Linear Equations in Two Variables
Exercise  3 
Q1. Draw the graph of each of the following linear equation in two variables:
(i) x + y = 4 (ii) x – y = 2 (iii) y = 3x (iv) 3 = 2x + y
Solution: (i) The given equation is
x + y = 4
y = 4 – x …………….equation (1)
Now , putting the value x = 0 in equation (1)
y = 4 – 0 = 4. So the solution is (0, 4)
Putting the value x = 1 in equation (1)
y = 4 – 1 = 3. So the solution is (1, 3)
Putting the value x = 2 in equation (1)
y = 4 – 2 = 2. So the solution is (2, 2)
So, the table of the different solutions of the equation is
(ii) The given equation is
x  y = 2
x = 2 + y …………….equation (1)
Now , putting the value y = 0 in equation (1)
x = 2 + 0 = 2. So the solution is (2, 0)
Putting the value y = 1 in equation (1)
x = 2 + 1 = 3. So the solution is (3, 1)
Putting the value y = 2 in equation (1)
x = 2 + 2 = 4. So the solution is (4, 2)
So, the table of the different solutions of the equation is
(iii) The given equation is
y = 3x
y = 3x …………….equation (1)
Now , putting the value x = 0 in equation (1)
y = 3 x 0 = 0. So the solution is (0, 0)
Putting the value x = 1 in equation (1)
y = 3 x 1 = 3. So the solution is (1, 3)
Putting the value x = 2 in equation (1)
y = 3 x 2 = 6. So the solution is (2, 6)
So, the table of the different solutions of the equation is
(iv) The given equation is
3 = 2 x + y
2 x + y = 3
y = 3 – 2x …………….equation (1)
Now, putting the value x = 0 in equation (1)
y = 3 – 2 x 0
y = 3 – 0 = 3. So the solution is (0, 3)
Putting the value x = 1 in equation (1)
y = 3 – 2 x 1
y = 3 – 2 = 1. So the solution is (1, 1)
Putting the value x = 2 in equation (1)
y = 3 – 2 x 2
y = 3 – 4 = 1. So the solution is (2, 1)
So, the table of the different solutions of the equation is
Q2. Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
Sol. Since the given solution is (2, 14)
Therefore, x = 2 and y = 14
Or, One equation is
x + y = 2 + 14 = 16
x + y = 16
Second equation is
x – y = 2 – 14 = 12
x  y = 12
Third equation is
y = 7x
0 = 7x – y
7x  y = 0
In fact we can find infinite equations because through one point infinite lines pass.
Q3. If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a?
Sol. The given equation is
3y = ax + 7 ……………….equation (1)
According to problem, point (3, 4) lie on it.
So, putting the value x = 3 and y = 4 in equation (1)
3 x 4 = a x 3 + 7
12 = 3a + 7
12 – 7 = 3a
5 = 3a
Q4. The taxi fare in a city is as follows: For the first kilometer, the fare is Rs. 8 and for the subsequent distance it is Rs. 5 per kilometer. Taking the distance covered as x km and total fares as Rs. y, write a linear equation for this information, and draw its graph.
Sol. Given,
Taxi fare for first kilometer = Rs. 8
Taxi fare for subsequent distance = Rs. 5
Total distance covered = x
Total fare = y
Since the fare for first kilometer = Rs. 8
According to problem,
Fare for (x – 1) kilometer = 5(x1)
So, the total fare
= 5(x1) + 8
y = 5(x1) + 8
y = 5x – 5 + 8
y = 5x + 3
Hence,
y = 5x + 3 is the required linear equation.
Now the equation is
y = 5x + 3 …………….equation (1)
Now, putting the value x = 0 in equation (1)
y = 5 x 0 + 3
y = 0 + 3 = 3 So the solution is (0, 3)
Putting the value x = 1 in equation (1)
y = 5 x 1 + 3
y = 5 + 3 = 8. So the solution is (1, 8)
Putting the value x = 2 in equation (1)
y = 5 x 2 + 3
y = 10 + 3 = 13. So the solution is (2, 13)
So, the table of the different solutions of the equation is
Q5. From the choices given below, choose the equation whose graphs are given in Fig. 1 and 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
Sol. From the given Fig. 1, the solutions of the equation are (1, 1), (0, 0) and (1, 1)
Therefore the equation which satisfies these solutions is the correct equation.
Equation (ii)
x + y = 0 , satisfies these solutions.
Proof:
Putting the value x = 1 and y = 1 in the equation x + y = 0
L.H.S =
x + y = 1 + 1 = 0 = R.H.S
Putting the value x = 0 and y = 0
L.H.S =
x + y = 0 + 0 = 0 = R.H.S
Putting the value x = 1 and y = 1
L.H.S =
x + y = 1 + (1) = 1 – 1 = 0 = R.H.S
Hence, option (ii) x + y = 0 is correct.
From the given Fig. 2, the solutions of the equation are (1, 3), (0, 2) and (2, 0)
Therefore the equation which satisfies these solutions is the correct equation.
Equation (i) y = x + 2 , satisfies these solutions.
Proof:
Putting the value x = 1 and y = 3 in the equation
y = x + 2
x + y = 2
L.H.S =
x + y = 1 + 3 = 2 = R.H.S
Putting the value x = 0 and y = 2
L.H.S =
x + y = 0 + 2 = 2 = R.H.S
Putting the value x = 2 and y = 0
L.H.S =
x + y = 2 + 0 = 2 = R.H.S
Hence, option (i) y = x + 2 is correct.
Q6. 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 is 5 units. Read from the graph the work done when the distance traveled by the body is
(i) 2 units (ii) 0 units
Sol. Let x units be the distance traveled by the body and y units be the work done by constant force.
According to problem, y = 5x …………….equation (1)
Now , putting the value x = 0 in equation (1)
y = 5 x 0 = 0. So the solution is (0, 0)
Putting the value x = 1 in equation (1)
y = 5 x 1 = 5. So the solution is (1, 5)
Putting the value x = 2 in equation (1)
y = 5 x 2 = 10. So the solution is (2, 10)
So, the table of the different solutions of the equation is
(i) When x = 2 units (distance)
Putting the value x in equation (1)
y = 5x
y = 5 x 2 = 10.
Hence, Work done = 10.
(ii) When x = 0 units (distance)
Putting the value x in equation (1)
y = 5x
y = 5 x 0 = 0.
Hence, Work done = 0.
Q7. 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 this data satisfies. (You may take their contributions as Rs. x and Rs. y). Draw the graph of the same.
Sol. Let the contribution of Yamini be x and that of Fatima be y.
According to problem, x + y = 100 …………..(1)
Now , putting the value x = 0 in equation (1)
0 + y = 100
y = 100 . So the solution is (0, 100)
Putting the value x = 50 in equation (1)
50 + y = 100,
y = 100 – 50
y = 50 . So the solution is (50, 50)
Putting the value x = 100 in equation (1)
100 + y = 100,
Or, y = 100 – 100
y = 0 . So the solution is (100, 0)
So, the table of the different solutions of the equation is
Q8. In countries like the 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 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.
Sol. Given equation is
(i) We have to take Celsius along xaxis and Fahrenheit along yaxis
Let C be x and F be y
So, the table of the different solutions of the equation is
(ii) When C = 30°,
(iii) When F = 95°,
(iv) When C = 0,
When x° F = x° C
Exercise 4
Q1. Give the geometric representations of y = 3 as an equation.
(i) in one variable (ii) in two variables.
Sol. (i) y = 3
(ii) y = 3
Or, 0x + y = 3
Or, 0x + y – 3 = 0
which is in fact y = 3
It is a line parallel to xaxis at a positive distance of 3 from it. We have two solution for it. i.e. (0, 3), (1, 3).
Q2. Give the geometric representation of 2x + 9 = 0 as and equation,
(i) in one variable (ii) in two variables.
Sol. (i)
2x + 9 = 0
Or, 2x = 9
Or, x =  9/2 =  4.5
(ii) Given equation is
2x + 9 = 0
2x + 0y + 9 = 0 {we know that it is actually 2x + 9 = 0}
Or, x =  9/2 =  4.5
It is line parallel to yaxis at a negative distance we have the two points lying it, the points are A(4.5, 0), B(4.5, 2).
