Chapter 3 : Pair of Linear Equations in Two Variables

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Exercise 1 ( Page No. : 44 )

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.

Let the present age of Aftab and his father be x years and y years respectively.

According to question,

7 years ago, we have

x – 7 = 7 (y - 7)

Or, x – 7 = 7y – 49

Or, x – 7y = - 42 …………… (1)

3 years from now, we have

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

Or, x + 3 = 3y + 9

Or, x – 3y = 6 ……………. (2)

Graphical Representation

From equation (1), x – 7y = -42

Table value of x and y

x:	0	-42	-35
y:	6	0	1

From equation (2), x – 3y = 6

Table value of x and y

x:	0	9	6
y:	-2	1	0

Plotting the tables on the graph:

Exercise 1 ( Page No. : 44 )

The coach of a cricket team buys 3 bats and 6 balls for ` 3900. Later, she buys another bat and 3 more balls of the same kind for ` 1300. Represent this situation algebraically and geometrically.

Let the cost of one bat and one ball be x Rupees and y Rupees respectively.

According to first condition,

3x + 6y = ₨ 3900 …………….(1)

According to second condition,

x + 3y = ₨ 1300 …………….(2)

Graphical Representation

From equation (1), 3x + 6y = 3900

Table value of x and y

x:	100	300	700
y:	600	500	300

From equation (2), x + 3y = 1300

Table value of x and y

x:	100	700	400
y:	400	200	300

1 Unit = 100

Exercise 1 ( Page No. : 44 )

The cost of 2 kg of apples and 1kg of grapes on a day was found to be ` 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ` 300. Represent the situation algebraically and geometrically.

Let the cost of one kg apple be x ₨ and 1 kg grapes be y ₨.

According to first condition,

2x + y = 160 ₨ …………..(1)

According to second condition,

4x + 2y = 300

2x + y = 150 ……………….(2)

Graphical Representation

Table for equation (1), 2x + y = 160

Table value of x and y

x:	40	60	80
y:	60	40	0

Table for equation (2), 2x + y = 150

Table value of x and y

x:	40	60	20
y:	70	30	110

Exercise 2 ( Page No. : 50 )

Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is
36 m. Find the dimensions of the garden.

Exercise 2 ( Page No. : 50 )

Given the linear equation 2x + 3y – 8 = 0, write 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

Exercise 2 ( Page No. : 50 )

Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Given equations,

x + y + 1 = 0 …………………. (1)

3x + 2y + 2 = 0 ………………….. (2)

From the equation (1), we get

x	0	1	2
y	1	2	3

From the equation (2), we get

x	4	3	0
y	0	3	6

The coordinates of vertices of triangle formed by these lines and the x- axis are (-1, 0), (4, 0), (2, 3).

Exercise 3 ( Page No. : 53 )

Solve the following pair of linear equations by the substitution method.

(i) x + y = 14 …………….(1)

x – y = 4 …………….(2)

From the equation (1), we get

x = 14 - y …………….(3)

Putting the value of x in equation (2), we get

(14 - y) – y = 4

14 – y – y = 4

- 2y = - 10

Putting the value of y in equation (3),

x = 14 – 5

x = 9

Hence, x = 9 and y = 5

(iii) 3x - y = 3 …………….(1)

9x – 3y = 9…………….(2)

From the equation (1), we get

Putting the value of y in equation (2), we get

9x – 3 (3x - 3) = 9

9x – 9x + 9 = 9

9 = 9, which is true.

Therefore, pair of linear equation has infinite many solutions.

Exercise 3 ( Page No. : 53 )

Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

Exercise 3 ( Page No. : 53 )

Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and 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 the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes Find the fraction.

(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?

(i) Let the numbers be x and y, such that x > y

Therefore, according to question

x - y = 26 …………….(1)

x = 3y…………….(2)

Putting the value of x from equation (2) to equation (1), we get

3y – y = 26

2y = 26

y = 13

Putting the value in equation (2), we get

x = 3 x 13

x = 39

Hence, the numbers are 39 and 13.

(ii) Let one be x◦ and other be y◦ such that (x◦ > y◦)

Therefore, according to question

x◦ + y◦ = 180◦…………….(1) (Supplementary angles)

x◦ = 18 + y◦…………….(2)

Putting the value of x from equation (2) to equation (1), we get

18 + y◦ + y◦ = 180◦

2y◦ = 162◦

Putting the value of y in equation (2), we get

x◦ = 18 + 81

x = 99

(iv) Let the fixed charge be = ₨ x

Let the charge for 1 km distance be = ₨ y

According to first condition,

x + 10y = ₨ 105

x = 105 – 10y …………….(1)

According to second condition,

x + 15y = 155 ………………(2)

Putting the value of x in equation (2), we get

105 – 10y + 15y = 155

5y = 50

y = 10

Putting the value of y in equation (2), we get

x + 15 x 10 = 155

x = 5

Hence, the fixed charge for taxi is ₨ 5 and, the charge for one km distance is ₨ 50.

Charge for 25 km distance

= 25 x 10 + 5

= ₨ 255

(vi) Let the age of Jacob be = x years

Let the age of Jacob’s father be = y years

After 5 years,

Jacob’s age x + 5 years

Son’s age y + 5 years

According to question,

x + 5 = 3 (y + 5)

x + 5 = 3y + 15

x = 3y + 10………………(1)

Five years ago,

(x- 5) = 7 (y - 5)

x – 5 = 7y – 35

x – 7y = -30……………….(2)

Putting the value of x in equation (2), we get

3y + 10 - 7y = -30

-4y = -40

y = 10

Putting the value of y in equation (1), we get

x = 3 (10) + 10

x = 40

Hence, the present age of Jacob is 40 years and the age of his son is 10 years.

Chapter 3 : Pair of Linear Equations in Two Variables

Exercise 1 ( Page No. : 44 )

Exercise 2 ( Page No. : 50 )

Exercise 2 ( Page No. : 50 )

Exercise 3 ( Page No. : 53 )

Exercise 3 ( Page No. : 53 )

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