Haryana State Board HBSE 10th Class Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 Textbook Exercise Questions and Answers.
Haryana Board 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 the age of Aftab be x years and age of his daughter be y years.
Seven years ago Aftab’s age = (x – 7) years
Seven years ago his daughter’s age = (y – 7) years
According to question,
x – 7 = 7 (y – 7)
⇒ x – 7 = 7y – 49
⇒ x – 7y = – 49 + 7 = – 42 ………………(1)
After three years Aftab’s age = (x + 3) years
After three years his daughter’s age = (y + 3) years
According to question
x + 3 = 3(y + 3)
⇒ x + 3 = 3y + 9
⇒ 3y = 9 – 3 = 6 …………..(2)
The algebraic representation of this situation is :
x – 7y = – 42 and x – 3y = 6
For representation of these equations graphically, we draw the graphs of these equations as follows :
x – 7y = – 42
⇒ – 7y = – 42 – x
⇒ y = \(\frac{42+x}{7}\)
We put the different values of x in this equation then we get different values of y and we prepare the table of x, y for the equation x – 7y = – 42.

and x – 3y = 6
⇒ 3y = x – 6
⇒ y = \(\frac{x-6}{3}\)
We put the different values of x in this equation then we get different values of y and we prepare the table of x, y for the equation x – 3y = 6.

Now, we plot the values of x and y from table 1 and 2 on the graph paper and we draw the graphs of the equations 1 and 2, those passes through these values.

Observe that we get two straight lines which intersect each other at point (42, 12).
![]()
Question 2.
The coach of a cricket team buys 3 bats and 6 balls for ‘ 3900. Later, he buys another bat and 3 more balls of the same kind for 1300. Represent this situation algebraically and geometrically.
Solution :
Let the price of 1 bat be ₹ x and price of 1 ball be ₹ y.
According to question,
3x + 6y = 3900
x + 2y = 1300 ………………(1)
and x + 3y = 1300 ……………..(2)
The algebraic representation of this situation is :
x + 2y = 1300
and x + 3y = 1300
For representation of these equations graphi-cally, we draw the graphs of these equations as follows :
x + 2y = 1300
⇒ 2y = 1300 – x
⇒ y = \(\frac{1300-x}{2}\)
We put the different values of x in this equation then we get different values of y and we prepare the table of x, y for the equation x + 2y = 1300.

and x + 3y = 1300
⇒ 3y = 1300 – x
⇒ y = \(\frac{1300-x}{3}\)
We put the different values ofx in this equation then we get different values of y and we prepare the table of x, y for the equation x + 3y = 1300.

Now, we plot the values of x andy on the graph paper from table 1 and 2 and we draw the graphs of equations 1, 2, those passes through these values.

Observe that we get two straight lines which intersect each other at point (1300, 0).
![]()
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 the cost of 1 kg. apples be ₹ x and cost of 1 kg. of grapes be ₹ y
According to question,
2x + y = 160 ……………….(1)
and 4x + 2y = 300
⇒ 2x + y = 150 ……………..(2)
The algebraic representation of this situation
2x + y = 160
and 2x + y = 150
For representation of these equations graphically, we draw the graphs of these equations as follows:
2x + y = 160
y = 160 – 2x
We put the different values of x in this equation then we get different values of y and we prepare the table of x, y for the equation 2x + y = 160.

2x + y = 150
⇒ y = 150 – 2x
We put the different values of x in this equation then we get different values of y and we prepare the table of x, y for the equation 2x + y = 150.

Now, we plot the values of x, y on the graph paper from table 1 and 2 and we draw the graphs of equations (1) and (2) which passes through these values.

Observe that we get two straight lines which are parallel to each other.