RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1

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

RBSE Class 9 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 notebook be ₹ x and that of a pen be ₹ y.
Since the cost of a notebook is twice the cost of a pen, so the required linear equation in two variables to represent the above statement is given by x = 2y.

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 = 9.35̄
Answer:
2x + 3y = 9.35̄ can be written as 2x + 3y – 9.35̄ = 0.
On comparing it with ax + by + c = 0,
a = 2, b = 3 and c = – 9.35̄.

(ii) x – \(\frac{y}{5}\) – 10 = 0
Answer:
On comparing x – – 10 = 0 with ax + by + c = 0,
a = 1, b = –\(\frac{1}{5}\) and c = – 10.

(iii) – 2x + 3y = 6
Answer:
– 2x + 3y = 6 can be written as – 2x + 3y – 6 = 0.
On comparing it with ax + by + c = 0,
a = – 2, b = 3 and c = – 6.

(iv) x = 3y
Answer:
x = 3y can be written as x – 3y = 0.
On comparing it with ax + by + c = 0,
a = 1, b = – 3 and c = 0.

(v) 2x = – 5y
Answer:
2x = – 5y can be written as 2x + 5y = 0.
On comparing it with ax + by + c = 0,
a – 2, b = 5 and c = 0.

(vi) 3x + 2 = 0
Answer:
On comparing 3x + 2 = 0 with ax + by + c = 0,
a = 3, b = 0 and c = 2.

(vii) y – 2 = 0
Answer:
On comparing y – 2-0 with ax + by + c = 0,
a = 0, b = 1 and c = – 2.

(viii) 5 = 2x
Answer:
5 = 2x can be written as 2x – 5 = 0.
On comparing it with ax + by + c = 0,
a = 2, b = 0 and c = – 5.