Example (Revised) - Chapter 4 - Linear Equations In Two Variables - Ncert Solutions class 9 - Maths

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Chapter 4 - Linear Equations in Two Variables - NCERT Solutions Class 9 Maths

Example 1 :

Write each of the following equations in the form $a x+b y+c=0$ and indicate the values of $a, b$ and $c$ in each case:
(i) $2 x+3 y=4.37$
(ii) $x-4=\sqrt{3} y$
(iii) $4=5 x-3 y$
(iv) $2 x=y$

Solution ;

(i) $2 x+3 y=4.37$ can be written as $2 x+3 y-4.37=0$. Here $a=2, b=3$ and $c=-4.37$.
(ii) The equation $x-4=\sqrt{3} y$ can be written as $x-\sqrt{3} y-4=0$. Here $a=1$, $b=-\sqrt{3}$ and $c=-4$.
(iii) The equation $4=5 x-3 y$ can be written as $5 x-3 y-4=0$. Here $a=5, b=-3$ and $c=-4$. Do you agree that it can also be written as $-5 x+3 y+4=0$ ? In this case $a=-5, b=3$ and $c=4$.

(iv) The equation $2 x=y$ can be written as $2 x-y+0=0$. Here $a=2, b=-1$ and $c=0$.

Equations of the type $a x+b=0$ are also examples of linear equations in two variables because they can be expressed as
$
a x+0 . y+b=0
$

For example, $4-3 x=0$ can be written as $-3 x+0 . y+4=0$.
Example 2 :

Write each of the following as an equation in two variables:
(i) $x=-5$
(ii) $y=2$
(iii) $2 x=3$
(iv) $5 y=2$

Solution :

(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 $0 \cdot x+1 \cdot y=2$, or $0 . x+1 \cdot y-2=0$.
(iii) $2 x=3$ can be written as $2 x+0 \cdot y-3=0$.
(iv) $5 y=2$ can be written as $0 . x+5 y-2=0$.

Example 3 :

Find four different solutions of the equation $x+2 y=6$.
Solution :

By inspection, $x=2, y=2$ is a solution because for $x=2, y=2$
$
x+2 y=2+4=6
$

Now, let us choose $x=0$. With this value of $x$, the given equation reduces to $2 y=6$ which has the unique solution $y=3$. So $x=0, y=3$ is also a solution of $x+2 y=6$. Similarly, taking $y=0$, the given equation reduces to $x=6$. So, $x=6, y=0$ is a solution of $x+2 y=6$ 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:
$
(2,2),(0,3),(6,0) \text { and }(4,1) \text {. }
$

Example 4 :

Find two solutions for each of the following equations:
(i) $4 x+3 y=12$
(ii) $2 x+5 y=0$
(iii) $3 y+4=0$

Solution :

(i) Taking $x=0$, we get $3 y=12$, 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 $5 y=0$, 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 $-\frac{2}{5}$. So $\left(1,-\frac{2}{5}\right)$ is another solution of $2 x+5 y=0$.
(iii) Writing the equation $3 y+4=0$ as $0 . x+3 y+4=0$, you will find that $y=-\frac{4}{3}$ for any value of $x$. Thus, two solutions can be given as $\left(0,-\frac{4}{3}\right)$ and $\left(1,-\frac{4}{3}\right)$.