Lesson: Pair of Linear Equations in Two
Variables
Exercise 3.1 (3)
Question: 1
Aftab tells his daughter, “Seven years ago, I was
seven times as old as you were then. Also, three
years from now, I shall be three times as old as you
will be.” (Isn’t this interesting?) Represent this
situation algebraically and graphically.
Solution:
Let Aftab’s present age be years.
x
And, let the present age of his daughter be years.
y
Seven years ago,
The age of Aftab’s
x 7
And, the age of his daughter
y 7
Now, according to the question,
x 7 7 y 7
x 7 7y 49
x 7y 7 49
x 7y 42 ......(1)
Three years from now,
The age of Aftab’s .
x 3
And, the age of his daughter .
y 3
Now, according to the question,
x 3 3 y 3
x 3 3y 9
x 3y 9 3
x 3y 6 .....(2)
To represent these equations graphically, you must
have at least two solutions for each equation.
For equation 1:
The points are , and
0,6
7,7
x 42 7y
x
7
0
7
y
5
6
7
For equation 2:
The points are and .
6,0 ,
3, 1
x 6 3y
x
6
3
0
y
0
1
2
The graphical representation is as follows:
Fig. Exc_3.1_1
Question: 2
The coach of a cricket team buys bats and balls
3
6
for Rs . Later, she buys another bat and more
3900
2
balls of the same kind for Rs . Represent this
1300
situation algebraically and geometrically.
Solution:
Let the cost of a bat and a ball be Rs .and Rs
x
y
respectively.
Now, according to the question,
3x 6y 3900 ..... 1
x 2y 1300 ..... 2
To represent these equations graphically, you must
have at least two solutions for each equation.
For equation 1:
The points are and
0,650
1300,0
3900 3x
y
6
x
0
1300
y
650
0
For equation 2:
The points are .
0,650
and
1300,0
1300 x
y
2
x
0
1300
y
650
0
The graphical representation is as follows:
Fig. Exc_3.1_2
Question: 3
The cost of kg of apples and kg of grapes on a
2
1
day was found to be Rs . After a month, the cost
160
of kg of apples and kg of grapes is Rs .
4
2
300
Represent the situation algebraically and
geometrically.
Solution:
Let the cost of 1 kg of apples be Rs .
x
And, let the cost of 1 kg of grapes be Rs .
y
According to the question,
2x y 160 ..... 1
4x 2y 300 ..... 2
To represent these equations graphically, you must
have at least two solutions for each equation.
For equation 1:
The points are and
80,0
60,40
y 160 2x
x
80
60
y
0
40
For equation 2:
The points are and
60,30
75,0
300 4x
y
2
x
60
75
y
30
0
The graphical representation is as follows:
Fig. Exc_3.1_3
Exercise 3.2 (7)
Question: 1
Form the pair of linear equations in the following
problems, and find their solutions graphically.
(i) students of Class X took part in a
10
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.
(ii) pencils and pens together cost Rs ,
5
7
50
whereas pencils and pens together cost Rs
7
5
. Find the cost of one pencil and that of one
46
pen.
Solution:
(i) Let the number of girls and boys be and
x
y
respectively.
According to the question,
.
x y 10 ..... 1
x y 4 ..... 2
To represent these equations graphically, you
must have at least two solutions for each
equation.
For equation 1:
The points are .and
0,10
10,0
y 10 x
x
0
10
y
10
0
For equation 2:
The points are .and
4,0
0, 4
y x 4
x
4
0
y
0
4
The graphical representation is as follows:
Fig. Exc_3.2_1(i)
From the graph, it can be observed that these
lines intersect each other at point .
7,3
Therefore, the number of girls and boys in the
class are and respectively.
7
3
(ii) Let the cost of pencil be Rs and the cost of
1
x
pen be Rs .
1
y
According to the question,
5x 7y 50 ..... 1
7x 5y 46 ..... 2
For equation 1:
The points are , and
3,5 ,
10,0
4,10
50 7y
x
5
x
3
10
4
y
5
0
10
For equation 2:
The points are , and
3,5
2,12
46 5y
x
7
x
8
3
2
y
2
5
12
Hence, the graphical representation is as
follows:
Fig. Exc_3.2_1(ii)
From the graph, it can be observed, that these
lines intersect each other at point .
3,5
Therefore, the cost of a pencil and a pen are Rs
and Rs respectively.
3
5
Question: 2
On comparing the ratios, , and find out
1
2
a
a
1
2
b
b
1
2
c
c
whether the lines representing the following pairs
of linear equations at a point, are parallel or
coincident:
(i)
5x 4y 8 0
7x 6y 9 0
(ii)
9x 3y 12 0
18x 6y 24 0
(iii)
6x 3y 10 0
2x y 9 0
Solution:
(i) Given equations:
5x 4y 8 0
7x 6y 9 0
Now, compare these equations with
and ,
1 1 1
a x b y c 0
2 2 2
a x b y c 0
1
a 5,
1
b 4,
1
c 8
2
b 6,
2
c 9
and
1
2
a 5
a 7
1
2
b 4
b 6
1
2
b 2
b 3
Since,
1 1
2 2
a b
a b
Thus, the lines that represent the given pair of
linear equations intersect at one point exactly.
Also, a unique solution exists for the equations.
(ii) Given equations:
9x 3y 12 0
18x 6y 24 0
Now, compare these equations with
and ,
1 1 1
a x b y c 0
2 2 2
a x b y c 0
1
a 9,
1
b 3,
1
c 12
2
a 18,
2
b 6,
2
c 24
1
2
a 9
a 18
1
2
a 1
a 2
1
2
b 3
b 6
1
2
b 1
b 2
1
2
c 12
c 24
1
2
c 1
c 2
Since,
1 1 1
2 2 2
a b c
a b c
Thus, the lines that represent the given pair of
linear equations. are coincident.
Also, infinite number of solutions are possible
for the equations.
(iii) Given equations:
6x 3y 10 0
2x y 9 0
Now, compare these equations with
and ,
1 1 1
a x b y c 0
2 2 2
a x b y c 0
1
a 6,
1
b 3,
1
c 10
2
b 1,
2
c 9
1
2
a 6
a 2
1
2
a 3
a 1
1
2
b 3
b 1
1
2
b 3
b 1
1
2
c 10
c 9
Since,
1 1 1
2 2 2
a b c
a b c
Thus, the lines that represent the given pair of
linear equations are parallel and will not
intersect each other
Also, no solution is possible for the equations.
Question: 3
On comparing the ratios, and find out
1
2
a
,
a
1
2
b
b
1
2
c
c
whether the following pairs of linear equations are
consistent or inconsistent:
(i)
3x 2y 5
2x 3y 7
(ii)
2x 3y 8
4x 6y 9
(iii)
3 5
x y 7
2 3
9x 10y 14
(iv)
5x 3y 11
10x 6y 22
(v)
4
x 2y 8
3
2x 3y 12
Solution:
(i) Given equations:
3x 2y 5
2x 3y 7
Now, compare these equations with
and ,
1 1 1
a x b y c 0
2 2 2
a x b y c 0
1
a 3,
1
b 2,
1
c 5
2
b 3,
2
c 7
,
1
2
a 3
a 2
1
2
b 2
b 3
1
2
b 2
b 3
1
2
c 5
c 7
Since, .
1 1
2 2
a b
a b
Thus, the lines that represent the given pair of
linear equations intersect at one point.
Also, there is only one possible solution and
the pair of equations is consistent.
(ii) Given equations:
2x 3y 8
4x 6y 9
Now, compare these equations with
and ,
1 1 1
a x b y c 0
2 2 2
a x b y c 0
1
a 2,
1
b 3,
1
c 8
2
b 6,
2
c 9
1
2
a 2
a 4
1
2
a 1
a 2
1
2
b 3
b 6
1
2
b 1
b 2
1
2
c 8
c 9
1
2
c 8
c 9
Since,
1 1 1
2 2 2
a b c
a b c
Therefore, these linear equations are parallel to
each other and thus have no possible solution.
Hence, the pair of linear equations is
inconsistent.
(iii) Given equations:
3 5
x y 7
2 3
9x 10y 14
Now, compare these equations with
and ,
1 1 1
a x b y c 0
2 2 2
a x b y c 0
1
3
a ,
2
1
5
b ,
3
1
c 7
2
b 10,
2
c 14
1
2
3
a
2
a 9
1
2
a 1
a 6
1
2
5
b
3
b 10
1
2
b 1
b 6
1
2
c 7
c 14
1
2
c 1
c 2
Since,
1 1
2 2
a b
a b
Thus, the lines that represent the given pair of
linear equations intersect at one point.
Also, there is only one possible solution and
the pair of equations is consistent.
(iv) Given equations:
5x 3y 11
10x 6y 22
Now, compare these equations with
and ,
1 1 1
a x b y c 0
2 2 2
a x b y c 0
1
a 5,
1
b 3,
1
c 11
2
a 10,
2
b 6,
2
c 22
1
2
a 5
a 10
1
2
a 1
a 2
1
2
b 3
b 6
1
2
b 1
b 2
1
2
c 11
c 22
1
2
c 1
c 2
Since,
1 1 1
2 2 2
a b c
a b c
Thus, the lines that represent the given pair of
linear equations are coincident.
Also, infinite number of solutions are possible
and the pair of equations is consistent.
(v) Given equations:
4
x 2y 8
3
2x 3y 12
Now, compare these equations with
and ,
1 1 1
a x b y c 0
2 2 2
a x b y c 0
1
4
a ,
3
1
b 2,
1
c 8
2
b 3,
2
c 12
1
2
4
a
3
a 2
,
1
2
a 2
a 3
1
2
b 2
b 3
1
2
c 8
c 12
1
2
c 2
c 3
Since,
1 1 1
2 2 2
a b c
a b c
Thus, the lines that represent the given pair of
linear equations are coincident.
Also, infinite number of solutions are possible
and the pair of equations is consistent.
Question: 4
Which of the following pairs of linear equations are
consistent/ inconsistent? If consistent, obtain the
solution graphically:
(i)
x y 5
2x 2y 10
(ii)
x y 8
3x 3y 16
(iii)
2x y 6 0
4x 2y 4 0
(iv)
2x 2y 2 0
4x 4y 5 0
Solution:
(i) Given equations:
x y 5 ..... 1
2x 2y 10 ..... 2
Now, compare these equations with
and .
1 1 1
a x b y c 0
2 2 2
a x b y c 0
1
a 1,
1
b 1,
1
c 5
2
b 2,
2
c 10
1
2
a 1
a 2
1
2
b 1
b 2
1
2
c 5
c 10
1
2
c 1
c 2
Since,
1 1 1
2 2 2
a b c
a b c
Thus, the lines that represent the given pair of
linear equations are coincident.
Also, infinite number of solutions are possible
and the pair of equations is consistent.
For equation 1:
The points are .and
0,5
5,0
y 5 x
x
0
5
y
5
0
For equation 2:
The points are and .
0,5
5,0
10 2x
y
2
x
0
5
y
5
0
Hence, the graphical representation is as
follows:
Fig. Exc_3.2_4(i)
The overlapping of lines can be observed from
the above graph.
Thus, infinite number of solutions are possible
for the equations.
(ii) Given equations:
x y 8
3x 3y 16
Now, compare these equations with
and .
1 1 1
a x b y c 0
2 2 2
a x b y c 0
1
a 1,
1
b 1,
1
c 8
2
b 3,
2
c 16
,
1
2
a 1
a 3
1
2
b 1
b 3
1
2
b 1
b 3
1
2
c 8
c 16
1
2
c 1
c 2
Since,
1 1 1
2 2 2
a b c
a b c
Thus, the lines that represent the given pair of
linear equations are parallel and will not
intersect each other
Also, no solution is possible for the equations
and the pair of equations is inconsistent.
(iii) Given equations:
2x y 6 0 ..... 1
4x 2y 4 0 ..... 2
Now, compare these equations with
and .
1 1 1
a x b y c 0
2 2 2
a x b y c 0
1
a 2,
1
b 1,
1
c 6
2
b 2,
2
c 4
1
2
a 2
a 4
1
2
a 1
a 2
1
2
b 1
b 2
1
2
c 6
c 4
1
2
c 3
c 2
Since,
1 1
2 2
a b
a b
Thus, the lines that represent the given pair of
linear equations intersect at one point exactly.
Also, a unique solution exists for the equations
and the pair of equations is consistent.
For equation 1:
The points are .and
0,6
3,0
y 6 2x
x
0
3
y
6
0
For equation 2:
The points are .and
1,0
4x 4
y
2
y 2 x 1
x
1
0
y
0
2
Hence, the graphical representation is as
follows:
Fig. Exc_3.2_4(iii)
The intersection of lines can be observed from
the above graph.
The intersection point is the solution for
2,2
the pair of equations.
(iv) Given equations:
2x 2y 2 0
4x 4y 5 0
Compare these equations with
and .
1 1 1
a x b y c 0
2 2 2
a x b y c 0
1
a 2,
1
b 2,
1
c 2
2
b 4,
2
c 5
1
2
a 2
a 4
1
2
a 1
a 2
1
2
b 2
b 4
1
2
b 1
b 2
1
2
c 2
c 5
1
2
c 2
c 5
Since,
1 1 1
2 2 2
a b c
a b c
Thus, the lines that represent the given pair of
linear equations are parallel and will not
intersect each other
Also, no solution is possible for the equations
and the pair of equations is inconsistent.
Question: 5
Half the perimeter of a rectangular garden, whose
length is more than its width, is . Find the
4m
36m
dimensions of the garden.
Solution:
Let the width of the garden be and its length be .
x
y
According to the question,
y x 4 ..... 1
y x 36 ..... 2
For equation 1:
The points are .and
0,4
8,12
y 4 x
x
0
8
y
4
12
For equation 2:
The points are .
0,36 and 36,0
y 36 x
x
0
36
y
36
0
Hence, the graphical representation is as follows:
Fig. Exc_3.2_5
The intersection of lines can be observed from the
above graph.
The intersection point is the solution for the
16,20
pair of equations.
Thus, the length of the garden is and the
20m
width of the garden is .
16m
Question: 6
Given the linear equation , write
2x 3y 8 0
another linear equation in two variables such that
the geometrical representation of the pair so
formed is:
(i) Intersecting lines
(ii) Parallel lines
(iii) Coincident lines
Solution:
(i) Intersecting lines
For this condition,
1 1
2 2
a b
a b
Now, the given equation is .
2x 3y 8 0
So, another linear equation of the line
intersecting the given line can be
as,
2x 4y 6 0
1
2
a 2
a 2
1
2
b 3
b 4
So,
1 1
2 2
a b
a b
(ii) Parallel lines
For this condition,
1 1 1
2 2 2
a b c
a b c
Now, the given equation is .
2x 3y 8 0
So, another linear equation of the line parallel
to the given line can be as,
4x 6y 8 0
1
2
a 2
a 4
1
2
a 1
a 2
1
2
b 3
b 6
1
2
b 1
b 2
1
2
c 8
c 8
1
2
c 1
c 1
So,
1 1 1
2 2 2
a b c
a b c
(iii) Coincident lines
For this condition,
1 1 1
2 2 2
a b c
a b c
Now, the given equation is .
2x 3y 8 0
So, another linear equation of the line
coincident to the given line can be
as,
6x 9y 24 0
1
2
a 2
a 6
1
2
a 1
a 3
1
2
b 3
b 9
1
2
b 1
b 3
1
2
c 8
c 24
So,
1 1 1
2 2 2
a b c
a b c
Question: 7
Draw the graphs of the equations and
x y 1 0
Determine the coordinates of the
3x 2y 12 0
vertices of the triangle formed by these lines and
the -axis, and shade the triangular region.
x
Solution:
Given:
x y 1 0 ..... 1
3x 2y 12 0 .... 2
For equation 1:
The points are .and
0,1
1,0
y x 1
x
0
1
y
1
0
For equation 2:
The points are .and
4,0
0,6
12 3x
y
2
x
4
0
y
0
6
Hence, the graphical representation is as follows:
Fig. Exc_3.2_7
The intersection of the lines at the point can
2,3
be observed from the above graph.
Thus, the coordinates of the vertices of the triangle
formed by the lines are and
2,3 ,
1,0
4,0
Exercise 3.3 (3)
Question: 1
Solve the following pair of linear equations by the
substitution method.
(i)
x y 14
x y 4
(ii)
s t 3
 
s t
6
3 2
(iii)
3x y 3
9x 3y 9
(iv)
0.2x 0.3y 1.3
0.4x 0.5y 2.3
(v)
2x 3y 0
3x 8y 0
(vi)
3x 5y
2
2 3
x y 13
3 2 6
Solution:
(i) Given equations:
x y 14
x y 4
x y 14 ..... 1
x y 4 ..... 2
Now, from equation ,
1
x 14 y ..... 3
Put the value of in equation .
x
2
14 y y 4
14 2y 4
2y 14 4
2y 10
y 5
Now, put the value of in equation .
y
3
x 14 5
x 9
Thus,
x 9,
y 5
(ii) Given equations:
s t 3
 
s t
6
3 2
s t 3 ..... 1
 
s t
6 ..... 2
3 2
Now, from equation ,
1
s t 3 ..... 3
Put the value of in equation .
x
2
 
t 3 t
6
3 2
2 t 3 3t
6
6
2t 6 3t
6
6
5t 6 6 6
5t 36 6
5t 30
t 6
Now, put the value of in equation .
t
3
s t 3
s 6 3
s 9
Thus,
s 9
t 6
(iii) Given equations:
3x y 3
9x 3y 9
3x y 3 ..... 1
9x 3y 9 ..... 2
Now, from equation ,
1
3 y
x ..... 3
3
Put the value of in equation .
x
2
3 y
9 3y 9
3
3 3 y 3y 9
9 3y 3y 9
9 9
This is always true.
Thus, infinite number of solutions are possible
for the given pair of equations.
Also, the variables are related to each other by
the relation given by .
y 3x 3
Thus, one of its possible solution is ,
x 1
y 0
(iv) Given equations:
0.2x 0.3y 1.3
0.4x 0.5y 2.3
0.2x 0.3y 1.3 ..... 1
0.4x 0.5y 2.3 ..... 2
Now, from equation ,
1
1.3 0.3y
x ..... 3
0.2
Put the value of in equation .
x
2
1.3 0.3y
0.4 0.5y 2.3
0.2
0.2 1.3 0.3y 0.5y 2.3
2.6 0.6y 0.5y 2.3
2.6 2.3 0.1y
0.3 0.1y
y 3
Now, put the value of in equation .
y
3
1.3 0.3 3
x
0.2
1.3 0.9
0.2
0.4
0.2
2
Thus,
x 2,
y 3
(v) Given equations:
2x 3y 0
3x 8y 0
2x 3y 0 ..... 1
3x 8y 0 ..... 2
Now, from equation ,
1
Put
3y
x ..... 3
2
the value of in equation .
x
2
3y
3 8y 0
2
3y
2 2y 0
2
3
y 2 2 0
2
y 0
Now, put the value of in equation .
y
3
x 0
Thus,
x 0,
y 0
(vi) Given equations:
3x 5y
2
2 3
x y 13
3 2 6
3x 5y
2 ..... 1
2 3
x y 13
..... 2
3 2 6
Now, from equation ,
1
3x 5y
2
2 3
3 3x 2 5y
2
6
9x 10y 2 6
9x 10y 12
12 10y
x ..... 3
9
Put the value of in equation .
x
2
12 10y
y 13
9
3 2 6
12 10y y 13
9 3 2 6
12 10y y 13
27 2 6
Take LCM on left hand side.
2 12 10y 27y
13
54 6
24 20y 27y 13
54 6
47y 24 13
54 6
13 54
47y 24
6
47y 13 9 24
47y 117 24
47y 141
y 3
Now, put the value of in equation .
y
3
12 10 3
x
9
12 30
x
9
18
x
9
x 2
Thus,
x 2,
y 3
Question: 2
Solve and and hence find
2x 3y 11
2x 4y 24
the value of for which .
'm'
y mx 3
Solution:
Given equations:
2x 3y 11 ..... 1
2x 4y 24 ..... 2
Now, from equation ,
1
11 3y
x .... 3
2
Put the value of in equation .
x
2
11 3y
2 4y 24
2
11 3y 4y 24
11 24 7y
7y 35
y 5
Now, put the value of in equation .
y
3
11 3 5
x
2
11 15
x
2
4
x
2
x 2
Thus, ,
x 2
y 5
Now, find the value of .
m
y mx 3
5 2m 3
2m 5 3
2m 2
2
m
2
m 1
Thus, .
m 1
Question: 3
Form the pair of linear equations for the following
problems and find their solution by substitution
method.
(i) The difference between two numbers is and
26
one number is three times the other. Find
them.
(ii) The larger of two supplementary angles
exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6
balls for Rs 3800. Later, she buys 3 bats and 5
balls for Rs 1750. Find the cost of each bat and
each ball.
(iv) The taxi charges in a city consist of a fixed
charge together with the charge for the
distance covered. For a distance of 10 km, the
charge paid is Rs 105 and for a journey of 15
km, the charge paid is Rs 155. What are the
fixed charges and the charge per km? How
much does a person have to pay for travelling
a distance of 25 km.
(v) A fraction becomes , if 2 is added to both
8
11
the numerator and the denominator. If, 3 is
added to both the numerator and the
denominator it becomes . Find the fraction.
5
6
(vi) Five years hence, the age of Jacob will be three
times that of his son. Five years ago, Jacob’s
age was seven times that of his son. What are
their present ages?
Solution:
(i) Let be the first number and be the other
x
y
number such that .
y x
According to the question,
y 3x ..... 1
y x 26 ..... 2
Now, put the value of from equation in
y
1
equation .
2
3x x 26
2x 26
x 13
Now, put the value of in equation .
x
1
y 3x
y 3 13
y 39
Hence, the numbers are and
13
39
(ii) Let be the larger angle and be the smaller
x
y
angle.
The sum of the measures of the pair of
supplementary angles is always 180º.
According to the question,
x y 180 ..... 1
x y 18 ..... 2
From equation ,
1
x 180 y ..... 3
Now, put the value of in equation .
x
2
180 y y 18
2y 180 18
2y 162
162
y
2
y 81
Now, put the value of in equation .
y
3
x 180 y
x 180 81
x 99
Hence, the angles are and
81
99
(iii) Let the cost of a bat be and the cost of ball
x
be .
y
According to the question,
7x 6y 3800 ..... 1
3x 5y 1750 .... 2
Now, from equation ,
1
3800 7x
y ..... 3
6
Now, put the value of in equation .
y
2
3800 7x
3x 5 1750
6
9500 35x
3x 1750
3 6
35x 9500
3x 1750
6 3
18x 35x 5250 9500
6 3
17x 4250
6 3
4250 6
17x
3
17x 8500
x 500
Now, substitute the value of in equation .
x
3
3800 7 500
y
6
3800 3500
y
6
300
y
6
y 50
Hence, the cost of a bat is Rs 500 and that of a
ball is Rs 50.
(iv) Let the fixed charge and per km charge be Rs
x
and Rs respectively.
y
According to the question,
x 10y 105 ..... 1
x 15y 155 ..... 2
Now, from equation ,
1
x 105 10y ..... 3
Now, put the value of in equation .
x
2
105 10y 15y 155
5y 155 105
5y 50
50
y
5
y 10
Now, put the value of in equation .
y
3
x 105 10 10
x 105 100
x 5
Thus, fixed charge is and per km charge is
Rs.5
Rs 10.
Now, the charge for 25 km
x 25y
5 25 10
5 250
Rs.255
Thus, the charge of 25 km is Rs. 255.
(v) Let the fraction be .
x
y
According to the first condition of the
question,
x 2 9
y 2 11
11 x 2 9 y 2
11x 22 9y 18
11x 9y 18 22
11x 9y 4 ..... 1
According to the second condition of the
question,
x 3 5
y 3 6
6 x 3 5 y 3
6x 18 5y 15
6x 5y 15 18
6x 5y 3 ..... 2
Now, from equation ,
1
4 9y
x ..... 3
11
Put the value of in equation .
x
2
4 9y
6 5y 3
11
6 4 9y 55y 3 11
24 54y 55y 33
54y 55y 33 24
y 9
Now, put the value of in equation .
y
3
4 9 9
x
11
4 81
x
11
x 7
Hence, the fraction is .
7
9
(vi) Let Jacob’s age be and his son’s age be .
x
y
After five years:
Jacob’s age
x 5
His son’s age
y 5
According to the question,
x 5 3 y 5
x 5 3y 15
x 3y 15 5
x 3y 10 ..... 1
Five years ago:
Jacob’s age
x 5
His son’s age
y 5
According to the question,
x 5 7 y 5
x 5 7y 35
x 7y 35 5
x 7y 30 ..... 2
Now, from equation ,
1
x 10 3y .... 3
Put the value of in equation .
x
2
10 3y 7y 30
3y 7y 30 10
4y 40
40
y
4
y 10
Now, put the value of in equation .
y
3
x 10 3 10
x 10 30
x 40
Thus,
The present age of Jacob years and the
40
present age of his son years.
10
Exercise 3.4 (2)
Question: 1
Solve the following pair of linear equations by the
elimination method and the substitution method:
(i) and
x y 5
2x 3y 4
(ii) and
3x 4y 10
2x 2y 2
(iii) and
3x 5y 4 0
9x 2y 7
(iv) and
x 2y
1
2 3
y
x 3
3
Solution:
(i) Given equations:
x y 5 ..... 1
2x 3y 4 ..... 2
By elimination method:
Multiply equation .
1 by 2
2x 2y 10 ..... 3
Subtract equation .
2 from 3
2x 2y 2x 3y 10 4
2x 2y 2x 3y 10 4
5y 6
6
y
5
Now, put the value of .
y in equation 1
 
6
x 5
5
5x 6 5 5
5x 6 25
5x 25 6
5x 19
19
x
5
Thus, and
19
x
5
6
y
5
By substitution method:
From equation ,
1
Put
x 5 y ..... 4
the value of in the equation .
x
2
2 5 y 3y 4
10 2y 3y 4
5y 4 10
5y 6
6
y
5
Now, put the value of .
y in equation 4
6
x 5
5
25 6
x
5
19
x
5
Thus, and
19
x
5
6
y
5
(ii) Given equations:
3x 4y 10 ..... 1
2x 2y 2 .... 2
By elimination method:
Multiply equation ,
2 by 2
4x 4y 4 .... 3
1 and 3
3x 4y 4x 4y 10 4
7x 14
14
x
7
x 2
Now, put the value of .
xin equation 1
3 2 4y 10
6 4y 10
4y 10 6
4y 4
4
y
4
y 1
Thus, .and
x 2
y 1
By substitution method:
From equation ,
2
Put the
x 1 y ..... 4
value of in equation .
x
1
3 1 y 4y 10
3 3y 4y 10
7y 10 3
7y 7
7
y
7
y 1
Now, put the value of .
y in equation 4
x 1 y
x 1 1
x 2
Thus, .and
x 2
y 1
(iii) Given equations:
3x 5y 4 0 .... 1
9x 2y 7
9x 2y 7 0 .... 2
By elimination method:
Multiply equation .
1 by 3
9x 15y 12 0 ..... 3
Now, subtract equation .
3 from 2
9x 2y 7 9x 15y 12 0
9x 2y 7 9x 15y 12 0
13y 5 0
13y 5
5
y
13
Put the value of .
y in equation 1
5
3x 5 4 0
13
25
3x 4
13
25
3x 4
13
52 25
3x
13
27
3x
13
27
x
13 3
9
x
13
Thus, and
9
x
13
5
y
13
By substitution method:
From equation ,
1
5y 4
x ..... 4
3
Put the value of in equation .
x
2
5y 4
9 2y 7 0
3
3 5y 4 2y 7 0
15y 12 2y 7 0
13y 12 7
13y 5
5
y
13
Now, put the value of .
y in equation 4
5
5 4
13
x
3
25
4
13
x
3
25 52
13
x
3
27
x
13 3
9
x
13
Thus, .and
9
x
13
5
y
13
(iv) Given equations:
x 2y
1
2 3
3x 4y 6 ..... 1
y
x 3
3
3x y 9 ..... 2
By elimination method:
Subtract equation .
2 from 1
3x 4y 3x y 6 9
3x 4y 3x y 6 9
5y 15
15
y
5
Put the value of .
y in equation 1
3x 4 3 6
3x 12 6
3x 6 12
3x 6
6
x
3
x 2
Thus, and
x 2
By substitution method:
From equation ,
2
y 9
x ..... 3
3
Put the value of in equation .
x
1
y 9
3 4y 6
3
y 9 4y 6
5y 6 9
5y 15
15
y
5
Now, put the value of .
y in equation 3
y 9
x
3
3 9
x
3
6
x
3
x 2
Thus, and
x 2
Question: 2
Form the pair of linear equations in the following
problems, and find their solutions (if they exist) by
the elimination method:
(i) If we add 1 to the numerator and subtract 1
from the denominator, a fraction reduces to 1.
It becomes if we only add 1 to the
1
2
denominator. What is the fraction?
(ii) Five years ago, Nuri was thrice as old as Sonu.
Ten years later, Nuri will be twice as old as
Sonu. How old are Nuri and Sonu?
(iii) The sum of the digits of a two-digit number is
9. Also, nine times this number is twice the
number obtained by reversing the order of the
digits. Find the number.
(iv) Meena went to bank to withdraw Rs 2000. She
asked the cashier to give her Rs 50 and Rs 100
notes only. Meena got 25 notes in all. Find how
many notes of Rs 50 and Rs 100 she received.
(v) A lending library has a fixed charge for the
first three days and an additional charge for
each day thereafter. Saritha paid Rs 27 for a
book kept for seven days, while Susy paid Rs
21 for the book she kept for five days. Find the
fixed charge and the charge for each extra day.
Solution:
(i) Let us consider the fraction be .
x
y
According to the question,
x 1
1
y 1
x 1 y 1
x y 1 1
x y 2 ..... 1
x 1
y 1 2
2x y 1
2x y 1 ..... 2
Subtract equation from equation .
1
2
2x y x y 1 2
x 3 .... 3
Now, put the value of in equation .
x
1
x y 2
3 y 2
y 3 2
y 5
Hence, the fraction is .
3
5
(ii) Let present age of Nuri be and present age of
x
Sonu be .
y
Five years ago:
Age of Nuri
x 5
Age of Sonu
y 5
According to the first condition of the
question,
x 5 3 y 5
x 5 3y 15
x 3y 5 15
x 3y 10 ..... 1
After ten years:
Age of Nuri
x 10
Age of Sonu
y 10
According to the second condition of the
question,
x 10 2 y 10
x 10 2y 20
x 2y 20 10
x 2y 10 ..... 2
Subtract equation from .
1
2
x 2y x 3y 10 10
y 20 ..... 3
Put the value of in equation .
y
1
x 3y 10
x 3 20 10
x 60 10
x 60 10
x 50
Hence,
Nuri’s present age years
50
Sonu’s present age years.
20
(iii) Let be the unit’s place digit and be the
x
y
tens place digit.
Then, the number .
10y x
Number after reversing the digits .
10x y
Now, according to the question,
x y 9 1
9 10y x 2 10x y
90y 9x 20x 2y
90y 2y 20x 9x 0
88y 11x 0
x 8y 0 ..... 2
1 and 2
x y x 8y 9 0
9y 9
9
y
9
y 1 ..... 3
Now, put the value of in equation .
y
1
x y 9
x 1 9
x 9 1
x 8
Hence, the number is
10y x 10 1 8
10y x 18
(iv) Let be the number of Rs 50 notes and be
x
y
the number of Rs 100 notes.
According to the question,
x y 25 ..... 1
50x 100y 2000 ..... 2
Multiply equation by .
1
50
50x 50y 1250 ..... 3
Subtract equation from .
3
2
50x 100y 50x 50y 2000 1250
50y 750
750
y
50
y 15
Put the value of in equation .
y
1
x y 25
x 15 25
x 25 15
x 10
Thus, Meena has notes of and
10
Rs.50
15
notes of .
Rs.100
(v) Let the fixed charge for first three days be
x
and each day charge thereafter be Rs .
y
According to the question,
x 4y 27
1
x 2y 21
2
Subtract equation from .
2
1
x 4y x 2y 27 21
x 4y x 2y 27 21
2y 6
6
y
2
y 3 ..... 3
Now, put the value of in equation .
y
1
x 4 3 27
x 12 27
x 27 12
x 15
Thus, fixed charge
Rs.15
and charge per day .
EXERCISE 3.5 (4)
Question: 1
Which of the following pairs of linear equations has
unique solution, no solution or infinitely many
solutions? In case there is a unique solution, find it
by using cross multiplication method.
(i)
x 3y 3 0
3x 9y 2 0
(ii)
2x y 5
3x 2y 8
(iii)
3x 5y 20
6x 10y 40
(iv)
x 3y 7 0
3x 3y 15 0
Solution
(i)
x 3y 3 0
3x 9y 2 0
Compare the given equations with the general
form for a pair of linear equations.
1 1 1
a x b y c 0
2 2 2
a x b y c 0
The values of the coefficients of the given
equations are:
1
a 1,
1
b 3,
1
c 3
2
b 9,
2
c 2
1
2
a 1
,
a 3
1
2
b 3
,
b 9
1
2
c 3
c 2
1
2
a 1
a 3
1
2
b 1
, ,
b 3
1
2
c 3
..... 1
c 2
From equation ,
1
1 1 1
2 2 2
a b c
a b c
Hence, the given lines do not intersect each
other and the pair of linear equations has no
solution.
(ii)
2x y 5
3x 2y 8
We can rewrite it as,
2x y 5 0
3x 2y 8 0
Compare the given equations with the general
form for a pair of linear equations.
1 1 1
a x b y c 0
2 2 2
a x b y c 0
The values of the coefficients of the given
equations are:
1
a 2,
1
b 1,
1
c 5
2
b 2,
2
c 8
1
2
a 2
,
a 3
1
2
b 1
,
b 2
1
2
c 5
c 8
1
2
a 2
,
a 3
1
2
b 1
,
b 2
1
2
c 5
..... 1
c 8
From equation ,
1
1 1
2 2
a b
a b
Hence, the given lines intersect at a unique
point and the pair of linear equations has a
unique solution.
By cross multiplication method,
1 2 2 1 1 2 2 1 1 2 2 1
x y 1
b c b c ca c a a b a b
Put the values of the coefficients.
x y 1
8 10 15 16 4 3
 
x y 1
1 8 2 5 5 3 8 2 2 2 3 1
x
y 1
2
Take first and third part.
x
1
2
x 2
Take second and third part.
y 1
Hence, .and
x 2,
y 1
(iii)
3x 5y 20
6x 10y 40
We can rewrite it as,
3x 5y 20 0
6x 10y 40 0
Compare the given equations with the general
form for a pair of linear equations.
1 1 1
a x b y c 0
2 2 2
a x b y c 0
The values of the coefficients of the given
equations are:
1
a 3,
1
b 5,
1
c 20
2
b 10,
2
c 40
1
2
a 3
,
a 6
1
2
b 5
,
b 10
1
2
c 20
c 40
1
2
a 1
,
a 2
1
2
b 1
b 2
1
2
c 1
, ..... 1
c 2
From equation ,
1
1 1 1
2 2 2
a b c
a b c
Hence, the given lines are coincident lines and
the pair of linear equations has infinitely many
solutions.
(iv)
x 3y 7 0
3x 3y 15 0
Compare the given equations with the general
form for a pair of linear equations.
1 1 1
a x b y c 0
2 2 2
a x b y c 0
The values of the coefficients of the given
equations are:
1
a 1,
1
b 3,
1
c 7
2
b 3,
2
c 15
1
2
a 1
,
a 3
1
2
b 3
,
b 3
1
2
c 7
c 15
1
2
a 1
,
a 3
1
2
b 1
,
b 1
1
2
c 7
c 15
1
From equation ,
1
1 1
2 2
a b
a b
Hence, the given lines intersect at a unique
point and the pair of linear equations has a
unique solution.
By cross multiplication method,
1 2 2 1 1 2 2 1 1 2 2 1
x y 1
b c b c ca c a a b a b
Put the values of the coefficients.
x y 1
3 15 3 7 7 3 15 1 1 3 3 3
x y 1
45 21 21 15 3 9
x y 1
24 6 6
Take first and third part.
x 1
24 6
x 4
Take second and third part.
y 1
6 6
6
y
6
y 1
Hence,
x 4,
y 1
Question: 2
(i) For which values of will the following
a andb
pair of linear equations have an infinite
number of solutions?
2x 3y 7
a b x a b y 3a b 2
(ii) For which value of will the following pair of
k
linear equations have no solution?
3x y 1
2k 1 x k 1 y 2k 1
Solution
(i) Given equations:
2x 3y 7
a b x a b y 3a b 2
We can rewrite it as,
2x 3y 7 0
a b x a b y 3a b 2 0
Compare the given equations with the general
form for a pair of linear equations.
1 1 1
a x b y c 0
2 2 2
a x b y c 0
The values of the coefficients of the given
equations are:
1
a 2,
1
b 3,
1
c 7
2
a a b ,
2
b a b ,
2
c 3a b 2
1
2
a 2
,
a a b
1
2
b 3
,
b a b
1
2
c 7
c 3a b 2
To have infinitely many solutions,
Put the
1 1 1
2 2 2
a b c
..... 1
a b c
values of in equation .
1 2 1 2
a ,a ,b ,b
1
2 3
a b a b
2 a b 3 a b
2a 2b 3a 3b
2a 3a 2b 3b 0
a 5b 0
a 5b 0
a 5b 0 ..... 2
Put the values of in equation
1 2 1 2
a ,a ,c ,c
1
2 7
a b 3a b 2
2 3a b 2 7 a b
6a 2b 4 7a 7b
6a 7a 2b 7b 4 0
a 9b 4 0
a 9b 4 0
a 9b 4 ..... 3
Subtract equation (3) from (2).
a 5b a 9b 0 4
a 5b a 9b 4
4b 4
b 1
Put the value of in equation (2).
b
a 5b 0
a 5 1 0
a 5
Hence, the given pair of linear equations has
an infinite number of solutions for and
a 5
b 1
(ii) Given equations:
3x y 1
2k 1 x k 1 y 2k 1
We can rewrite it as,
3x y 1 0
2k 1 x k 1 y 2k 1 0
Compare the given equations with the general
form for a pair of linear equations.
1 1 1
a x b y c 0
2 2 2
a x b y c 0
The values of the coefficients of the given
equations are:
1
a 3,
1
b 1,
1
c 1
,
2
a 2k 1 ,
2
c 2k 1
2
b k 1 ,
1
2
a 3
,
a 2k 1
1
2
b 1
,
b k 1
1
2
c 1
..... 1
c 2k 1
To have no solutions,
1 1 1
2 2 2
a b c
a b c
Take first and second part using equation .
1
3 1
2k 1 k 1
3 k 1 2k 1
3k 3 2k 1
3k 2k 3 1 0
k 2 0
k 2
Hence, the given pair of linear equations has
no solution for .
k 2
Question: 3
Solve the following pair of linear equations by the
substitution and cross-multiplication methods:
8x 5y 9
3x 2y 4
Solution
Substitution method:
8x 5y 9 ..... 1
3x 2y 4 ..... 2
From equation ,
2
3x 2y 4
3x 4 2y
4 2y
x
3
Put the value of in equation .
x
1
4 2y
8 5y 9
3
32 16y
5y 9
3
32 16y 15y 9 3
y 32 27 0
y 5 0
y 5
Put the value of in equation .
y
2
3x 2 5 4
3x 10 4
3x 4 10
6
x
3
x 2
Hence, .and
x 2
y 5
Cross multiplication method:
8x 5y 9
3x 2y 4
We can rewrite it as,
8x 5y 9 0
3x 2y 4 0
Compare the given equations with the general
form for a pair of linear equations.
1 1 1
a x b y c 0
2 2 2
a x b y c 0
The values of the coefficients of the given
equations are:
1
a 8,
1
b 5,
1
c 9
2
b 2,
2
c 4
By cross multiplication method.
1 2 2 1 1 2 2 1 1 2 2 1
x y 1
b c b c ca c a a b a b
Put the values of .
1
a ,
2
a ,
1
b ,
2
b ,
1
c ,
2
c
x y 1
5 4 2 9 9 3 4 8 8 2 3 5
x y 1
20 18 27 32 16 15
x y 1
2 5 1
Take first and third part.
x
1
2
x 2
Take second and third part.
y
1
5
y 5
Hence, . and .
x 2
y 5
Question: 4
Form the pair of linear equations in the following
problems and find their solutions (if they exist) by
any algebraic method:
(i) A part of monthly hostel charges is fixed and
the remaining depends on the number of days
one has taken food in the mess. When a
student A takes food for 20 days she has to pay
Rs 1000 as hostel charges whereas a student B,
who takes food for 26 days, pays Rs 1180 as
hostel charges. Find the fixed charges and the
cost of food per day.
(ii) A fraction becomes when 1 is subtracted
1
3
from the numerator and it becomes when 8
1
4
is added to its denominator. Find the fraction.
(iii) Yash scored 40 marks in a test, getting 3 marks
for each right answer and losing 1 mark for
awarded for each correct answer and 2 marks
been deducted for each incorrect answer, then
Yash would have scored 50 marks. How many
questions were there in the test?
(iv) Places A and B are 100 km apart on a highway.
One car starts from A and another from B at
the same time. If the cars travel in the same
direction at different speeds, they meet in 5
hours. If they travel towards each other, they
meet in 1 hour. What are the speeds of the two
cars?
(v) The area of a rectangle gets reduced by 9
square units, if its length is reduced by 5 units
and breadth is increased by 3 units. If we
increase the length by 3 units and the breadth
by 2 units, the area increases by 67 square
units. Find the dimensions of the rectangle.
Solution
(i) Let us consider that the fixed charge and the
charge for food per day are
x and y
respectively.
According to the question,
x 20y 1000 ..... 1
x 26y 1180 ..... 2
Subtract equation .
1 from 2
x 26y x 20y 1180 1000
x 26y x 20y 180
6y 180
180
y
6
y 3
Put the value of in equation .
y
1
x 20 30 1000
x 600 1000
x 1000 600
x 400
Hence, the fixed charge and the charge for
food per day are and respectively.
400
30
(ii) Let us consider that the fraction is .
x
y
According to the first condition in the
question,
x 1 1
y 3
3 x 1 y
3x 3 y
3x y 3 ..... 1
According to the second condition in the
question,
x 1
y 8 4
4x y 8
4x y 8 .... 2
Subtract equation .
1 from 2
4x y 3x y 8 3
4x y 3x y 5
x 5
Put the value of in equation .
x
1
3 5 y 3
15 y 3
y 15 3
y 12
Hence, the required fraction is .
5
12
(iii) Let Yash wrote right answers and wrong
x
y
According to the question,
3x y 40 .... 1
4x 2y 50 ..... 2
Divide equation by 2.
2
2x y 25 ..... 3
Subtract equation from .
3
1
3x y 2x y 40 25
3x y 2x y 15
x 15
Put the value of in equation .
x
1
3 15 y 40
45 y 40
y 45 40
y 5
Therefore, .and
x 15
y 5
Hence, Yash wrote 15 right answers and 5
Total number of questions .
15 5 20
(iv) Let us consider,
Speed of one car .
ukm / h
Speed of other car .
vkm / h
By the formula of speed, distance and time,
Distance
Speed
Time
When travelling in same direction,
Respective speed of cars
u v km / h
Therefore,
100
u v
5
u v 20 ..... 1
When travelling in opposite direction,
Respective speed of cars .
u v km / h
Therefore,
100
u v
1
u v 100 ..... 2
1 and 2
u v u v 20 100
2u 120
120
u
2
u 60
Put the value of in equation .
u
1
60 v 20
v 60 20
v 40
Thus,
Speed of one car
60km / h
and speed of the other car
40km / h
(v) Let the length of the rectangle is unit and
x
y
Area of the rectangle square unit.
xy
Length of the rectangle after reducing by 5
square units .
x 5
Breadth of the rectangle after increasing by 3
square units .
y 3
Area of the rectangle square
x 5 y 3
units.
According to the question,
x 5 y 3 xy 9
xy 3x 5y 15 xy 9
3x 5y 6 0 ..... 1
Length of the rectangle after increasing by 3
square units
x 3
Breadth of the rectangle after increasing by 2
square units
y 2
Area of the rectangle square
x 3 y 2
units.
According to the question,
x 3 y 2 xy 67
xy 2x 3y 6 xy 67
2x 3y 61 0 ..... 2
Compare equations with the
1 and 2
general form of linear equations.
1 1 1
a x b y c 0
2 2 2
a x b y c 0
The values of the coefficients of the given
equations are:
1
a 3,
1
b 5,
1
c 6
2
b 3,
2
c 61
By cross multiplication method,
1 2 2 1 1 2 2 1 1 2 2 1
x y 1
b c b c ca c a a b a b
Put the values of the coefficients.
x y 1
5 61 3 6 6 2 61 3 3 3 2 5
x y 1
305 18 12 183 9 10
x y 1
323 171 19
Take first and third part.
x 1
323 19
323
x
19
x 17
Take second and third part.
y 1
171 19
171
y
19
y 9
Thus, the length of the rectangle units
17
9
Exercise 3.6 (2)
Question: 1
Solve the following pairs of equations by reducing
them to a pair of linear equations:
(i)
1 1
2
2x 3y
1 1 13
3x 2y 6
(ii)
2 3
2
x y
4 9
1
x y
(iii)
4
3y 14
x
3
4y 23
x
(iv)
5 1
2
x 1 y 2
6 3
1
x 1 y 2
(v)
7x 2y
5
xy
8x 7y
15
xy
(vi)
6x 3y 6xy
2x 4y 5xy
(vii)
10 2
4
x y x y
15 5
2
x y x y
(viii)
1 1 3
3x y 3x y 4
1 1 1
2 3x y 2 3x y 8
Solution
(i) Given equations:
1 1
2 .... 1
2x 3y
1 1 13
.... 2
3x 2y 6
Let and and put in equations
1
p
x
1
q
y
1
and .
2
From equation
1
p q
2
2 3
3p 2q
2
6
3p 2q 12 .... 3
From equation (2)
p q 13
3 2 6
2p 3q 13
6 6
2p 3q 13 4
Compare the equations and with the
3
4
general form of linear equations.
1 1 1
a p b q c 0
2 2 2
a p b q c 0
The values of the coefficients are:
1
a 3,
1
b 2
1
c 12
2
b 3,
2
c 13
By cross multiplication method,
1 2 2 1 1 2 2 1 1 2 2 1
p q 1
b c b c ca c a a b a b
Put the values of the coefficients.
p q 1
2 13 3 12 12 2 13 3 3 3 2 2
p q 1
26 36 24 39 9 4
p q 1
10 15 5
Take first and third part.
p 1
10 5
10
p
5
p 2
Take second and third part.
q 1
15 5
15
q
5
q 3
We know that .and
1
p
x
1
q
y
Thus,
and
1
2
x
1
3
y
and
1
x
2
1
y
3
Hence, and
1
x
2
1
y
3
(ii) Given equations:
2 3
2
x y
4 9
1
x y
Let and put in given equations.
1
p
x
1
q
y
2p 3q 2 ..... 1
4p 9q 1 ..... 2
Multiply equation by .
1
3
6p 9q 6 ..... 3
2
3
4p 9q 6p 9q 1 6
10p 5
5
p
10
1
p
2
Put the value of in equation .
p
1
1
2 3q 2
2
1 3q 2
3q 2 1
1
q
3
We know that and .
1
p
x
1
q
y
Thus,
and
1 1
2
x
1 1
3
y
and
1 1
x 4
1 1
y 9
and
x 4
y 9
Hence, and
x 4
y 9
(iii)
4
3y 14
x
3
4y 23
x
Let and put it in the given equations.
1
p
x
4p 3y 14 0 ..... 1
3p 4y 23 0 ..... 2
Compare the equations with the
1 and 2
general form of linear equations.
1 1 1
a p b y c 0
2 2 2
a p b y c 0
The values of the coefficients are:
1
a 4
1
,b 3,
1
c 14
2
a 3
2
,b 4,
2
c 23
By cross multiplication method,
1 2 2 1 1 2 2 1 1 2 2 1
p y 1
b c b c ca c a a b a b
Put the values of the coefficients.
p y 1
3 23 4 14 14 3 23 4 4 4 3 3
p y 1
69 56 42 92 16 9
p y 1
125 50 25
Take first and third part.
p 1
125 25
125
p
25
p 5
We know that .
1
p
x
Thus,
1
5
x
1
x
5
Take second and third part.
y 1
50 25
50
y
25
Hence, and
1
x
5
(iv)
5 1
2
x 1 y 2
6 3
1
x 1 y 2
Let and ,and put in the given
1
p
x 1
1
q
y 2
equations.
5p q 2 ..... 1
6p 3q 1 ..... 2
Multiply equation .by
1
3
15p 3q 6 ..... 3
2
3
6p 3q 15p 3q 1 6
21p 7
7
p
21
1
p
3
Put the value of in equation .
p
1
1
5 q 2
3
5
q 2
3
6 5
q
3
1
q
3
We know that and
1
p
x 1
1
q
y 2
Thus,
and
1 1
x 1 3
1 1
y 2 3
and
x 1 3
y 2 3
and
x 3 1
y 3 2
and
x 4
y 5
Hence, and
x 4
y 5
(v)
7x 2y
5
xy
8x 7y
15
xy
Simplify the given equations.
7x 2y
5
xy
7x 2y
5
xy xy
 
7 2
5 ..... 1
y x
Now,
8x 7y
15
xy
8x 7y
15
xy xy
8 7
15 .... 2
y x
Let and and put them in equations
1
p
x
1
q
y
and .
1
2
2p 7q 5 0 ..... 3
7p 8q 15 0 ..... 4
Compare the equations and with the
3
4
general form of linear equations.
1 1 1
a p b q c 0
2 2 2
a p b q c 0
The values of the coefficients are:
1
a 2
1
b 7,
1
c 5
2
b 8,
2
c 15
By cross multiplication method,
1 2 2 1 1 2 2 1 1 2 2 1
p q 1
b c b c ca c a a b a b
Put the values of the coefficients.
p q 1
7 15 8 5 5 7 15 2 2 8 7 7
p q 1
105 40 35 30 16 49
p q 1
65 65 65
Take first and third part.
p 1
65 65
p 1
Take second and third part.
q 1
65 65
q 1
We know that and .
1
p
x
1
q
y
Thus,
and
1
1
x
1
1
y
and
x 1
y 1
Hence,. and
x 1
y 1
(vi)
6x 3y 6xy
2x 4y 5xy
Divide the given equations by .
xy
6x 3y 6xy
xy xy xy
6 3
6 .... 1
y x
2x 4y 5xy
xy xy xy
2 4
5 ..... 2
y x
Let and , and put them in
1
p
x
1
q
y
equations and respectively.
1
2
3p 6q 6 0 ..... 3
4p 2q 5 0 .... 4
Compare the equations and with the
3
4
general form of linear equations.
1 1 1
a p b q c 0
2 2 2
a p b q c 0
The values of the coefficients are:
, ,
1
a 3
1
b 6
1
c 6
2
b 2,
2
c 5
By cross multiplication method,
1 2 2 1 1 2 2 1 1 2 2 1
p q 1
b c b c ca c a a b a b
Put the values of the coefficients.
p q 1
6 5 2 6 6 4 5 3 3 2 4 6
p q 1
30 12 24 15 6 24
p q 1
18 9 18
Take first and third part.
p 1
18 18
p 1
Now, take second and third part
q 1
9 18
9
q
18
1
q
2
We know that. and
1
p
x
1
q
y
Thus,
and
1
1
x
1 1
y 2
and
x 1
y 2
Hence, and .
x 1
y 2
(vii)
10 2
4
x y x y
15 5
2
x y x y
Let and and put them in the
1
p
x y
1
q
x y
given equations.
10p 2q 4 0 ..... 1
15p 5q 2 0 ..... 2
Compare the equations and with the
1
2
general form of linear equations.
1 1 1
a p b q c 0
2 2 2
a p b q c 0
The values of the coefficients are:
,
1
a 10
1
b 2,
1
c 4
2
a 15,
2
b 5,
2
c 2
By cross multiplication method,
1 2 2 1 1 2 2 1 1 2 2 1
p q 1
b c b c ca c a a b a b
Put the values of the coefficients.
p q 1
2 2 5 4 4 15 2 10 10 5 15 2
p q 1
4 20 60 20 50 30
p q 1
16 80 80
Take first and third part.
p 1
16 80
1
p
5
Take second and third part.
q 1
80 80
q 1
We know that and .
1
p
x y
1
q
x y
Thus,
and
1 1
x y 5
1
1
x y
and
x y 5
x y 1
x y 5 .... 3
x y 1 ..... 4
3
4
x y x y 5 1
2x 6
6
x
2
x 3
Put the value of in equation .
x
3
3 y 5
y 5 3
Hence, and .
x 3
y 2
(viii)
1 1 3
3x y 3x y 4
1 1 1
2 3x y 2 3x y 8
Let and and put them in
1
p
3x y
1
q
3x y
the given equations.
3
p q ..... 1
4
p q 1
2 2 8
1
p q ..... 2
4
1
2
3 1
p q p q
4 4
2
2p
4
2
p
4 2
1
p
4
Put the value of in equation .
p
1
1 3
q
4 4
3 1
q
4 4
2
q
4
1
q
2
We know that and .
1
p
3x y
1
q
3x y
Thus,
and
1 1
3x y 4
1 1
3x y 2
and
3x y 4
3x y 2
3x y 4 ..... 3
3x y 2 ..... 4
3
4
3x y 3x y 4 2
6x 6
6
x
6
x 1
Put the value of in equation .
x
3
3 1 y 4
y 4 3
y 1
Hence, .and
x 1
y 1
Question: 2
Formulate the following problems as a pair of
equations, and hence find their solutions:
(i) Ritu can row downstream 20 km in 2 hours,
and upstream 4 km in 2 hours. Find her speed
of rowing in still water and the speed of the
current.
(ii) 2 women and 5 men can together finish an
embroidery work in 4 days, while 3 women and
6 men can finish it in 3 days. Find the time
taken by 1 woman alone to finish the work,
and also that taken by 1 man alone.
(iii) Roohi travels 300 km to her home partly by
train and partly by bus. She takes 4 hours if
she travels 60 km by train and the remaining
by bus. If she travels 100 km by train and the
remaining by bus, she takes 10 minutes longer.
Find the speed of the train and the bus
separately.
Solution
(i) Let us consider,
Speed of Ritu in still water
xkm / h
Speed of the current
y km / h
In case of rowing upstream,
Speed of Ritu
x y km / h
In case of rowing downstream,
Speed of Ritu
x y km / h
From the formula of speed, distance and time,
Distance
Speed
Time
According to the question,
20
x y
2
x y 10 ..... 1
4
x y
2
x y 2 ..... 2
1
2
x y x y 10 2
2x 12
12
x
2
x 6
Put the value of in equation .
x
1
6 y 10
y 10 6
y 4
So, .and
x 6
y 4
Thus, the speed of Ritu in still water is
6km / h
and the speed of the current is .
4km / h
(ii) Let a woman takes days and a man takes
x
y
days to finish the work alone.
Work of a woman completed in one day
1
x
Work of a man completed in one day
1
y
Work done by 2 women and 5 men in 4 days
2 5
4
x y
Work done by 3 women and 6 men in 3 days
3 6
3
x y
According to the question,
2 5
4 1
x y
2 5 1
..... 1
x y 4
3 6
3 1
x y
3 6 1
..... 2
x y 3
Let and
and put in equations
1
p
x
1
q
y
1
and
2
1
2p 5q
4
8p 20q 1 ..... 3
1
3p 6q
3
9p 18q 1 ..... 4
Compare the equations and with the
3
4
general form of linear equations.
1 1 1
a p b q c 0
2 2 2
a p b q c 0
The values of the coefficients are:
1 1 1
a 8,b 20,c 1
2 2 2
a 9,b 18,c 1
By cross multiplication method,
1 2 2 1 1 2 2 1 1 2 2 1
p q 1
b c b c ca c a a b a b
Put the values of the coefficients.
 
p q 1
20 1 18 1 1 9 1 8 8 18 9 20
p q 1
20 18 9 8 144 180
p q 1
2 1 36
Take first and third part.
p 1
2 36
1
p
18
Take second and third part.
q 1
1 36
We know that .and
1
p
x
1
q
y
Thus,
and
1 1
x 18
1 1
y 36
and
x 18
y 36
Thus, and
x 18
y 36
Hence, the time taken by one woman is 18
days and the time taken by one man is 36
days.
(iii) Let the speed of the rain
ukm / h
Speed of the bus
vkm / h
Total distance covered km
300
From the formula of speed, distance and time,
Distance
Time
Speed
Time taken in covering 60 km by train
60
u
Time taken in covering remaining 240 km by
bus
240
v
According to the question, time taken in
covering complete 300 km is 4 hours.
Thus,
60 240
4 ..... 1
u v
Time taken in covering 100 km by train
100
u
Time taken in covering remaining 200 km by
bus
200
v
According to the question, time taken in
covering complete 300 km is 4 hours and 10
minutes.
Thus,
100 200 10
4
u v 60
100 200 1
4
u v 6
100 200 25
..... 2
u v 6
Let and and put them in equations
1
p
u
1
q
v
.and
1
2
60p 240q 4 ..... 3
25
100p 200q
6
600p 1200q 25 ..... 4
Multiply equation (3) by 10.
600p 2400q 40 ..... 5
Subtract equation (4) from (5).
600p 2400q 600p 1200q 40 25
1200q 15
15
q
1200
Put the value of in equation (3).
q
240
60p 4
80
60p 3 4
60p 4 3
We know that and .
1
p
u
1
q
v
Thus,
and
1 1
u 60
1 1
v 80
and
u 60
v 80
Thus, and
u 60
v 80
Hence, the speed of the train is and
60km / h
the speed of the bus is .
80km / h
Exercise 3.7 (8) (Optional)
Question: 1
The ages of two friends Ani and Biju differ by 3
years. Ani’s father Dharam is twice as old as Ani
and Biju is twice as old as his sister Cathy. The ages
of Cathy and Dharam differs by 30 years. Find the
ages of Ani and Biju.
Solution:
The difference between the ages of Biju and
Ani .
3years
There can be two conditions.
Biju is 3 years older than Ani.
Ani is 3 years older than Biju.
However, in both the cases, Ani’s father
Dharam will be 30 years older than Cathy.
Let us consider, the age of Ani be and the
x
age of Biju be years.
y
Thus, b the age of Dharam years.
2x
And the age of Biju’s sister Cathy years.
y
2
According to the question,
Case Ani is older than Biju by years,
1:
3
x y 3 ......(1)
y
2x 30
2
4x y 60 ......(2)
Subtract equation .
1 from 2
4x y x y 60 3
4x y x y 60 3
3x 57
x 19
Therefore,
Age of Ani years
19
Age of Biju years.
19 3 16
Case Biju is older than Ani by years,
2:
3
y x 3 ...... 1
y
2x 30
2
4x y 60 ......(2)
1 and 2
y x 4x y 3 60
3x 63
x 21
Therefore,
Age of Ani years
21
Age of Biju years.
21 3 24
Question: 2
One says, “Give me a hundred, friend! I shall then
become twice as rich as you”. The other replies, “If
you give me ten, I shall be six times as rich as you”.
Tell me what is the amount of their (respective)
capital?
Solution:
Let the two friends have Rs and Rs with them.
x
y
Now according to the question,
x 100 2 y 100
x 100 2y 200
x 2y 200 100
x 2y 300 ......(1)
And,
6 x 10 y 10
6x 60 y 10
6x y 10 60
6x y 70 ......(2)
Multiply equation by .
2
2
12x 2y 140 ......(3)
Now, subtract equation .
1 from 3
12x 2y x 2y 140 300
12x 2y x 2y 140 300
11x 440
x 40
Put the value of .
xin equation 1
x 2y 300
40 2y 300
2y 340
y 170
Thus, the two friends had Rs 40 and Rs 170 with
them respectively.
Question: 3
A train covered a certain distance at a uniform
speed. If the train would have been faster,
10Km/h
it would have taken hours less than the scheduled
2
time. And if the train were slower by ; it
10Km/h
would have taken 3 hours more than the scheduled
time. Find the distance covered by the train.
Solution:
Let the speed of train km/h.
x
The time taken by the train to cover the given
distance hours.
t
And the distance travelled km.
d
Now,
Distance
Speed
Time
Thus,
d
x
t
d xt ...... 1
According to the first condition of the question,
d
x 10
t 2
x 10 t 2 d
xt 2x 10t 20 d
Put the value of from equation .
d
1
xt 10t 2x 20 xt
2x 10t 20 ......(2)
According to the second condition of the question
d
x 10
t 3
x 10 t 3 d
xt 10t 3x 30 d
Put the value of from equation .
d
1
xt 10t 3x 30 xt
3x 10t 30 ......(3)
2 and 3
2x 10t 3x 10t 20 30
x 50
Put the value of
x
in equation
2
.
2 10 20x t
2 50 10 20t
10 20 100t
10 120t
12t hours
Now, put the value of
and
x t
in equation
1
.
50 12d
600d km
Hence, the distance travelled by the train is 600 km.
Question: 4
The students of a class are made to stand in rows. If
3 students are extra in a row, there would be 1 row
less. If 3 students are less in a row, there would be 2
rows more. Find the number of students in the
class.
Solution:
Let the number of rows
x
and number of
students in each row
y
.
Number of
Total number of Number
students in
students in the class of rows
a row
xy
According to the question,
Case
1
:
1 3x y
Total number of
studentsin the class
1 3xy x y
3 3xy xy x y
3 3 ......(1)x y
Case
2
:
2 3x y
Total number of
studentsin the class
2 3 6xy xy y x
3 2 6 ......(2)x y
Subtract equation
from1 2
.
3 2 3 6 3x y x y
3 2 3 6 3x y x y
9y
9y
Now, put the value of
in equation 1
y
.
3 3x y
3 9 3x
3 12x
4x
Number of rows
4
x
 
.
Number of students in a row
9
y
 
.
Thus, the total number of students in a class
4 9 36
xy
.
Question: 5
In a
ABC
,
3C
,
2B A B
. Find the
three angles.
Solution:
Given:
C 3 B 2 A B
Take second and third part.
3 2B A B
3 2 2B A B
2B A
2 0 ..... 1A B
By the angle sum property,
A B C 180
A B 3 B 180
A 4 B 180 ......(2)
Multiply equation
by1 4
.
8 A 4 B 0 ......(3)
and2 3
.
A 4 B 8 A 4 B 180 0
9 A 180
A 20
Put the value of
A
in equation
2
.
A 4 B 180
20 4 B 180
4 B 180 20
4 B 160
B 40
Now, find the value of
C
.
C 3 B
C 3 40
C 120
Thus, , and are , and
A
B
C
20
40
120
respectively.
Question: 6
Draw the graphs of the equations
5 5
x y
and
3 3
x y
. Determine the coordinates of the
vertices of the triangle formed by these lines and
the
y
-axis.
Solution:
Given equations:
5x y 5 ......(1)
)
3x y 3 ......(2)
To represent these equations graphically, you must
have at least two solutions for each equation.
For equation 1:
The points are , and .
1,0
2,5
5 5
y x
x
0
1
2
y
5
0
5
For equation 2:
The points are , and .
1,0
2,3
3 3
y x
The solution table is given below,
x
0
1
2
y
3
0
3
The graphical representation is as follows:
Fig. Exc_3.7_6
The formation of
ABC
by the given lines and the
axis
y
can be observed from the above graph. The
vertices of the triangle have coordinates A
1,0
, B
0, 3
, and C
0, 5
.
Question: 7
Solve the following pair of linear equations.
(i)
,
px qy p q
qx py p q
(ii)
,
ax by c
bx ay 1 c
(iii)
,
x y
0
a b
2 2
ax by a b
(iv)
2 2
a b x a b y a 2ab b
2 2
a b x y a b
(v)
,
152x 378y 74
378x 152y 604
Solution:
(i) Given equations:
px qy p q ......(1)
qx py p q ......(2)
Multiply equation
by and by1 2
p q
.
2 2
p x pqy p pq ......(3)
2 2
q x pqy pq q ......(4)
and3 4
.
2 2 2 2
p x q x p q
2 2 2 2
x p q p q
2 2
2 2
p q
x
p q
x 1
Put value of
in equation 1
x
.
px qy p q
p qy p q
qy q
y 1
Thus,
and1 1
x y
.
(ii) Given equations:
ax by c ......(1)
bx ay 1 c ......(2)
Multiply equation
by and by1 2
a b
.
2
a x aby ca ......(3)
2
b x aby b bc ......(4)
Subtract equation from .
4
3
2 2
a x aby b x aby ca b bc
2 2
a x b x ca b bc
2 2
x a b c a b b
2 2
c a b b
x
a b
Put the value of
x
in equation
1
.
2 2
c a b b
a by c
a b
2 2
ac a b ab
by c
a b
That is,
2 2
ac a b ab
by c
a b
2 2 2
2 2
a c b c a c abc ab
by
a b
2
2 2
abc b c ab
by
a b
2 2
bc a b ab
by
a b
2 2
c a b a
y
a b
Thus, and
2 2
c a b b
x
a b
2 2
c a b a
y
a b
(iii) Given equations:
x y
0
a b
bx ay 0 ......(1)
2 2
ax by a b ......(2)
Multiply equation
and by and1 2
b a
respectively.
2
b x aby 0 ......(3)
2 3 2
a x aby a ab ......(4)
3
4
2 2 3 2
b x aby a x aby 0 a ab
2 2 3 2
a x b x a ab
2 2 2 2
x a b a a b
x a
Put the value of
x
in equation
1
.
bx ay 0
b a ay 0
ay ab
y b
Thus, . and
x a
y b
(iv) Given equations:
2 2
a b x a b y a 2ab b ......(1)
2 2
a b x y a b
2 2
a b x a b y a b ......(2)
Subtract equation from .
2
1
2 2 2 2
a b x a b x a 2ab b a b
2
a b a b x 2ab 2b
2bx 2b a b
x a b
Put value of
x
in equation .
1
2 2
a b x a b y a 2ab b
2 2
a b a b a b y a 2ab b
2 2 2 2
a b a b y a 2ab b
a b y 2ab
2ab
y
a b
Thus, and .
x a b
2ab
y
a b
(v) Given:
152x 378y 74
76x 189y 37
189y 37
x ......(1)
76
378x 152y 604
189x 76y 302 ......(2)
Put the value of
x
from equation in to .
1
2
189y 37
189 76y 302
76
2 2
189 y 189 37 76 y 302 76
2 2
189 37 302 76 189 y 76 y
6993 22952 189 76 189 76 y
29945 113 265 y
29945 29945y
y 1
Now, out the value of
y
in equation
1
.
189y 37
x
76
189 1 37
x
76
x 2
Thus, , .
x 2
y 1
Question: 8
ABCD is a cyclic quadrilateral (see Fig. 3.7). Find
the angles of the cyclic quadrilateral.
Fig. Exc_3.7_8
Solution:
Given: ABCD is a cyclic quadrilateral.
In a cyclic quadrilateral, the sum of the measures of
opposite angles is
180
.
Therefore,
A C 180
.
A C 180
4y 20 4x 180
4x 4y 160
x y 40 ......(1)
Also,
B D 180
B D 180
3y 5 7x 5 180
7x 3y 180 ......(2)
Multiply equation
1
by
3
.
3x 3y 120 ......(3)
2
3
7x 3y 3x 3y 180 120
7x 3x 180 120
4x 60
x 15
Put the value of
x
in equation
1
.
x y 40
15 y 40
y 25
 
A 4y 20 4 25 20 120
 
B 3y 5 3 25 5 70
C 4x 4 15 60
. Thus,
D 7x 5 7 15 5 110
A,
and are , , , and
B,
C
D
120
70
60
110
respectively.