Haryana State Board HBSE 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.1 Textbook Exercise Questions and Answers.

## Haryana Board 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Exercise 4.1

Question 1.

The cost of a note-book is twice the cost of a pen. Write a linear equation in two variables to represent this statement.

[Hint : Take the cost of a note-book to be ₹ x and that of a pen to be ₹ y.]

Solution:

Let the cost of a note-book be. ₹x and cost of a pen be ₹ y.

According to statement, the cost of a notebook is twice the cost of a pen.

So, the required linear equation in two variables to represent above statement is

x = 2y or x – 2y = 0

Question 2.

Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case :

(i) 2x + 3y = \(9 \cdot 3 \overline{5}\)

(ii) x – \(\frac {y}{5}\) – 10 = 0

(iii) – 2x + 3y = 6

(iv) x = 3y

(v) 2x = – 5y

(vi) 3x + 2 = 0

(vii) y – 2 = 0

(viii) 5 = 2x.

Solution :

(i) Writing the given equation 2x + 3y = \(9 \cdot 3 \overline{5}\) in the form ax + by + c = 0, we get

2x + 3y – \(9 \cdot 3 \overline{5}\) = 0 …(i)

Comparing the equation (i) with the standard form of the linear equation ax + by + c = 0, we get

a = 2, b = 3 and c= – \(9 \cdot 3 \overline{5}\).

(ii) Writing the given equation x – \(\frac {y}{5}\) – 10 = 0 in the form ax + by + c = 0, we get

1.x – \(\frac {1}{5}\)y – 10 = 0 ……..(i)

Comparing the equation (i) with the standard form of the linear equation ax + by + c = 0, we get

a = 1, b = – \(\frac {1}{5}\) and c = – 10.

(iii) Writing the given equation – 2x + 3y = 6 in the form ax + by + c = 0, we get

– 2x + 3y – 6 = 0 …(i)

Comparing the equation (i) with the standard form of the linear equation ax + by + c = 0, we get

a = – 2, b = 3 and c =- 6.

(iv) Writing the given equation x = 3y in the form ax + by + c = 0, we get

1.x – 3y + 0 = 0 …….(i)

Comparing the equation (i) with the standard form of the linear equation ax + by + c = 0, we get

a = 1, b = -3 and c = 0.

(v) Writing the given equation 2x = – 5y in the form ax + by + c = 0, we get

2x + 5y + 0 = 0 …….(i)

Comparing the equation (i) with the standard form of the linear equation ax + by + c = 0, we get

a = 2, b = 5 and c = 0.

(vi) Writing the given equation 3x + 2 = 0 in the form ax + by + c = 0, we get

3x + 0.y + 2 = 0 ……..(i)

Comparing the equation (i) with the standard form of the linear equation ax + by + c = 0, we get

a = 3, b = 0 and c = 2.

(vii) Writing the given equation y – 2 = 0 in the form ax + by + c = 0, we get

0.x + 1.y – 2 = 0 …(i)

Comparing the equation (i) with the standard form of the linear equation ax + by + c = 0, we get

a = 0, b = 1 and c = -2.

(viii) Writing the given equation 5 = 2x in the form ax + by + c = 0, we get

– 2x + 0.y + 5 = 0 …..(i)

Comparing the equation (i) with the standard form of the linear equation ax + by + c = 0, we get

a = – 2, b = 0 and c = 5.