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.