NCERT Solutions Class 9 Maths Chapter 4 Linear Equations In Two Variable Exercise 4.1
Introduction:
In this exercise/article we will learn about Linear Equations In Two Variable. Can you write down a linear equations in one variable? We may say that x + 1 = 0, x + √2 = 0 and √2y + √3 = 0 are examples of linear equations in one variable. You also know that such equations have a unique (i.e., one and only one) solution. You may also remember how to represent the solution on a number line. You shall also use the concepts you studied in chapter 3 to answer these questions.
NCERT Class 9 Maths Chapter 4 Coordinate Geometry :
- NCERT Class 9 Maths Chapter 4 Linear Equations In Two Variable Exercise 4.1
- NCERT Class 9 Maths Chapter 4 Linear Equations In Two Variable Exercise 4.2
Class 9 Maths Exercise 4.1 (Page-57)
Q1. 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 Rs. x and that of a pen to be Rs. y ).
Solution :
Let the cost of a notebook = x
Let the cost of a pen = y
According to the question,
The cost of a notebook is twice the cost of a pen
⇒ The cost of a notebook = 2 × the cost of a pen
⇒ x = 2 × y
⇒ x = 2y
⇒ x - 2y = 0
∴ The required linear equation = x - 2y = 0 .
Q2. 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\(\displaystyle \overline{5}\)
Solution :
We have 2x + 3y = 9.3\(\displaystyle \overline{5}\)
Re-arranging the equation, we get
2x + 3y - c9.3\(\displaystyle \overline{5}\) = 0
Now, comparing with ax + by + c = 0
we get,
⇒ a2 + b3 - c9.3\(\displaystyle \overline{5}\) = 0
⇒ a = 2 , b = 3 and c = -9.3\(\displaystyle \overline{5}\)
(ii) x - \(\displaystyle \frac{y}{5}\) - 10 = 0
Solution :
We have x - \(\displaystyle \frac{y}{5}\) - 10 = 0
Now, comparing with ax + by + c = 0
we get,
⇒ ax - b\(\displaystyle \frac{y}{5}\) - c10 = 0
⇒ a = 1 , b = -\(\displaystyle \frac{1}{5}\) and c = - 10
(iii) -2x + 3y = 6
Solution :
We have -2x + 3y = 6
Re-arranging the equation, we get
-2x + 3y - 6 = 0
Now, comparing with ax + by + c = 0
we get,
⇒ -a2 + b3 - 6 = 0
⇒ a = -2 , b = 3 and c = -6
(iv) x = 3y
Solution :
We have x = 3y
Re-arranging the equation, we get
x - 3y + 0 = 0
Now, comparing with ax + by + c = 0
we get,
⇒ ax - b3 + c0 = 0
⇒ a = 1 , b = -3 and c = 0
(v) 2x = -5y
Solution :
We have 2x = -5y
Re-arranging the equation, we get
2x + 5y + 0 = 0
Now, comparing with ax + by + c = 0
we get,
⇒ a2 + b5 + c0 = 0
⇒ a = 2 , b = 5 and c = 0
(vi) 3x + 2 = 0
Solution :
We have 3x + 2 = 0
Re-arranging the equation, we get
3x + 0 + 2 = 0
Now, comparing with ax + by + c = 0
we get,
⇒ a3 + b0 + c2 = 0
⇒ a = 3 , b = 0 and c = 2
(vii) y - 2 = 0
Solution :
We have y - 2 = 0
Re-arranging the equation, we get
0 + y - 2 = 0
Now, comparing with ax + by + c = 0
we get,
⇒ a0 + b1 - c2 = 0
⇒ a = 0 , b = 1 and c = -2
(viii) 5 = 2x
Solution :
We have 5 = 2x
Re-arranging the equation, we get
-2x + 0 + 5 = 0
Now, comparing with ax + by + c = 0
we get,
⇒ -a2 + b0 + c5 = 0
⇒ a = -2 , b = 0 and c = 5