Page 1
24 MATHEMA TICS
3
3.1 Introduction
You must have come across situations like the one given below :
Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel
and play Hoopla (a game in which you throw a ring on the items kept in a stall, and if
the ring covers any object completely, you get it). The number of times she played
Hoopla is half the number of rides she had on the Giant Wheel. If each ride costs
` 3, and a game of Hoopla costs ` 4, how would you find out the number of rides she
had and how many times she played Hoopla, provided she spent ` 20.
May be you will try it by considering different cases. If she has one ride, is it
possible? Is it possible to have two rides? And so on. Or you may use the knowledge
of Class IX, to represent such situations as linear equations in two variables.
PAIR OF LINEAR EQUATIONS
IN TWO VARIABLES
Rationalised 202324
Page 2
24 MATHEMA TICS
3
3.1 Introduction
You must have come across situations like the one given below :
Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel
and play Hoopla (a game in which you throw a ring on the items kept in a stall, and if
the ring covers any object completely, you get it). The number of times she played
Hoopla is half the number of rides she had on the Giant Wheel. If each ride costs
` 3, and a game of Hoopla costs ` 4, how would you find out the number of rides she
had and how many times she played Hoopla, provided she spent ` 20.
May be you will try it by considering different cases. If she has one ride, is it
possible? Is it possible to have two rides? And so on. Or you may use the knowledge
of Class IX, to represent such situations as linear equations in two variables.
PAIR OF LINEAR EQUATIONS
IN TWO VARIABLES
Rationalised 202324
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 25
Let us try this approach.
Denote the number of rides that Akhila had by x, and the number of times she
played Hoopla by y. Now the situation can be represented by the two equations:
y =
1
2
x (1)
3x + 4y = 20 (2)
Can we find the solutions of this pair of equations? There are several ways of
finding these, which we will study in this chapter.
3.2 Graphical Method of Solution of a Pair of Linear Equations
A pair of linear equations which has no solution, is called an inconsistent pair of linear
equations. A pair of linear equations in two variables, which has a solution, is called a
consistent pair of linear equations. A pair of linear equations which are equivalent has
infinitely many distinct common solutions. Such a pair is called a dependent pair of
linear equations in two variables. Note that a dependent pair of linear equations is
always consistent.
W e can now summarise the behaviour of lines representing a pair of linear equations
in two variables and the existence of solutions as follows:
(i) the lines may intersect in a single point. In this case, the pair of equations
has a unique solution (consistent pair of equations).
(ii) the lines may be parallel. In this case, the equations have no solution
(inconsistent pair of equations).
(iii) the lines may be coincident. In this case, the equations have infinitely many
solutions [dependent (consistent) pair of equations].
Consider the following three pairs of equations.
(i) x – 2y = 0 and 3x + 4y – 20 = 0 (The lines intersect)
(ii) 2x + 3y – 9 = 0 and 4x + 6y – 18 = 0 (The lines coincide)
(iii) x + 2y – 4 = 0 and 2x + 4y – 12 = 0 (The lines are parallel)
Let us now write down, and compare, the values of
11 1
2 2
2
,
and
ab c
c a
b
in all the
three examples. Here, a
1
, b
1
, c
1
and a
2
, b
2
, c
2
denote the coefficents of equations
given in the general form in Section 3.2.
Rationalised 202324
Page 3
24 MATHEMA TICS
3
3.1 Introduction
You must have come across situations like the one given below :
Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel
and play Hoopla (a game in which you throw a ring on the items kept in a stall, and if
the ring covers any object completely, you get it). The number of times she played
Hoopla is half the number of rides she had on the Giant Wheel. If each ride costs
` 3, and a game of Hoopla costs ` 4, how would you find out the number of rides she
had and how many times she played Hoopla, provided she spent ` 20.
May be you will try it by considering different cases. If she has one ride, is it
possible? Is it possible to have two rides? And so on. Or you may use the knowledge
of Class IX, to represent such situations as linear equations in two variables.
PAIR OF LINEAR EQUATIONS
IN TWO VARIABLES
Rationalised 202324
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 25
Let us try this approach.
Denote the number of rides that Akhila had by x, and the number of times she
played Hoopla by y. Now the situation can be represented by the two equations:
y =
1
2
x (1)
3x + 4y = 20 (2)
Can we find the solutions of this pair of equations? There are several ways of
finding these, which we will study in this chapter.
3.2 Graphical Method of Solution of a Pair of Linear Equations
A pair of linear equations which has no solution, is called an inconsistent pair of linear
equations. A pair of linear equations in two variables, which has a solution, is called a
consistent pair of linear equations. A pair of linear equations which are equivalent has
infinitely many distinct common solutions. Such a pair is called a dependent pair of
linear equations in two variables. Note that a dependent pair of linear equations is
always consistent.
W e can now summarise the behaviour of lines representing a pair of linear equations
in two variables and the existence of solutions as follows:
(i) the lines may intersect in a single point. In this case, the pair of equations
has a unique solution (consistent pair of equations).
(ii) the lines may be parallel. In this case, the equations have no solution
(inconsistent pair of equations).
(iii) the lines may be coincident. In this case, the equations have infinitely many
solutions [dependent (consistent) pair of equations].
Consider the following three pairs of equations.
(i) x – 2y = 0 and 3x + 4y – 20 = 0 (The lines intersect)
(ii) 2x + 3y – 9 = 0 and 4x + 6y – 18 = 0 (The lines coincide)
(iii) x + 2y – 4 = 0 and 2x + 4y – 12 = 0 (The lines are parallel)
Let us now write down, and compare, the values of
11 1
2 2
2
,
and
ab c
c a
b
in all the
three examples. Here, a
1
, b
1
, c
1
and a
2
, b
2
, c
2
denote the coefficents of equations
given in the general form in Section 3.2.
Rationalised 202324
26 MATHEMA TICS
Table 3.1
Sl Pair of lines
1
2
a
a
1
2
b
b
1
2
c
c
Compare the Graphical Algebraic
No. ratios representation interpretation
1. x – 2y = 0
1
3
2
4
 0
20 
1 1
2 2
a b
a b
?
Intersecting Exactly one
3x + 4y – 20 = 0
lines solution
(unique)
2. 2x + 3y – 9 = 0
2
4
3
6
9
18


1 1 1
2 2 2
a b c
a b c
= =
Coincident Infinitely
4x + 6y – 18 = 0
lines many solutions
3. x + 2y – 4 = 0
1
2
2
4
4
12


1 1 1
2 2 2
a b c
a b c
= ?
Parallel lines No solution
2x + 4y – 12 = 0
From the table above, you can observe that if the lines represented by the equation
a
1
x + b
1
y + c
1
= 0
and a
2
x + b
2
y + c
2
= 0
are (i) intersecting, then
1 1
2 2
a b
a b
? ·
(ii) coincident, then
1 1 1
2 2 2
a b c
a b c
= = ·
(iii) parallel, then
1 1 1
2 2 2
a b c
a b c
= ? ·
In fact, the converse is also true for any pair of lines. You can verify them by
considering some more examples by yourself.
Let us now consider some more examples to illustrate it.
Example 1 : Check graphically whether the pair of equations
x + 3y = 6 (1)
and 2x – 3y = 12 (2)
is consistent. If so, solve them graphically.
Solution : Let us draw the graphs of the Equations (1) and (2). For this, we find two
solutions of each of the equations, which are given in Table 3.2
Rationalised 202324
Page 4
24 MATHEMA TICS
3
3.1 Introduction
You must have come across situations like the one given below :
Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel
and play Hoopla (a game in which you throw a ring on the items kept in a stall, and if
the ring covers any object completely, you get it). The number of times she played
Hoopla is half the number of rides she had on the Giant Wheel. If each ride costs
` 3, and a game of Hoopla costs ` 4, how would you find out the number of rides she
had and how many times she played Hoopla, provided she spent ` 20.
May be you will try it by considering different cases. If she has one ride, is it
possible? Is it possible to have two rides? And so on. Or you may use the knowledge
of Class IX, to represent such situations as linear equations in two variables.
PAIR OF LINEAR EQUATIONS
IN TWO VARIABLES
Rationalised 202324
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 25
Let us try this approach.
Denote the number of rides that Akhila had by x, and the number of times she
played Hoopla by y. Now the situation can be represented by the two equations:
y =
1
2
x (1)
3x + 4y = 20 (2)
Can we find the solutions of this pair of equations? There are several ways of
finding these, which we will study in this chapter.
3.2 Graphical Method of Solution of a Pair of Linear Equations
A pair of linear equations which has no solution, is called an inconsistent pair of linear
equations. A pair of linear equations in two variables, which has a solution, is called a
consistent pair of linear equations. A pair of linear equations which are equivalent has
infinitely many distinct common solutions. Such a pair is called a dependent pair of
linear equations in two variables. Note that a dependent pair of linear equations is
always consistent.
W e can now summarise the behaviour of lines representing a pair of linear equations
in two variables and the existence of solutions as follows:
(i) the lines may intersect in a single point. In this case, the pair of equations
has a unique solution (consistent pair of equations).
(ii) the lines may be parallel. In this case, the equations have no solution
(inconsistent pair of equations).
(iii) the lines may be coincident. In this case, the equations have infinitely many
solutions [dependent (consistent) pair of equations].
Consider the following three pairs of equations.
(i) x – 2y = 0 and 3x + 4y – 20 = 0 (The lines intersect)
(ii) 2x + 3y – 9 = 0 and 4x + 6y – 18 = 0 (The lines coincide)
(iii) x + 2y – 4 = 0 and 2x + 4y – 12 = 0 (The lines are parallel)
Let us now write down, and compare, the values of
11 1
2 2
2
,
and
ab c
c a
b
in all the
three examples. Here, a
1
, b
1
, c
1
and a
2
, b
2
, c
2
denote the coefficents of equations
given in the general form in Section 3.2.
Rationalised 202324
26 MATHEMA TICS
Table 3.1
Sl Pair of lines
1
2
a
a
1
2
b
b
1
2
c
c
Compare the Graphical Algebraic
No. ratios representation interpretation
1. x – 2y = 0
1
3
2
4
 0
20 
1 1
2 2
a b
a b
?
Intersecting Exactly one
3x + 4y – 20 = 0
lines solution
(unique)
2. 2x + 3y – 9 = 0
2
4
3
6
9
18


1 1 1
2 2 2
a b c
a b c
= =
Coincident Infinitely
4x + 6y – 18 = 0
lines many solutions
3. x + 2y – 4 = 0
1
2
2
4
4
12


1 1 1
2 2 2
a b c
a b c
= ?
Parallel lines No solution
2x + 4y – 12 = 0
From the table above, you can observe that if the lines represented by the equation
a
1
x + b
1
y + c
1
= 0
and a
2
x + b
2
y + c
2
= 0
are (i) intersecting, then
1 1
2 2
a b
a b
? ·
(ii) coincident, then
1 1 1
2 2 2
a b c
a b c
= = ·
(iii) parallel, then
1 1 1
2 2 2
a b c
a b c
= ? ·
In fact, the converse is also true for any pair of lines. You can verify them by
considering some more examples by yourself.
Let us now consider some more examples to illustrate it.
Example 1 : Check graphically whether the pair of equations
x + 3y = 6 (1)
and 2x – 3y = 12 (2)
is consistent. If so, solve them graphically.
Solution : Let us draw the graphs of the Equations (1) and (2). For this, we find two
solutions of each of the equations, which are given in Table 3.2
Rationalised 202324
PAIR OF LINEAR EQUA TIONS IN TWO VARIABLES 27
Fig. 3.1
Table 3.2
x 0 6 x 0 3
y =
6
3
x 
2 0 y =
2 12
3
x 
– 4 –2
Plot the points A(0, 2), B(6, 0),
P(0, – 4) and Q(3, – 2) on graph
paper, and join the points to form the
lines AB and PQ as shown in
Fig. 3.1.
W e observe that there is a point
B (6, 0) common to both the lines
AB and PQ. So, the solution of the
pair of linear equations is x = 6 and
y = 0, i.e., the given pair of equations
is consistent.
Example 2 : Graphically, find whether the following pair of equations has no solution,
unique solution or infinitely many solutions:
5x – 8y + 1 = 0 (1)
3x –
24
5
y
+
3
5
= 0 (2)
Solution : Multiplying Equation (2) by
5
,
3
we get
5x – 8y + 1 = 0
But, this is the same as Equation (1). Hence the lines represented by Equations (1)
and (2) are coincident. Therefore, Equations (1) and (2) have infinitely many solutions.
Plot few points on the graph and verify it yourself.
Example 3 : Champa went to a ‘Sale’ to purchase some pants and skirts. When her
friends asked her how many of each she had bought, she answered, “The number of
skirts is two less than twice the number of pants purchased. Also, the number of skirts
is four less than four times the number of pants purchased”. Help her friends to find
how many pants and skirts Champa bought.
Rationalised 202324
Page 5
24 MATHEMA TICS
3
3.1 Introduction
You must have come across situations like the one given below :
Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel
and play Hoopla (a game in which you throw a ring on the items kept in a stall, and if
the ring covers any object completely, you get it). The number of times she played
Hoopla is half the number of rides she had on the Giant Wheel. If each ride costs
` 3, and a game of Hoopla costs ` 4, how would you find out the number of rides she
had and how many times she played Hoopla, provided she spent ` 20.
May be you will try it by considering different cases. If she has one ride, is it
possible? Is it possible to have two rides? And so on. Or you may use the knowledge
of Class IX, to represent such situations as linear equations in two variables.
PAIR OF LINEAR EQUATIONS
IN TWO VARIABLES
Rationalised 202324
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 25
Let us try this approach.
Denote the number of rides that Akhila had by x, and the number of times she
played Hoopla by y. Now the situation can be represented by the two equations:
y =
1
2
x (1)
3x + 4y = 20 (2)
Can we find the solutions of this pair of equations? There are several ways of
finding these, which we will study in this chapter.
3.2 Graphical Method of Solution of a Pair of Linear Equations
A pair of linear equations which has no solution, is called an inconsistent pair of linear
equations. A pair of linear equations in two variables, which has a solution, is called a
consistent pair of linear equations. A pair of linear equations which are equivalent has
infinitely many distinct common solutions. Such a pair is called a dependent pair of
linear equations in two variables. Note that a dependent pair of linear equations is
always consistent.
W e can now summarise the behaviour of lines representing a pair of linear equations
in two variables and the existence of solutions as follows:
(i) the lines may intersect in a single point. In this case, the pair of equations
has a unique solution (consistent pair of equations).
(ii) the lines may be parallel. In this case, the equations have no solution
(inconsistent pair of equations).
(iii) the lines may be coincident. In this case, the equations have infinitely many
solutions [dependent (consistent) pair of equations].
Consider the following three pairs of equations.
(i) x – 2y = 0 and 3x + 4y – 20 = 0 (The lines intersect)
(ii) 2x + 3y – 9 = 0 and 4x + 6y – 18 = 0 (The lines coincide)
(iii) x + 2y – 4 = 0 and 2x + 4y – 12 = 0 (The lines are parallel)
Let us now write down, and compare, the values of
11 1
2 2
2
,
and
ab c
c a
b
in all the
three examples. Here, a
1
, b
1
, c
1
and a
2
, b
2
, c
2
denote the coefficents of equations
given in the general form in Section 3.2.
Rationalised 202324
26 MATHEMA TICS
Table 3.1
Sl Pair of lines
1
2
a
a
1
2
b
b
1
2
c
c
Compare the Graphical Algebraic
No. ratios representation interpretation
1. x – 2y = 0
1
3
2
4
 0
20 
1 1
2 2
a b
a b
?
Intersecting Exactly one
3x + 4y – 20 = 0
lines solution
(unique)
2. 2x + 3y – 9 = 0
2
4
3
6
9
18


1 1 1
2 2 2
a b c
a b c
= =
Coincident Infinitely
4x + 6y – 18 = 0
lines many solutions
3. x + 2y – 4 = 0
1
2
2
4
4
12


1 1 1
2 2 2
a b c
a b c
= ?
Parallel lines No solution
2x + 4y – 12 = 0
From the table above, you can observe that if the lines represented by the equation
a
1
x + b
1
y + c
1
= 0
and a
2
x + b
2
y + c
2
= 0
are (i) intersecting, then
1 1
2 2
a b
a b
? ·
(ii) coincident, then
1 1 1
2 2 2
a b c
a b c
= = ·
(iii) parallel, then
1 1 1
2 2 2
a b c
a b c
= ? ·
In fact, the converse is also true for any pair of lines. You can verify them by
considering some more examples by yourself.
Let us now consider some more examples to illustrate it.
Example 1 : Check graphically whether the pair of equations
x + 3y = 6 (1)
and 2x – 3y = 12 (2)
is consistent. If so, solve them graphically.
Solution : Let us draw the graphs of the Equations (1) and (2). For this, we find two
solutions of each of the equations, which are given in Table 3.2
Rationalised 202324
PAIR OF LINEAR EQUA TIONS IN TWO VARIABLES 27
Fig. 3.1
Table 3.2
x 0 6 x 0 3
y =
6
3
x 
2 0 y =
2 12
3
x 
– 4 –2
Plot the points A(0, 2), B(6, 0),
P(0, – 4) and Q(3, – 2) on graph
paper, and join the points to form the
lines AB and PQ as shown in
Fig. 3.1.
W e observe that there is a point
B (6, 0) common to both the lines
AB and PQ. So, the solution of the
pair of linear equations is x = 6 and
y = 0, i.e., the given pair of equations
is consistent.
Example 2 : Graphically, find whether the following pair of equations has no solution,
unique solution or infinitely many solutions:
5x – 8y + 1 = 0 (1)
3x –
24
5
y
+
3
5
= 0 (2)
Solution : Multiplying Equation (2) by
5
,
3
we get
5x – 8y + 1 = 0
But, this is the same as Equation (1). Hence the lines represented by Equations (1)
and (2) are coincident. Therefore, Equations (1) and (2) have infinitely many solutions.
Plot few points on the graph and verify it yourself.
Example 3 : Champa went to a ‘Sale’ to purchase some pants and skirts. When her
friends asked her how many of each she had bought, she answered, “The number of
skirts is two less than twice the number of pants purchased. Also, the number of skirts
is four less than four times the number of pants purchased”. Help her friends to find
how many pants and skirts Champa bought.
Rationalised 202324
28 MATHEMA TICS
Solution : Let us denote the number of pants by x and the number of skirts by y. Then
the equations formed are :
y = 2x – 2 (1)
and y = 4x – 4 (2)
Let us draw the graphs of
Equations (1) and (2) by finding two
solutions for each of the equations.
They are given in Table 3.3.
Table 3.3
x 2 0
y = 2x – 2 2 – 2
x 0 1
y = 4x – 4 – 4 0
Plot the points and draw the lines passing through them to represent the equations,
as shown in Fig. 3.2.
The two lines intersect at the point (1, 0). So, x = 1, y = 0 is the required solution
of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did
not buy any skirt.
Verify the answer by checking whether it satisfies the conditions of the given
problem.
EXERCISE 3.1
1. Form the pair of linear equations in the following problems, and find their solutions
graphically.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4
more than the number of boys, find the number of boys and girls who took part in
the quiz.
Fig. 3.2
Rationalised 202324
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