Previous exercise – Linear equations in two variables – Exercise 4.2 – Class IX
 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
For x = 0 ; y = 4
For x = 1 ; y = 3
Therefore, the two coordinates are (0, 4) and (1, 3)
(ii) x – y = 2
For x = 0 ; y = 2
For x = 1 ; y = 2 +1 = 1
Therefore, the two coordinates are (0, 2) and (1, 1)
(iii) y = 3x
If x = 0 then y = 0
If x = 1 then y = 3
Therefore, the two coordinates are (0, 0) and (1, 3)
(iv) 3 = 2x + y
If x = 0 then y = 3
If x = 1 then y = 3 – 2 = 1
Therefore, the two coordinates are (0, 3) and ( 1, 1)
 Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
Solution:
Let x + y = k be a such line
2 + 14 = k
Thus, k = 16
Therefore, (x + y) = 16 passes through (2, 14)
Let 2x + 3y = k* be another line passes through (2, 14)
2×2 + 3×14 = k*
4 + 42 = k*
46 = k*
Therefore, 2x + 3y = k* passes through (2, 14)
There are infinitely many such straight lines can be drawn through a point.
 If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
Solution:
We have (3, 4) lies on 3y = ax + 7
Substitute the values of x and y in 3y = ax + 7
3×4 = a x 3 + 7
12 – 7 = 3a
5 = 3a
a = ^{5}/_{3}
 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:
Fare for the first km = Rs. 8
Total distance covered = x km
After travelling first km, remaining distance = (x – 1) km
Rate of fare = Rs. 5 per km
Fare for (x – 1) km = 5 *(x – 1)
Total fare = Rs y
Fare for the first km + fare for the remaining distance = y
8 + (x – 1) 5 = y
8 + 5x – 5 = y
5x – y + 3 = 0
If x = 1 then 5 – y + 3 = 0
y = 3 – 5 = – 8
y = 8
If x = 2 then 5(2) – y + 3 = 0
10 – y + 3 = 0
13 – y = 0
y = 13
So the points are (1, 8) and (2, 13)
 From the choices given below, choose the equation whose graphs are given in Fig. 4.6 and Fig. 4.7.
For Fig. 4. 6  For Fig. 4.7

(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:
Fig A
(ii) x + y = 0 is the right solution.
if x = 1 then y = x =(1) = 1
If x = 1 then y = x = 1
If x = 0 then y = 0
In fig B
(iii) y = x + 2 is the right solution.
if x = 0 then y = 0 + 2 = 2
if x = 2 then y = 2 + 2 = 0
if x = 1 then y = (1) + 2 = 3
 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:
Let the distance travelled and the work done by the body be x and y respectively.
Work done α distance travelled
y α x
y = kx, where k is a constant
If constant force is 5 units, then work done y = 5x
It can be observed that point (1, 5) and (1 , 5) satisfy the above equation. Therefore, these are the solutions of this equation. The graph of this equation is constructed as follows:
(i) From the graph, it is observed that the value of y corresponding to x = 2 is 10. This implies that the work done by the body is 10 units when the distance travelled by its 2 units.
(ii) From the graph it is observed that the value of y corresponding to x = 0 is 0. This implies that the work done by the body is 0 units when the distance travelled by it is 0 unit.
 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:
x + y = 100 is the linear equation satisfying the given data.
For x = 0 ; y = 100
x = 10 ; y = 90
x = 20 then y = 80
So, the points are (0, 100) , (10, 90), (20, 80)
 In countries like USA and Canada, temperature is measured in Fahrenheit, whereas I 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 xaxis and Fahrenheit for yaxis.
(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:
F = (^{9}/_{5})C + 32
(i) For C = 0 ; F = 32
C = 5 ; F = 41
C = 10 ; F = 50
(ii) If the temperature is 30°C, The corresponding temperature in Fahrenheit is
F = (^{9}/_{5})*30 + 32
= 54 + 32
= 86˚F
(iii) If the temperature is 95°F, the corresponding temperature in Celsius is
95 = (^{9}/_{5})C + 32
95 – 32 = (^{9}/_{5})C
63 = (^{9}/_{5})C
C = 63*(^{5}/_{9}) = 7*5 = 35
C = 35˚C
(iv) If the temperature is 0°C,the corresponding temperature in Fahrenheit
For C = 0˚C then
F = (^{9}/_{5})(0) + 32
F = 32˚F
if the temperature is 0°F, the corresponding temperature in Celsius
F = (^{9}/_{5})C + 32
0 = (^{9}/_{5})C + 32
(^{9}/_{5})C = – 32
C = – 32 *^{5}/_{9}
= – ^{160}/_{9} = 17.8˚C
(v) Let F = x , C = x
Then x = (^{9}/_{5})x + 32
x – ^{9}/_{5}x = 32
x(1 – ^{9}/_{5}) = 32
x(^{4}/_{5}) = 32
x = ‑^{32*5}/_{4} = – 40˚C
Thus, 40˚C and 40˚F are equal temperatures.
Next exercise – Linear equations in two variables – Exercise 4.4 – Class IX
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