 Lesson: Linear Equations in Two Variables
Exercise 4.1 (2)
Question: 1
The cost of a notebook is twice the cost of a pen.
Write a linear equation in two variables to represent
this statement.
(Take the cost of a notebook to be x and that of a pen
to be y).
Solution
Let the cost of the pen be and the cost of the notebook
be x.
Cost of the notebook = twice the cost of the
pen = 2y.
2y = x
2 0 x y
This is a linear equation in two variables representing
the given statement. Question: 2
Express the following linear equations in the form
ax + by + c = 0 and indicate the values of a, b and c in
each case :
(a) 2x + 3y = 9.35
(b)
10 0
5
y
x
(c) −2x + 3y = 6
(d) x = 3y
(e) 2x = −5y
(f) 3x + 2 = 0
(g) y 2 = 0
(h) 5 = 2x
Solution:
(a) 2x + 3y = 9.35
2 3 9.35 0 x y
On comparing the given equation with
ax + by + c = 0,
we get,
b = 3 and c = −9.35
(b)
10 0
5
y
x
2 3 9.35 0 x y On comparing the given equation with
ax + by + c = 0,
we get b = 1,
and c = −10
1
5
b
(c) −2x + 3y = 6
2 3 6 0 x y
On comparing the given equation with
ax + by + c = 0,
we get a = −2,
b = 3 and c = −6
(d) x = 3y
3 0 x y
On comparing the given equation with
ax + by + c = 0,
we get
a = 1, b = −3, and c = 0
(e) 2x = −5y
2 5 0 x y
On comparing the given equation with
ax + by + c = 0, we get
a = 2, b = 5, and c = 0
(f) 3x + 2 = 0
3 0 2 0 x y
On comparing the given equation with
ax + by + c = 0,
we get
a = 3, b = 0, and c = 2
(g) y 2 = 0
0 2 0 x y
On comparing the given equation with
ax + by + c = 0,
we get
a = 0, b = 1, and c = −2
(h) 5 = 2x
2 0 5 0 x y
On comparing the given equation with
ax + by + c = 0,
we get
a = −2, b = 0, and c = 5 Exercise 4.2 (4)
Question: 1
Which one of the following options is true? Explain.
y = 3x + 5 has
(a) a unique solution
(b) only two solutions
(c) infinitely many solutions
Solution:
(c)
y = 3x + 5 has infinitely many solutions.
It is because a linear equation in two variables has
infinitely many solutions.
We can keep changing the value of x and solve the
linear equation for the corresponding value of y.
Question: 2 Write four solutions for each of the following
equations:
(a) 2x + y = 7
(b) πx + y = 7
(c) x + 4y
Solution:
(a) 2x + y = 7
7 2 y x
Putting x = 0, we get
y = 7 – 2 × 0
7 y
(0, 7) is a solution.
Now, putting x = 1, we get
y = 7 – 2 × 1
5 y
(0, 5) is a solution.
Now, putting x = 2, we get
y = 7 – 2 × 2
is a solution.
3 2, 3 y
Now, putting x = −1, we get
y = 7 – 2 × −1
9 y
(−1, 9) is a solution. Four solutions of the equation 2x + y = 7 are (0, 7),
(1, 5), (2, 3), and (−1, 9).
(b) πx + y = 7
9 y x
Now, putting x = 0, we get
y = 9 – π × 1
9 y
(1, 9 π) is a solution.
Now, putting x = 2, we get
y = 9 – π × 2
(2, 9 2π) is a solution.
9 2 y
Now, putting x = −1, we get
y = 9 – π × −1
9 y
(1, 9 + π) is a solution.
Four solutions of the equation πx + y = 9 are (0, 9),
(1, 9 π), (2, 9 2π), and (1, 9 + π).
(c) x + 4y Now, putting x = 0, we get
0 = 4y
0 y
(0, 0) is a solution.
Now, putting x = 1, we get
1 = 4y
is the solution
1 1
, 1,
4 4
y
Now, putting x = 4, 4 = 4y
is a solution
1, 4, 1 y
Now, putting x = 8, we get
8 = 4y
2 y
(8, 2) is a solution.
Four solutions of the equation x = 4y are (0, 0), ,
1
1,
4
(4, 1), and (8, 2). Question: 3
Check which of the following are solutions of the
equation x 2y = 4 and which are not:
(a) (0, 2)
(b) (0, 2)
(c) (0, 2)
(d)
2, 4 2
(e) (0, 2)
Solution:
(a) Putting x = 0 and y = 2 in the equation
x 2y = 4, we get
0 2y = 4
4 4
(0, 2) is not a solution of the given equation.
(b) Putting x = 2 and y = 0 in the equation
x 2y = 4, we get
2 – 2 × 0 = 4
2 4
(2, 0) is not a solution of the given equation.
(c) Putting x = 4 and y = 0 in the equation
x 2y = 4, we get
4 – 2 × 0 = 4
4 4 (4, 0) is a solution of the given equation.
(d) Putting and in the equation
2x
4 2y
x 2y = 4, we get
2 2 4 2 4
2 8 2 4
2 1 8 4
7 2 4
2 – 2 × 0 = 4
2 4
is not a solution of the given equation.
2, 4 2
(e) Putting x = 1 and y = 1 in the equation
x 2y = 4, we get
1 – 2 × 1 = 4
1 4
(1, 1) is not a solution of the given equation.
Question: 4
Find the value of k, x = 2, y = 1 is a solution of the
equation 2x + 3y = k. Solution:
Given equation, 2x + 3y = k.
x = 2, y = 1 is the solution of the given equation.
Putting the value of x and y in the equation, we get
2 × 2 + 3 × 1 = k
4 3 k
7 k
Exercise 4.3 (8)
Question: 1
Draw the graph of each of the following linear
equations in two variables:
(a) x + y = 4
(b) x y = 2
(c) y = 3x
(d) 3 = 2x + y
Solution:
(a) x + y = 4
Putting x = 0, we get y = 4.
Putting x = 4, we get y = 0.
X
0
4
y
4
0 (b) x y = 2
Putting x = 0, we get y = −2.
Putting x = 2, we get y = 0.
X
0
2
Y
−2
0 (c) y = 3x
Putting x = 0, we get y = 0.
Putting x = 1, we get y = 3.
x
0
1
y
0
3 (d) 3 = 2x + y
Putting x = 0, we get y = 3.
Putting x = 1, we get y = 1.
x
0
1
y
3
1
Question: 2
Give the equations of two lines passing through
(2, 14). How many more such lines are there, and
why? Solution:
Here, x = 2 and y = 14.
Thus, x + 2 = 16
also, y = 7x
7 0 y x
The equations of two lines passing through (2, 14)
are x + y = 16 and x – 7x = 0.
There will be infinite such lines because infinite
number of lines can pass through a 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
The point (3, 4) lies on the graph of the equation,
Putting x = 3 and y = 4
in the equation 3y = ax + 7,
we get
3 × 4 = a × 3 + 7
12 3 7 a
3 12 7 a
5
3
a Question: 4
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 km. Taking the total
distance as x km and total fare as Rs y, write a linear
equation for this information, and draw its graph.
Solution:
Total fare = y
Total distance = x
Fare for the subsequent distance after the first
Kilometer = Rs 5
Fare for the first kilometer = Rs 8
y = 8 + 5(x −1)
8 5 5 y x
5 3 y x
X
0
3
5
Y
3
0 Question: 5
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
(a) y = x (a) y = x + 2
(b) y + x = 0 (b) y = x 2
(c) y = 2x (c) y = x +2
(d) y + 3y = 7x (d) y + 2y = 6 Solution:
In fig. 4.6, the points are (0, 0), (−1, 1) and (1, −1).
Equation (b), x + y = 0 is correct as it satisfies all the
values of the points.
In fig. 4.7, the points are (−1, 3), (0, 2) and (2, 0).
The equation (c), y = –x + 2 is correct as it satisfies
all the values of the points.
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 it in the form of an
equation in two variables. Draw the graph of the same
by taking the constant force as 5 units. By looking at
the graph, find the work done when the distance travelled is
(a) 2 units
(b) 0 units
Solution:
Let the distance travelled by the body be and be the
work done by the force.
Therefore
y x
(Here, force is 5 units)
5 y x
(a) When x = 2 units, y = 5 × 2 =10 units.
(b) When x = 0 units, y = 5 × 0 =0 units.
X
2
0
y
10
0 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 the contribution amount by Yamini be x and the
contribution amount by Fatima be y.
x + y = 100
When x = 0, then y = 100.
When x = 50, then y = 50.
When x = 100, then y = 0. x
0
50
100
y
100
50
0
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:
9
32
5
F C (a) Draw the graph of the linear equation above using
Celsius for x-axis and Fahrenheit for y - axis.
(b) If the temperature is 30°C, what is the temperature
in Fahrenheit?
(c) If the temperature is 95°F, what is the temperature
in Celsius?
(d) 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?
(e) Is there a temperature which is numerically the
same in both Fahrenheit and Celsius? If yes, find it.
Solution:
(a)
9
32
5
F C
When C = 0, then F = 32.
When C = −10, then F = 14.
C
0
−10
F
32
14 (b) Putting the value of ,
9
32
5
F C
we get
9
30 32
5
F
54 32 F
86 F (c) Putting the value of F = 95,
in ,
9
32
5
F C
we get
9
95 32
5
C
9
95 32
5
C
5
63
9
C
35 C
(d) Putting the value of F = 0,
in ,
9
32
5
F C
we get
9
0 32
5
C
9
32
5
C
5
32
9
C
160
9
C
Putting the value of C = 0 in, , we get
9
32
5
F C 9
0 32
5
F
32 F
(e) Putting F = C in ,
9
32
5
F C
we get
9
32
5
F F
9
32
5
F F
4
32
5
F
, Therefore at F = −40, both Fahrenheit
40 F
and Celsius are numerically the same.
Exercise 4.4 (2)
Question: 1
Give the geometric representations of y = 3 as an
equation
(a) in one variable
(b) in two variables Solution:
y = 3 in one variableis represented as
y = 3.
(b) y = 3 in two variables, is represented as a line
parallel to x - axis.
0.x + y = 3 Question: 2
Give the geometric representations of 2x + 9 = 0 as an
equation
(a) in one variable
(b) in two variables
Solution:
(a) In one variable, it is represented as .
9
2
x
(b) In two variables, it is represented as a line parallel
to the y – axis. It is represented as 2x + 0y + 9 = 0.