pair of linear equations in two variables 1

1PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

HOTS

(1) On comparing the ratios 1 1 1

2 2 2

a b c, and

a b c find out whether the lines representing the following pair of

linear equations, intersect at a point, are parallel or coincident. If it has unique solution then find by graphical

method.

3x + 2y – 6 = 0, 4x + 6y + 5 = 0

Ans : 3x + 2y – 6 = 0, 4x + 6y + 5 = 0

From given equation,

a1 = 3, b1 = 2, c1 = – 6,

a2 = 4, b2 = 6, c2 = 5

1

2

a 3

a 4� ... ... (1)

1

2

b 2 1

b 6 3� � ... ... (2)

1

2

c 6

c 5

�� ... ... (3)

From equation (1) and (2),

1 1

2 2

a b

a b�

Given pair of linear equations intersect in a point.

From graph solution : 23 39

,5 10

��

(Note : Student have to draw a graph)

(2) On comparing the ratios 1 1 1

2 2 2

a b c, and

a b c find out whther equations are consistent or inconsistent. If it

is consistent then find by graphical method.

4x – 5y – 20 = 0, 3x + 5y – 15 = 0

Ans : 4x – 5y – 20 = 0, 3x + 5y – 15 = 0

From given equation,

a1 = 4, b1 = – 5, c1 = – 20,

a2 = 3, b2 = 5, c2 = – 15

1

2

a 4

a 3� ... ... (1)

03

Chapter

2 MATHEMATICS

1

2

b 5 1

b 5 1

� �� ... ... (2)

From equation (1) and (2),

1 1

2 2

a b

a b�

Given pair of euqations are consistent.

Solution from graph : (5, 0)

(Note : Student have to draw a graph)

(3) Find the solution of folloiwng pair of linear equation by substitution method :

(i) 2x + 3y = 10; 3x – y = 4

(ii) 9x – 4y = 14; 7x – 3y = 11

Ans : 2x + 3y = 10 ... ... (1)

3x – y = 4 ... ... (2)

From equation (1),

2x = 10 – 3y

x = 10 3y

2

�

Substitute x = 10 3y

2

�, in equation (2),

3 10 3y

2

� ��

– y = 4

30 9y

2

� – y = 4

30 – 9y – 2y = 8

– 11y = 8 – 30

– 11y = – 22

y = 22

11

�

�

y = 2

Substitute y = 2, in equation (1),

2x + 3 (2) = 10

2x + 6 = 10

2x = 10 – 6

2x = 4

x = 2

Solution : (x, y) = (2, 2)


(ii) 9x – 4y = 14; 7x – 3y = 11

Ans : 9x – 4y = 14 ... ... (1)

7x – 3y = 11 ... ... (2)

From euqation (1),

9x = 14 + 4y

x = 14 4y

9

�

Substitute x = 14 4y

9

�, in equation (2) t

7 14 4y

9

� ��

– 3y = 11

98 + 28y – 27y = 99

y = 1

Substitute y = 1, in equation (1)

9x – 4 (1) = 14

9x = 14 + 4

9x = 18

x = 189

x = 2

Solution : (x, y) = (2, 1)

(4) Find the solution of folloiwng pair of linear equation by elimination method :

(i) x + y = 12; 4x – 5y = 12

Ans : x + y = 12 ... ... (1)

4x – 5y = 12 ... ... (2)

Multiply equation (1) by 5 and add equation (1) and (2)

5x + 5y = 60

5x – 5y = 12

9x = 72

x = 729

x = 8

Substitute x = 8, in euqation (1)t8 + y = 12

y = 12 – 8

y = 4

Solution : (x, y) = (8, 4)

4 MATHEMATICS

(ii) 22x + 25y = 21000, 25x + 22y = 21300

Ans : 22x + 25y = 21000 ... ... (1)

25x + 22y = 21300 ... ... (2)

Here, coefficient of x in equation (1) and coefiicient of y in equation (2) are same, as well coefficient

of x in equation (2) and coefficient of y in equation (1) are same. Add equation (1) and (2) so we get

equation (3).

22x + 25y = 21000

25x + 22y = 21300

47x + 47y = 42300

47 (x + y) = 42300

(x + y) = 42300

47

x + y = 900 ... ... (3)

Now, subtract equation (2) from equation (1), so we get equation (4)

22x + 25y = 21000

25x + 22y = 21300

– – – .

– 3x + 3y = – 300

– 3 (x – y) = – 300

x – y = – 300

– 3

x – y = 100 ... ... (4)

Add equation (3) and equation (4)

x + y = 900

x – y = 100

2x = 1000

x = 10002

x = 500

Substitute x = 500, in equation (3)

500 + y = 900

y = 900 – 500

y = 400

Solution : (x, y) = (500, 400)

(iii) � � � � � �x y x y

1 0; 15 010 5 8 6

Ans : First convert the above two equations into linear equations.


x + 2y – 10 = 0 ... ... (1)

3x + 4y – 360 = 0 ... ... (2)

Multiply equation (1) by 3 and subtract equation (2) from equation (1)

3x + 6y – 30 = 0

3x + 4y – 360 = 0

– – + _ .

2y + 330 = 0

2y = – 330

y = 3302

�

y = – 165

Substitute y = – 165, in equation (1),

x + 2 (– 165) – 10 = 0

x – 330 – 10 = 10

x – 340 = 0

x = 340

Solution : (x, y) = (340, – 165)

(iv) 7x – 2y = 3; 11x – 3

2y = 8

Ans : Convert 11x – 3

2y = 8, into linear equation, we get 22x – 3y = 16

7x – 6y = 3 ... ... (1)

22x – 3y = 16 ... ... (2)

Multiply equation (1) by 3 and equation (2) by 2 and subtract equation (2) from equation (1)

21x – 6y = 9

44x – 6y = 32

– + – .

– 23x = – 23

x = 23

23

�

�

x = 1

Substitute x = 1, in equation (1),

7 (1) – 2y = 3

– 2y = 3 – 7

– 2y = – 4

6 MATHEMATICS

y = 4

2

�

�

y = 2

Solution : (x, y) = (1, 2)

(v) 29x + 37y = 103 ... ... (1)

37x + 29y = 95 ... ... (2)

Ans : Here, coefficient of x in equation (1) and coefiicient of y in equation (2) are same, as well coefficient

of x in equation (2) and coefficient of y in equation (1) are same. Add equation (1) and (2) so we get

equation (3).

29x + 37y = 103

37x + 29y = 95

66x + 66y = 198

66 (x + y) = 198

(x + y) = 198

66

x + y = 3 ... ... (3)

Now, subtract equation (2) from equation (1), so we get equation (4)

29x + 37y = 103

37x + 29y = 95

– – – .

– 8x + 8y = 8

– 8 (x – y) = 8

x – y = 8

x – y = 8

8�

x – y = – 1 ... ... (4)

Add equation (3) and (4),x + y = 3

x – y = – 1

2x = 2

x = 2

2

x = 1

Substitute x = 1, in equation (1)

1 + y = 3

y = 3 – 1


y = 2

Solution : (x, y) = (1, 2)

(vi) 2x + y = 2x – y = 8

Ans : 2x + y = 2x – y = 8

2x + y = 2x – y =

3

22

x + y = 3

2

2x + 2y = 3 ... ... (1)

x – y = 3

2

2x – 2y = 3 ... ... (2)

Add equation (1) and (2)

2x + 3y = 3

2x – 3y = 3

4x = 6

x = 6

4

x = 3

2

Substitute x = 3

2, in equation (1)

2 3

2

� �� + 2y = 3

3 + 2y = 3

2y = 3 – 3

2y = 0

y = 0

Solution : (x, y) = (3

2, 0)

(5) Solve the following system of linear equations graphically. Shade the region bounded by these lines

and Yaxis. Also find the area of the region bounded by these lines and the Yaxis.

5x – y = 7; x – y + 1 = 0

Ans : 5x – y = 7; x – y + 1 = 0

Solution from graph : (2, 3)

8 MATHEMATICS

Area = 8 Sq. unit

(6) Father tells his son, “Five years ago, I was seven times as old as you were. After five years, I will be

three times as old as you will be.” Find present ages of father and his son.

Ans : Let present age of father is x years and present age of son is years.

Before 5 years,

Father's age = x – 5 years

Son's age = y – 5 years

From the given information,

x – 5 = 7 (y – 5)

x – 5 = 7y – 35

x – 7y + 30 = 0 ... ... (1)

After 5 years,

Father's age = x + 5 years

Son's age = 3 (y + 5) years


x + 5 = 3 (y + 5)

x + 5 = 3y + 15

x – 3y – 10 = 0 ... ... (2)

Subtract equation (2) from equation (1),

x – 7y + 30 = 0

x – 3y – 10 = 0

– + + .

– 4y + 40 = 0

– 4y = – 40

y = 40

4

�

�y = 10


x – 7 (10) + 30 = 0

x – 70 + 30 = 0

x – 40 = 0

x = 40

Solution : Present age of father is 40 years and present age of son is 10 years.

(7) Two years ago, Salim was thrice as old as his daughter and six years later, he will be four years older than

twice her age. How old are they now ?

Ans : Let present age of Salim is x years and her daughter is y years.


Before 2 years,

Salim's age = x – 2 years

Daughter's age = y – 2 years


x – 2 = 3 (y – 2)

x – 2 = 3y – 6

x – 3y + 4 = 0 ... ... (1)

After six years,

Salim's age = x + 6 years

Daughter's age = y + 6 years


x + 6 = 2 (y + 6) + 4

x + 6 = 2y + 12 + 4

x – 2y – 10 = 0 ... ... (2)


x – 3y + 4 = 0

x – 2y – 10 = 0

– + + .

– y + 14 = 0

y = 14


x – 3 (14) + 4 = 0

x – 42 + 4 = 0

x – 38 = 0

x = 38

Solution : Salim's present age is 38 years and his son's present age is 14 years.

(8) Two numbers are in the ratio of 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5.

Find the numbers.

Ans : Let two numbers are x and y are in the ratio of 5 : 6.

x 5

y 6�

x = 5y

6... ... (1)

If 8 is subtracted from both numbers, then their ratio becomes 4 : 5.

x 8 4

y 8 5

��

�

5x – 40 = 4y – 32

10 MATHEMATICS

5x – 4y – 8 = 0 ... ... (2)

Substitute x = 5y

6, in equation (2),

5 5y

6

� �� – 4y – 8 = 0

25y – 24y – 48 = 0

y – 48 = 0

y = 48

Substitute y = 48, in euqaiton (1)

x = 5(48)

6

x = 40 , So, reuquired two numbers are 40 and 48.

(9) The cost of a table is thrice the cost of a chair. The total cost of 4 chairs and a table is ` 2100. Find

the cost of a table and the cost of a chair.

Ans : Let, cost of a table is ` x and cost of a chair is ` y.


x = 3y ... ... (1)

x + 4y = 2100 ... ... (2)

Substitute x = 3y, in equation (2),

3y + 4y = 2100

7y = 2100

y = 2100

700

y = 300


x = 3 (300)

x = 900

Solution : Cost of a table is ` 900 and cost of a chair is ` 300.

(10) A fraction becomes 4

5, if 3 is added to both the numerator and the denominator. If 5 is added to the

numerator and the denominator, fraction

becomes 5

6. Find the fraction.

Ans : Let in a fraction, numerator is x and denominator is y.



x 3 4

y 3 5

��

�

5x + 15 = 4y + 12

5x – 4y + 3 = 0 ... ... (1)

x 5 5

y 5 6

��

�

6x + 30 = 5y + 25

6x – 5y + 5 = 0 ... ... (2)

Multiply equation (1) by 6 and equation (2) by 5, and subtract equation (2) from equation (1),

30x – 24y + 18 = 0

30x – 25y + 25 = 0

– + – – .

y – 7 = 0

y = 7


5x – 4(7) + 3 = 0

5x – 28 + 3 = 0

5x – 25 = 0

5x = 25

x = 25

5

x = 5

Solution : required fraction is5

7.

(11) The sum of two numbers is 137, and their difference is 43. Find the numbers.

Ans : Let, required two numbers are x and y.


x + y = 137 ... ... (1)

x – y = 43 ... ... (2)

Add equation (1) and (2),

x + y = 137

x + y = 137

2x = 180

x = 180

2

12 MATHEMATICS

x = 90


90 + y = 137

y = 137 – 90

y = 47

Solution : required two mubers are 90 and 47.

(12) Taxi charges in a city consists of fixed charges and the remaining depending upto the distance travelled

in kilometres. If a person travels 70 km, he pays ` 500 and for travelling 100 km, he pays ` 680. Find

the fixed charge and the rate per km.

Ans : Let fixed charge is ` x and charge per km is ` y.


x + 70y = 500 ... ... (1)

x + 100y = 680 ... ... (2)


x + 70y = 500

x + 100y = 680

– – – .

– 30y = – 180

y = 180

30

�

�

y = 60


x + 70 (60) = 500

x + 420 = 500

x = 500 – 420

x = 80

Solution : Fixed charge is ` 80 and charge per km is ` 60.

(13) A lady has some 50 paisa coins, 25 paisa coins in her purse. If she has 50 coins in all totaling

` 19.50, how many of each kind does she have ?

Ans : Let, a lady has x coins of 50 paisa and y coins of 25 paisa.

` 19.50 = 1950 paisa


x + y = 50 ... ... (1)

50x + 25y = 1950 ... ... (2)

From equation (1),


x = 50 – y

Substitue x = 50 – y, in equation (2),

50 (50 – y) + 25y = 1950

2500 – 50y + 25y = 1950

– 25y = 1950 – 2500

– 25y = – 550

y = 550

25

�

�

y = 22


x + 22 = 50

x = 50 – 20

x = 28

Solution : 28 coins of 50 paisa and 22 coins of 25 paisa.

(14) The sum of the digits of a two digit number is 12. The number obtained by interchanging its digits

exceeds the given number by 18. Find the number.

Ans : Let required two digit number is xy.

Original number = 10x + y

New number obtain after the interchanging of digits = 10y + x


x + y = 12 ... ... (1)

10y + x = 10x + y + 18

9y – 9x = 18

y – x = 2

y = 2 + x ... ... (2)

Substitute y = 2 + x, in equation (1),

x + 2 + x = 12

2x + 2 = 12

2x = 10

x = 5


y = 2 + 5

y = 7

Solution : Required two digit number is 57.

(15) In a cyclic quadrilateral ABCD A = (4y + 20)o, B = (3y – 5)o, C = (– 4x)o and D = (– 7x + 5)o. Find

the angles of the cyclic quadrilateral.

14 MATHEMATICS

Ans : A + C = 1800 (In a cyclic quadrilateral, opposite angles are supplementary)

4y + 20 + (– 4x) = 1800

4y – 4x = 1600

4(y – x) = 1600

y – x = 400

y = 400 + x ... ... (1)

B + D = 1800 (In a cyclic quadrilateral, opposite angles are supplementary)

3y – 5 – 7x + 5 = 1800

3y – 7x = 1800 ... ... (2)

Substitute y = 400 + x, in equation (2),

3 (400 + x) – 7x = 1800

1200 + 3x – 7x = 1800

– 4x = 600

x = – 150

Substitute x = – 150, in equation (1),

y = 40 + (– 15)

y = 25

Now, A = 4y + 20,

= 4 (25) + 20

= 100 + 20

= 1200

B = 3y – 5

= 3 (25) – 5

= 75 – 5

= 700

C = – 4x

= – 4(– 15)

= 600

B = – 7x + 5

= – 7 (– 15) + 5

= 1100

Solution : A = 1200, B = 700, C = 600, D = 1100

(16) The ratio of incomes of two persons A and B is 5 : 4 and the ratio of their expenditures is 7 : 5. If

each of them manages to save ` 9000 per month, find their monthly incomes.


Ans : Let monthly income and expenditure of two persons A and B are ` x and ` y respectively.

Monthly income of A is 5x,

Monthly income of B is 4x,

Monthly expence of A is 7y,

Monthly expence of B is 5y.


5x – 7y = 9000 ... ... (1)

4x – 5y = 9000 ... ... (2)


20x – 28y = 36000

20x – 25y = 45000

– + – .

– 3y = – 9000

y = 3000


5x – 7(3000) = 9000

5x – 21000 = 9000

5x = 9000 + 21000

5x = 30000

x = 30000

5

x = 6000

Now, monthly income of A = 5x

= 5 (6000)

= ` 30000

monthly income of B = 4x

= 4 (6000)

= ` 24000

Solution : Monthly income of A is ` 30000 and monthly income of B is ` 24000.

(17) A father's present age is three times the sum of ages of his two children. After 5 years his age will

be twice the sum of ages of the two children. Find the present age of the father.

Ans : Let fathers present age is x years and sum of present ages of his two children is y years


x = 3y ... ... (1)

After five years,

Father's age = x + 5 years

16 MATHEMATICS

Sum of ages of two children = (y + 10) years

x + 5 = 2 (y + 10)

x + 5 = 2y + 20

x – 2y = 15 ... ... (2)

Substitute x = 3y, in equation (2)

3y – 2y = 15

y = 15


x = 3(15)

= 45

Solution : Father's present age is 45 years.

(18) Total cost of 2 table and 3 chair is ` 2000 and total cost of 3 table and 2 chair is ` 2500. Find cost

of 1 table and 1 chair.

Ans : Let cost of a table is ` x and cost of a chair is ` y.


2x + 3y = 2000 ... ... (1)

3x + 2y = 2500 ... ... (2)


6x + 9y = 6000

6x + 4y = 5000

– – – .

5y = 1000

y = 1000

5

y = 200


2x + 3(200) = 2000

2x + 600 = 2000

2x = 2000 – 600

2x = 1400

x = 1400

2

x = 700

Solution : Cost of a table is ` 700 and cost of a chair is ` 200.(19) In a fraction sum of numerator and denominator is 4 more then twice the numerator. if 3 is added to

both numerator and denominator, the ratio of numerator and denominator becomes 2 : 3. Find the

fraction.

Ans : Let required fraction is x

y.



x + y = 2x + 4

x = y – 4 ... ... (1)

x 3 2

y 3 3

��

�

3x + 9 = 2y + 6

3x – 2y = – 3 ... ... (2)

Substitute x = y – 4, in equation (2)

3(y – 4) – 2y = – 3

3y – 12 – 2y = – 3

y = 12 – 3

y = 9


x = 9 – 4

= 5

Solution : Required fraction is 5

9.

(20) In the given figure, PQRST is a pentagon with QT || RS and QR || ST. If the perimeter of PQRST is

21 cm, find x and y.

Ans : RS = QT (Opposite sides of rectangle are equal)

x + y = 5

x = 5 – y ... ... (1)

Now, perimeter of pentagon PQRST is 21 cm.

PQ + QR + RS + TS + PT = 21

3 + x – y + x + y + x – y + 3 = 21

3x + y = 15 ... ... (2)

Substitute x = 5 – y, in equation (2)

3(5 – y) + y = 15

15 – 3y + y = 15

– 2y = 0

y = 0

Substitute y = 0, in eqwuation (1)

x = 5 – 0

x = 5

Solution : x = 5, y = 0.

18 MATHEMATICS

(21) In given figure ABCD is a rectangle. Then find x and y.

Ans : ABCD is a rectangle. So, their opposite sides are equal.

AB = CD and BC = AD

x + y = 30 ... ... (1)

x – y = 14 ... ... (2)

Add euqation (1) and (2),

x + y = 30

x – y = 14

2x = 44

x = 44

2

x = 22

Substitute x = 22, in euation (1)

22 + y = 30

y = 30 – 22

y = 8

Solution : x = 5, y = 0.

��

D C

A B

14

30

x + y

x – y

pair of linear equations in two variables 1

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