NCERT Solution Class 9th Maths Chapter – 4 Linear Equations in Two Variables Examples

NCERT Solution Class 9th Maths Chapter – 4 Linear Equations in Two Variables Examples
Last Doubt

NCERT Solution Class 9th Maths Chapter – 4 Linear Equations in Two Variables

TextbookNCERT
Class9th
SubjectMathematics
Chapter4th
Chapter NameLinear Equations in Two Variables
CategoryClass 9th Math Solutions 
Medium English
SourceLast Doubt

NCERT Solution Class 9th Maths Chapter – 4 Linear Equations in Two Variables

Chapter – 4

Linear Equations in Two Variables

Examples

Example 1  Write each of the following equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
(i) 2x + 3y = 4.37 (ii) x – 4 = √3 y (iii) 4 = 5x – 3y (iv) 2x = ySolution – (i) 2x + 3y = 4.37 can be written as 2x + 3y – 4.37 = 0. Here a = 2, b = 3 and c = – 4.37.
(ii) The equation x – 4 = √3y can be written as x – √3y – 4 = 0. Here a = 1, b = – √3 and c = – 4.
(iii) The equation 4 = 5x – 3y can be written as 5x – 3y – 4 = 0. Here a = 5, b = –3 and c = – 4. Do you agree that it can also be written as –5x + 3y + 4 = 0 ? In this case a = –5, b = 3 and c = 4.(iv) The equation 2x = y can be written as 2x – y + 0 = 0. Here a = 2, b = –1 and c = 0.
Equations of the type ax + b = 0 are also examples of linear equations in two variables
because they can be expressed as
ax + 0.y + b = 0
For example, 4 – 3x = 0 can be written as –3x + 0.y + 4 = 0
Example 2 : Write each of the following as an equation in two variables:
(i) x = –5 (ii) y = 2 (iii) 2x = 3 (iv) 5y = 2Solution – (i) x = –5 can be written as 1.x + 0.y = –5, or 1.x + 0.y + 5 = 0.
(ii) y = 2 can be written as 0.x + 1.y = 2, or 0.x + 1.y – 2 = 0.
(iii) 2x = 3 can be written as 2x + 0.y – 3 = 0.
(iv) 5y = 2 can be written as 0.x + 5y – 2 = 0.
Example 3 : Find four different solutions of the equation x + 2y = 6.

Solution – By inspection, x = 2, y = 2 is a solution because for x = 2, y = 2
x + 2y = 2 + 4 = 6
Now, let us choose x = 0. With this value of x, the given equation reduces to 2y = 6 which has the unique solution y = 3. So x = 0, y = 3 is also a solution of x + 2y = 6. Similarly, taking y = 0, the given equation reduces to x = 6. So, x = 6, y = 0 is a solution of x + 2y = 6 as well. Finally, let us take y = 1. The given equation now reduces to x + 2 = 6, whose solution is given by x = 4. Therefore, (4, 1) is also a solution of the given equation. So four of the infinitely many solutions of the given equation are:

Example 4 : Find two solutions for each of the following equations:
(i) 4x + 3y = 12
(ii) 2x + 5y = 0
(iii) 3y + 4 = 0Solution – (i) Taking x = 0, we get 3y = 12, i.e., y = 4. So, (0, 4) is a solution of the
given equation. Similarly, by taking y = 0, we get x = 3. Thus, (3, 0) is also a solution

(ii) Taking x = 0, we get 5y = 0, i.e., y = 0. So (0, 0) is a solution of the given equation. Now, if you take y = 0, you again get (0, 0) as a solution, which is the same as the earlier one. To get another solution, take x = 1, say. Then you can check that the
corresponding value of y is 2/5 So 2 (1,5/2) is another solution of 2x + 5y = 0.
(iii) Writing the equation 3y + 4 = 0 as 0.x + 3y + 4 = 0, you will find that y = 4/3 for
any value of x. Thus, two solutions can be given as (0,-4/3) – and (1, –4/3)

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