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

7,5

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

0, 2

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

8, 2

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

a 7,

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

a 2,

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

a 2,

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

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

a 4,

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

a 9,

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

a 2,

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

a 2,

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

a 3,

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

a 4,

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

0, 2

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

a 4,

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

a 1

a 1

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

1

2

c 1

c 3

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

10

y

2

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

35

y

7

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

26

x

2

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

77

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

Add equation .

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

y 3

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

y 3

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

y 3

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

y 3

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

Add equations .

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 .

Rs.3

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

a 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

a 3,

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

a 6,

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

a 3,

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

24

x

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

a 3,

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

each wrong answer. Had 4 marks been

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

answers.

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

wrong answers.

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

Add equations .

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

the breadth is unit.

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

a 2,

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

and the breadth units.

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

a 2,

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

Add equation and .

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

y 2

Hence, and

1

x

5

y 2

(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

Add equations .and

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

a 7,

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

a 4,

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

16

p

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

Add equations and .

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

Add equations .and

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

Add equations .and

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

Add equations .and

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

2

p

36

1

p

18

Take second and third part.

q 1

1 36

1

q

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

1

q

80

Put the value of in equation (3).

q

240

60p 4

80

60p 3 4

60p 4 3

1

p

60

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)

Add equations ,

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)

Add equations .

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)

Now, add equations

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 .

0, 5

1,0

2,5

5 5

y x

x

0

1

2

y

5

0

5

For equation 2:

The points are , and .

0, 3

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)

Add equations

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)

Add equations and .

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)

Now, add equations and .

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.