Jharkhand Board JAC Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.1 Textbook Exercise Questions and Answers.

## JAC Board Class 9th Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.1

Question 1.

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

(Take the cost of a notebook to be ₹ x and that of a pen to be ₹ y).

Answer:

Let the cost of a pen be ₹ y and the cost of a notebook be ₹ x.

Cost of a notebook = twice the cost of a pen = 2y.

∴ x = 2y

⇒ x – 2y = 0

This is a linear equation in two variables to represent the given statement.

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 = \(9.3 \overline{5}\)

(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

Answer:

(i) 2x + 3y = \(9.3 \overline{5}\)

⇒ 2x + 3y – \(9.3 \overline{5}\) = 0

On comparing this equation with ax + by + c = 0, we get a = 2, b = 3 and c = –\(9.3 \overline{5}\)

(ii) x – \(\frac {y}{5}\) – 10 = 0

On comparing this equation with ax + by + c = 0, we get a = 1, b = –\(\frac {1}{5}\)and c = -10

(iii) -2x + 3y = 6

⇒ -2x + 3y – 6 = 0

On comparing this equation with ax + by + c = 0, we get a = -2, b = 3 and c = -6

(iv) x = 3y

⇒ x – 3y = 0

On comparing this equation with ax + by + c = 0, we get a = 1, b = -3 and c = 0

(v) 2x = -5y

⇒ 2x + 5y = 0

On comparing this equation with ax + by + c = 0, we get a = 2, b = 5 and c = 0

(vi) 3x + 2 = 0

⇒ 3x + Oy + 2 = 0

On comparing this equation with ax + by + c = 0, we get a = 3, b = 0 and c = 2

(vii) y – 2 = 0

⇒ 0x + y – 2= 0

On comparing this equation with ax + by + c = 0, we get a = 0, b = 1 and c = -2

(viii) 5 = 2x

⇒ 2x + 0y – 5 = 0

On comparing this equation with ax + by + c = 0, we get a = 2, b = 0 and c = -5