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

Ex - 4.3

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)  x + y = 4

We may observe that x = 0, y = 4 and x = 4, y = 0 are solutions of above equation. So our solution table is

                   

x	0	4
y	4	0

Now we may draw the graph of this equation as below.

(ii)   x - y = 2

We may observe that x = 4, y = 2 and x = 2, y = 0 are solutions of above equation. So our solution table is

x	4	2
y	2	0

Now we can draw the graph of above equation as following

(iii)  y = 3x

We may observe that x = - 1, y = - 3 and x = 1, y = 3 are solutions of above equation. So our solution table is
  

x	- 1	1
y	- 3	3

 
Now we may draw the graph of above equation as following

(iv)  3 = 2x + y

We may observe that x = 0, y = 3 and x = 1, y = 1 are solutions of above equation. So our solution table is

               

x	0	1
y	3	1

Now we may construct the graph of this equation as below

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

Solution

We may observe that point (2, 14) satisfies the equation 7x - y = 0 and
2x + y -18 = 0.

So, 7x - y = 0 and 2x + y - 18 = 0 are two lines passing through point (2, 14).
As we know that through one point, infinite number of lines can pass through.
So, there are infinite lines of such type passing through given point.

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

Solution

Putting x = 3, y = 4 in the given equation
    3y = ax + 7
    3 (4) = a (3) + 7  
    5 = 3a
    a = 

Question 4.  The taxi fare in a city is as follows: For the first kilometre, the fares 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

Total distance covered = x km.
    Fair for 1st kilometre = Rs .8
    Fair for rest distance = (x - 1) 5
    Total fair = 8 + (x - 1) 5
    y = 8 + 5x - 5
    y = 5x + 3
    5x - y + 3 = 0   

We may observe that point (0, 3) and  satisfies the above equation. So these are solutions of this equation.   

           

x	0
y	3	0

  Now we may draw the graph of this equation as below

                                    

Here we may find that variable x and y are representing the distance covered and fare paid for that distance respectively and these quantities may not be negative. Hence we will consider only those values of x and y which are lying in 1^st quadrant.

Question 5.  From the choices given below, choose the equation whose graphs are given in given figures 

For the first figure                                    For the second figure
(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

Solution

     

Points on the given line are (- 1, 1), (0, 0), (1, - 1)           
We may observe that coordinates of the points of the graph satisfy the equation x + y = 0.  
So, x + y = 0 is the equation corresponding to graph as shown in the first figure. 

Hence (ii) is the correct answer

Points on the given line are (- 1, 3), (0, 2) and (2, 0). We may observe that coordinates of the points of the graph satisfies the equation y = - x + 2.
So y = - x + 2 is the equation corresponding to the graph shown in the second figure.
Hence,

(iii) is the correct answer.

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 units

Solution

Let distance travelled by body and work done by body be d and w respectively.
Work done  distance traveled 
w  d
w = kd   
Where k is a constant
If constant force is 5 units, work done w = 5d   
Now we may observe that point (1, 5) and (-1, -5) satisfy the above equation. 
So (1, 5) and (-1, -5) are solutions of this equation. 
The graph can be drawn as follows:

(i)     From the graphs we may observe that the value of y corresponding to x = 2 is 10. Thus the work done by the body is 10 units when the distance traveled by it is 2 units.  

(ii)    From the graphs we may observe that the value of y corresponding to x = 0 is 0. Thus the work done by the body is 0 units when the distance traveled by it is 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 Yamini and Fatima contributed x and y repectively towards the Prime Minister's Relief fund.
Amount contributed by Yamini + amount contributed by Fatima = 100 
x + y = 100  
Now we observe that (100, 0) and (0, 100) satisfy the above equation.
So, (100, 0) and (0, 100) are solutions of above equation. 
The graph of equation x + y = 100 can be drawn as follows:

Here we may find that variable x and y are representing the amount contributed by Yamini and Fatima respectively and these quantities may not be negative. Hence we will consider only those values of x and y which are lying in 1^st quadrant.

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:

         
(i)    Draw the graph of the linear equation above using Celsius of x-axis and Fahrenheit for y-axis.
(ii)    If the temperature so 30^oC, what is the temperature in Fahrenheit?
(iii)    If the temperature is 95^oC F, what is the temperature is Celsius?
(iv)    IF the temperature is 0^oC, what is the temperature in Fahrenheit and if the temperature is 0^oF, 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 may observe that point (0, 32) and (-40, -40) satisfy the given equation. So these points are solutions of this equation.
Now we can draw the graph of above equation as below

(ii)    Temperature = 30^oC                                         

        

        So temperature is Fahrenheit is 86^o F.      

(iii)    Temperature = 95^o F    

          

    So temperature in Celsius is 35^oC.

(iv)

       

      So, if C = 0^oC, F = 32^o F   
      If F = 0^o

      

So if F = 0^oF, C = - 17.8^oC

(v)  

         

Yes there is a temperature - 40^o which is numerically same in both Fahrenheit and Celsius.

Ex-4.4