Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.4 Textbook Exercise Questions and Answers.

## RBSE Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.4

Question 1.

Give the geometric representations of y = 3 as an equation :

(i) in one variable,

Answer:

The representation of the solution on the number line, when y = 3 is treated as an equation in one variable, is as under :

(ii) in two variables.

Answer:

We know that y = 3 can be written as

0. x + y = 3 as a linear equation in two variables x and y.

Now all the values of x are permissible as 0.x is always 0. However, y must satisfy the relation y = 3.

Hence, three solutions of the given equation are :

x = 0, y = 3;x = 2, y = 3 and x = – 1, y = 3.

Plotting the points (0, 3), (2, 3) and (- 1, 3) and on joining them, we get the graph AB as a line parallel to x-axis at a distance of 3 units above it.

Question 2.

ive the geometric representations of 2x + 9 = 0 as an equation :

(i) in one variable,

Answer:

The representation of the solution of the equation 2x + 9 = 0, i.e., x

= – \(\frac{9}{2}\) is represented as under, by treating it as an equation in one variable :

(ii) in two variables.

Answer:

We know that 2x + 9 = 0 can be written as 2x + 0.y + 9 = 0 as a linear equation in two variables x and y. Now all the values of y are permissible as 0.y is always 0. However, x must satisfy the relation 2x + 9 = 0, i.e. 9

Hence, three solutions of the given equation are x = – \(\frac{9}{2}\) y = 0; x = – \(\frac{9}{2}\)

y = 2. and x = – \(\frac{9}{2}\) y = -2

Plotting the points (- \(\frac{9}{2}\), -2) (- \(\frac{9}{2}\), o) and (- \(\frac{9}{2}\) – 2) and on joining them, we get the graph AB as a line parallel toy-axis at a distance of \(\frac{9}{2}\)units on the left of y-axis.

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