NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1.
- Linear Equations in Two Variables Class 9 Ex 4.1
- Linear Equations in Two Variables Class 9 Ex 4.2
- Linear Equations in Two Variables Class 9 Ex 4.3
- Linear Equations in Two Variables Class 9 Ex 4.4
Board | CBSE |
Textbook | NCERT |
Class | Class 9 |
Subject | Maths |
Chapter | Chapter 4 |
Chapter Name | Linear Equations in Two Variables |
Exercise | Ex 4.1 |
Number of Questions Solved | 2 |
Category | NCERT Solutions |
NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1
Ex 4.1 Class 9 Maths 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 Rs. x and that of a pen to be Rs. y)
Solution:
Let the cost of a notebook =Rs.x
and the cost of a pen = Rs.
y According to question,
Cost of a notebook = 2 (cost of a pen)
x = 2y ⇒ x -2y = 0
Ex 4.1 Class 9 Maths 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 \(\bar { 5 } \)
(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) Given linear equation, 2x + 3y= 9.35
∴ (2)x + (3)y + (- 9.3 \(\bar { 5 } \)) = 0
On comparing with ax ‘+ bx + c = 0, we have
a = 2, b = 3 and c = – 9.3\(\bar { 5 } \)
(ii) Given linear equation x-\(\frac { y }{ 5 }\)-10 =0
⇒ x+(-\(\frac { y }{ 5 }\)) y+(-10)=0
On comparing with ax + by + c = 0, we get
a = 1, b = –\(\frac { 1 }{ 5 }\) and c = -10
(iii) Given linear equation -2x + 3y = 6
⇒ -2x + 3 y – 6 = 0
⇒ (-2)x + (3) y + (-6) = 0
On Comparing with ax + by + c – 0, we get
a = – 2, b = 3 and c = – 6
(iv) Given linear equation x = 3y
⇒ x-3y=0
⇒(1) x + (-3) y + 0 = 0
On comparing with ax + by + c = 0, we get
a = 1, b = -3 and c = 0.
(v) Given linear equation 2x = – 5y
⇒ 2x + 5y = 0
⇒ (2)x + (5)y + 0 = 0
On comparing with ax + by + c = 0, we get
a – 2, b = 5 and c = 0.
(vi) Given linear equation 3x + 2 = 0
⇒ 3x +2 + 0y = 0
⇒ (3)x + (0)y + (2) = 0
On comparing with ax + by + c = 0 , we get
a = 3, b = 0 and c = 2.
(vii) Given linear equation y – 2=0
⇒ (0)x + (l)y + (-2) = 0
On comparing with ax + by + c = 0, we get
a= 0, b-1 and c = – 2
(viii) Given linear equation
5= 2x
⇒ 5 – 2x = 0
⇒ -2x+0y + 5 = 0
⇒ (-2)x + (0)y + (5) = 0
On comparing with ax + by-c = 0, we get
a= – 2, b = 0 and c = 5.
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