Linear Equations in Two Variables – Exercise 4.3 – (MATHEMATICS) – 9th Class

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Exercise 4.3

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

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Sol –
Consider equation:  x + y = 4
⇒   y = 4 –  x
Some points on graph are

x024
y420

 

Plotting the points On graph paper and joining them, we get the required graph (line) as shown in figure.


(ii) x – y  = 2

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Sol –
Consider equation:
x – y  = 2
⇒  y = x  – 2
Some points 0n  graph are

x024
y-202

Plotting the points on graph paper and joining them, we get the required graph (line) as shown in the adjoining figure.


(iii) y = 3x

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Sol –
Consider equation: y = 3x Some points on graph are

x01-1
y03-1

Plotting the points on graph paper and joining them, we get the required graph (line) as shown the adjoining figure.


(iv) 3 = 2x + y.

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Sol –
Consider equation:
3 = 2x  + y
⇒  y = 3  – 2x
Some points on graph are

x012
y31-1

Plotting the points on graph paper and joining them, we get the required graph (line) as shown in the adjoining figure.


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

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Sol. Let equation of a line be x + y + c = 0.
If (2, 14) lies on it,
then 2 + 14 + c  = 0  ⇒  c =  – 16
∴ Equation of a line is x + y – 16 = 0.
Let equation of another line be x – 3y + c = 0.
If (2, 14) lies on it, then 2 – 42 + c = 0 ⇒  c = 40.
∴ Equation of another line is x – 3y + 40 = 0.
There can he infinitely many lines passing through the point (2, 14), because through a point infinitely many lines can be drawn.

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

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Sol. If point (3, 4) lies on the graph of the equation 3y = ax + 7, then the point satisfies this equation.
12 = 3a + 7 \Rightarrow 3a = 5 \Rightarrow a = {5 \over 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.

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Sol. Fare for one kilometre = Rs.8.
Fare for subsequent kilometre = Rs.5.
Total distance covered = x km; total fare = Rs.y.
∴  y = 8 + 5 (x – 1)
⇒   y =  5x + 3 is the required linear equation.
Some points on graph are

x0-11
y3-28

Plotting the points on graph paper and joining them, we get the required graph (line) as shown in the adjoining figure.


5. From the choices given below, choose the equation whose graphs are given in Fig. (1) and Fig. (2).
For Fig.(1)                                                For Fig. (2)
(i) y = x                                                    (i) y = x + 2
(ii) x + y = 0                                            (ii) y = x – 2
(iii) y = 2x                                               (iii) y =  – x + 2
(iv) 2 + 3y = 7x                                      (iv) x + 2y = 6

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Sol. (1) To choose the right equation, we test that the points on the graph satisfy which linear equation from given choices. We notice (-1, 1 ), (0, 0) and (1, – 1) satisfy  equation x + y = 0.
Hence, graph in Fig. (1) is of linear equation x + y = 0.

(2) To choose the right equation, we test, that the points on the graph satisfy which linear equation from given choices. We notice (-1, 3), (0, 2) and (2, 0) satisfy the equation y = – x + 2.
Hence, graph in Fig. (2) is of linear equation y = – x + 2


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 on 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.

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Sol. Let y units be the work done by a body on application of a constant force of 5 units and the distance travelled be x units.
Therefore, we have y = 5x as the required equation. Some points on graph are:

x01-1
y05-5

Plotting these points on the graph paper and joining them, we get the required graph (line) as shown in the adjoining figure.

From the graph
(i) When x = 2, y = 10,
∴ work done = 10 units.
(ii) When x = 0, y = 0,
∴ work done = 0 unit.


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.

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Sol. Let Yamini contributed Rs.x and Fatima contributed Rs.y.
According to the given condition,
x + y = 100
This is the required linear equation.
Some points on graph are:

x010050
y100050


Plotting these points on the graph paper and joining them, we get the required graph (line) as shown in the adjoining figure.


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( {{9 \over 5}} \right)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 temerature 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.

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Sol. –
Given equation is F = {9 \over 5}c + 32

(i) Some points on graph are

xC05-5
yF324123

Plotting these points on the graph paper and joining them, we get the required line as shown below.

(ii) When temperature is 30°C, i.e., C = 30
Substituting this value of C in the given equation., we get
F = {9 \over 5}  × 30 + 32 = 54 + 32 = 865
Hence, the required temperature is 86°F.

(iii) When temperature is 95°F, i.e., F = 95
Substituting this value of F in the given equation, we get
95 = {9 \over 5}c + 32     ⇒     {9 \over 5}c = 95 - 32 = 63
 \Rightarrow c = {{63 \times 5} \over 9} = 35
Hence, the required temperature is 35°C.

(iv) When C = 0, F = {9 \over 5}  ×  0 + 32 = 32
When F =  0, 0 = {9 \over 5} C + 32
 \Rightarrow C = - {{32 \times 5} \over 9} = - 17.8 (approx.):
Hence, the required temperatures arc 32°F and – 17.8°C (approx.).
(a) When F = C numerically, from given equation, we get
F = {9 \over 5}F + 32 \Rightarrow F - {9 \over 5}F = 32
Hence, there is a temperature which is numerically the same in both Fahrenheit and Celsius, i.e., – 40°F = – 40°C.


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