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