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Linear Equations in Two Variables : Exercise 4.4 (Mathematics NCERT Class 9th)

Q.1      Give the geometric representation of y = 3 as an equation.
(i) in one variable           (ii) in two variables
Sol.

(i) The representation of the solution on the  number when y = 3 is treated as an equation in one variable is as under (ii)  We know that y = 3 can be written as 0. x + y = 3 as a linear equation in 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  solution of the given equation are
x = 0, y = 3 ;      x = 2, y = 3   ;    x = - 2 , y = 3 Plotting the  points (0, 3) (2, 3) and  (–2, 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.

Q.2       Given the geometric representations of 2x + 9 = 0  as an equation.
(i)  in one  variable             (ii)  in two variables
Sol.

(i) The  representation of the solution on the number line 2x + 9 = 0 i.e., $x = -{9\over 2}$ is treated as an  equation in one variable is as under. (ii) We know  that 2x + 9 = 0 can be written as 2x + 0y + 9  = 0 as a linear equation in variables x and y. Now all the  values of x are  permissible  as 0. y is always 0. However, x must  satisfy the relation 2x + 9 = 0
i.e., $x = - {9\over 2}$.
Hence three solution of the  given  equation are  $x =- {9\over 2}$,y = 0 ; $x =- {9\over 2}$,y = 2 and $x = - {9\over 2}$,y =– 2.
Therefore  Plotting  the point $\left( { - {9 \over 2},0} \right),\left( { - {9 \over 2},2} \right)\,and\,\left( { - {9 \over 2}, - 2} \right)$and on  joining them we get the graph AB as a line  parallel to y-axis at a distance of ${{9 \over 2}}$ on the left of y-axis. • Very easy solution

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