Solution

$y=3 x+5$ is a linear equation in two variables and it has infinite possible solutions. As for every value of $x$, there will be a value of $y$ satisfying the above equation and vice-versa.

Hence, the correct answer is (iii).

Solution

For $x=0$,

$2(0)+y=7 \Rightarrow$

$y=7$

Therefore, $(0,7)$ is a solution of this equation. For $x=1,2(1)$

$+y=7 \Rightarrow y=$

5

Therefore, $(1,5)$ is a solution of this equation.

For $x=-1$,

$2(-1)+y=7 \Rightarrow$

$y=9$

Therefore, $(-1,9)$ is a solution of this equation.

For $x=2$,

$2(2)+y=7$

$\Rightarrow y=3$

Therefore, $(2,3)$ is a solution of this equation.

(ii) $n x+y=9$

For $x=0, n(0)$

$+y=9$

$\Rightarrow y=9$

Therefore, $(0,9)$ is a solution of this equation.

For $x=1, \pi(1)+y=9 \Rightarrow y=9-n$

Therefore, $(1,9-\pi)$ is a solution of this equation.

For $x=2, n(2)+y=9 \Rightarrow y=9-2 n$

Therefore, $(2,9-2 \pi)$ is a solution of this equation.

For $x=-1, \Pi(-1)+y=9 \Rightarrow y=9+n$

$\Rightarrow(-1,9+\pi)$ is a solution of this equation.

(iii) $x=4 y$

For $x=0$,

$0=4 y$

$\Rightarrow y=0$

Therefore, $(0,0)$ is a solution of this equation.

For $y=1, x=4(1)=4$

Therefore, $(4,1)$ is a solution of this equation.

For $y=-1, x=4(-1) \Rightarrow x=-4$

$ \begin{aligned} & \text{ For } x=2, \\ & 2=4 y \\ & \Rightarrow y=\frac{2}{4}=\frac{1}{2} \\ & \text{ Therefore, }(2, \frac{1}{2}) \end{aligned} $

Therefore, $(-4,-1)$ is a solution of this equation.

Solution

(i) $(0,2)$

Putting $x=0$ and $y=2$ in the L.H.S of the given equation, $x$

$-2 y=0-2 x-4 \neq 42=$

L.H.S $\neq$ R.H.S

Therefore, $(0,2)$ is not a solution of this equation.

(ii) $(2,0)$

Putting $x=2$ and $y=0$ in the L.H.S of the given equation, $x$

$-2 y 2-2 \times 0=2 \neq 4=$

$\neq$

L.H.S R.H.S

Therefore, $(2,0)$ is not a solution of this equation.

(iii) $(4,0)$

Putting $x=4$ and $y=0$ in the L.H.S of the given equation, $x$

$-2 y=4-2(0)$

$=4=$ R.H.S

Therefore, $(4,0)$ is a solution of this equation.

$ \begin{aligned} & \begin{matrix} x=\sqrt{2} \quad y=4 \sqrt{2} & (\sqrt{2}, 4 \sqrt{2}) \\ \text{ Puttina } \begin{matrix} \text{ and } \end{matrix} \\ x-2 y=\sqrt{2}-2(4 \sqrt{2}) & \text{ in the L.H.S of the given equation, } \end{matrix} \\ & =\sqrt{2}-8 \sqrt{2}=-7 \sqrt{2} \neq 4 \end{aligned} $

L.H.S $\neq$ R.H.S

Therefore,

$ (\sqrt{2}, 4 \sqrt{2}) $

(v) $(1,1)$

is not a solution of this equation.

Putting $x=1$ and $y=1$ in the L.H.S of the given equation, $x$

$-2 y 1-2(1)=1-2=-1 \neq 4=$

L.H.S $\neq$ R.H.S

Therefore, $(1,1)$ is not a solution of this equation.

Solution

Putting $x=2$ and $y=1$ in the given equation,

$ \begin{aligned} & 2 x+3 y=k \\ & \Rightarrow \quad 2(2)+3(1)=k \\ & \Rightarrow \quad 4+3=k \Rightarrow k \\ & =7 \end{aligned} $

Therefore, the value of $k$ is 7 .