# PSEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1

PSEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1

## PSEB 10th Class Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1

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 Aftabs present age = x years

and Aftab’s daughter’s present age = y years

Algebraical-Situation

According to 1st condition,

x – 7 = 7(y – 7)

or x – 7 = 7y – 49

or x – 7y + 42 = 0

According to 2nd condition,

x + 3 = 3(y + 3)

or x + 3 = 3y + 9

or x – 3y – 6 = 0

∴ Pair of Line’ar Equation in two variables are

x – 7y + 42 = 0

and x – 3y – 6 = 0

Graphical – Situation:

x – 7y + 42 = 0

x – 3y – 6 = 0

x = 7y – 42 ………….(1)

x = 3y + 6 …………..(2)

Putting y = 5 in (1), we get

Putting y = 0 in (2), we get

x = 7 × 5 – 42 = 35 – 42

x = -7

Putting y = 6 in (1), we get

x = 7 × 6 – 42 = 42 – 42 = 0

Putting y = 7 in (1), we get:

x = 7 × 7 – 42 = 49 – 42 = 7

Table

Plotting the points A (-7, 5), B (0, 6), C (7, 7) and drawing a line joining them, we get the graph of the equation x – 7y + 42 = 0

x = 3 × 0 + 6

x = 0 + 6 = 6

Putting y = 3 in (2), we get:

x = 3 × 3 + 6

= 9 + 6 = 15

Putting y = -2 in (2), we get:

x = 3 × -2 + 6

= -6 + 6 = 0

Table:

Plotting the points D (6, 0), E (15, 3), F (0, -2) and drawing a line joining them, we get the graph of the equation x – 3y – 6 = 0

From the graph it is clear that the two lines intersect at G (42, 12).

Hence, x = 42 and y = 12 is the solution of given pair of linear equations.

Question 2.

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. [Pb. 2019, Set-A, B, C]

Solution:

Let cost of one bat = x

Cost of one hail = y

Algehraical – Situation

According to 1st condition,

3x + 6y = 3900

or x + 2y = 1300

According to 2nd condition,

1x + 3y = 1300

∴ Pair of linear equations in two variables are:

x + 2y = 13001

and x + 3y = 1300

Graphical – Situation:

x + 2y = 1300

x = 1300 – 2y …………..(1)

Putting y = 0 in (1), we get

x = 1300 – 2 × 0

x= 1300

Putting y = 500 in (1), we get :

= 1300 – 2 × 500

= 1300 – 1000 = 300

Putting y = 650 in (1), we get :

x = 1300 – 2 × 650

1300 – 1300 = 0

Table:

Plotting the points A (1300, 0), B (300, 500) and C (0, 650) drawing a line joining them we get the graph of the equation x + 2y = 1300.

x + 3y = 1300

x = 1300 – 3y …………….(2)

Putting y = 0 in (2), we get :

x = 1300 – 3 × 0 = 1300

Putting y = 500 in (2), we get :

x = 1300 – 3 × 500

= 1300 – 1500 = – 200

Putting y = 300 in (2), we get :

x = 1300 – 3 × 300

= 1300 – 900 = 400

Table:

Plotting the points A (1300, 0), E (-200, 500), F (400, 300) and drawing a line joining them we get the graph of the

equation.

x + 3y = 1300

From the graph it is clear that the two lines intersect at A (1300, 0).

Hence x = 1300 and y = 0 is the solution of given pair of linear equations.

Question 3.

The cost of 2 kg of apples and 1 kg 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.

Solution:

Let cost of 1 kg apples = ₹ x

Cost of 1 kg grapes = ₹ y

Algebraical – Situation

According to 1st condition,

2x + 1y = 160

According to 2nd condition,

4x + 2y = 300

∴ Pair of linear equations in two variables

2x + y = 160

and 4x + 2y = 300

Graphical – Situation

2x + y = 160

2x = 160 – y

Plotting the points D (75, 0), E (50, 50), F (0, 150) and drawing a line joining them, we get the graph of equation

4x + 2y = 300

From the graph, it is clear that the two lines do not intersect anywhere i.e. they are parallel.

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