 as it is satisfied by the points (- 1, 1) and (1, – 1) given on the graph. For Fig. (b), the correct equation from the choices given is clearly y = – x + 2 as it is satisfied by the points (- 1, 3), (0, 2) and (2, 0) given on the graph.

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.
Answer:
Let x be the distance and y be continue the work done. Therefore, according to the problem the equation will be
y = 5x.

To draw its graph :
When x = 0, we have : y = 5(0) = 0
When x = 1, we have : y = 5(1) = 5
When x = – 1, we have : y = 5(- 1) = – 5

∴ The table is as under :

Plotting (0, 0), (1, 5) and (- 1, – 5) on the graph paper and drawing a line joining them, we obtain the graph of the equation y = 5x as shown.
From the graph, clearly :
(i) when distance travelled is 2 units, i.e, x = 2, then y = 10.
∴ Work done = 10 units.
(ii) when distance travelled is 0 emit, i.e. x = 0, then y = 0.
∴ Work done = 0 unit.

Question 7.
Yamini and Fatima, two students of Class IX of a school, together contributed ₹ 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 ₹ x and ₹ y.) Draw the graph of the same.
Answer:
Let Yamini and Fatima respectively contributed ₹ x and ₹ y towards the P.M.’s Relief Fund totalling X 100.
∴ The linear equation using the above data is x + y = 100, i.e. y = 100 – x.

To draw its graph :
When x = 0, we have : y = 100 – 0 = 100
When x = 100, we have : y = 100 – 100 = 0
When x = 50, we have : y = 100 – 50 = 50

The table for these values is as under :

Plotting the points (0,100), (100, 0) and (50, 50) on the graph paper and drawing a line joining them, we obtain the graph of the equation x + y = 100 as shown.

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 = \(\left(\frac{9}{5}\right)\)C + 32
(i) Draw the graph of the linear equation above using Celsius for .v-axis and Fahrenheit for y-axis.
Answer:
We have F= \(\left(\frac{9}{5}\right)\)C + 32 ⇒ C = \(\frac{5}{9}\)(F – 32)
We calculate the values of F for different values of C.
When C = – 40, F = \(\frac{9}{5}\) × (- 40) + 32 = – 72 + 32 = – 40
When C = 10, F = \(\frac{9}{5}\) × 10 + 32 = 18 + 32 = 50
The table of values is as under :

We choose a suitable scale on the x-axis (for Celsius) and the same scale on the y-axis (forFahrenheit):

Plotting the points (- 40, – 40) and (10, 50) on the graph paper and joining these points, we obtain the graph of F = \(\frac{9}{5}\)C + 32 as shown in the figure.

(ii) If the temperature is 30 °C, what is the temperature in Fahrenheit ?
Answer:
From the graph, we see that when C = 30° shown by C₁ on the v-axis in the positive direction, then F = 86° shown by the point F₁ on the y-axis in the positive direction.
Hence, 30 °C = 86 °F.

(iii) If the temperature is 95 °F, what is the temperature in Celsius ?
Answer:
From the graph, we see that, when F = 95° shown by F₂ on the y-axis in the positive direction, then C = 35° shown by the point C₂ on the x-axis in the positive direction.
Hence, 95 °F = 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 ?
Answer:
From the graph, clearly 0 °C = 32 °F and 0 °F = – 17.8 °C.

(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
Answer:
Clearly, from the graph, the temperature which is numerically the same in both Fahrenheit and Celsius is – 40°, i.e. – 40 °C = – 40 °F.