NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.4 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 4.4.
- 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 4.4 |
| Number of Questions Solved | 2 |
| Category | NCERT Solutions |
NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.4
Ex 4.4 Class 9 Maths Question 1.
Give the geometric representations of y – 3 as an equation
(i) in one variable.
(ii) in two variables.
Solution:
The given linear equation is
y = 3 …(i)
(i) The representation of the solution on the number line is shown in the figure below, where y = 3 is treated as an equation in one variable.
(ii) We know that y = 3 can be written as
0. x + y = 3
Which is a linear equation in the variables x and y. This is represented by a line. Now, all the values of x are permissible because 0.x is always 0. However, y must satisfy the equation y = 3.
Note that, the graph AB is a line parallel to the x-axis and at a distance of 3 units of the upper side of it.
Ex 4.4 Class 9 Maths Question 2.
Give the geometric representations of 2x + 9 = 0 as an equation
(i) in one variable
(ii) in two variables
Solution:
Given linear equation is 2x + 9 = 0 i.e., x = – \(\cfrac { 9 }{ 2 } \)
(i) If x = – \(\cfrac { 9 }{ 2 } \) is treated as an equation in one variable, then it has a unique
solution x = – \(\cfrac { 9 }{ 2 } \).
So, it is a point on the number line as shown below :
(ii) Given equation 2x + 9 = 0 can be written as 2x + 0 . y + 9 = 0, which is a linear equation in two variables x and y.
We can write the given equation as,
\(x=\cfrac { -9+0.y }{ 2 } \)
When y=1, then \(x=\cfrac { -9+0.(1) }{ 2 } \) = – \(\frac { 9 }{ 2 } \)
When y=2, then \(x=\cfrac { -9+0.(2) }{ 2 } \) = – \(\frac { 9 }{ 2 } \)
When y=3, then \(x=\cfrac { -9+0.(3) }{ 2 } \) = – \(\frac { 9 }{ 2 } \)
| X | – \(\frac { 9 }{ 2 } \) | – \(\frac { 9 }{ 2 } \) | – \(\frac { 9 }{ 2 } \) |
| y | 1 | 2 | 3 |
Now Playing the points – \((\cfrac { 9 }{ 2 },1 )\), – \((\cfrac { 9 }{ 2 },2 )\) – \((\cfrac { 9 }{ 2 },3 )\) paper and joining then ,we get a line PQ as a solution of 2x+9=0
Thus, the graph PQ is a line parallel to the y-axis at a distance \(\frac { 9 }{ 2 }=4.5 \)of units in the direction of negative X-axis.
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