Solution

$y=3x+5$ 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) $nx+y=9$

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

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

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

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

(iii) $x=4y$

For $x=0$,

$0=4y$

$\Rightarrow y=0$

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

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

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, $x$

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

L.H.S $\ne$ 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, $x$

$-2y2-2\times 0=2\ne 4=$

$\ne$

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, $x$

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

$=4=$ R.H.S

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

$\begin{array}{rl}& \begin{array}{cc}x=\sqrt{2}y=4\sqrt{2}& (\sqrt{2},4\sqrt{2})\\ \text{ Puttina }\begin{array}{c}\text{ and }\end{array}\\ x-2y=\sqrt{2}-2(4\sqrt{2})& \text{ in the L.H.S of the given equation, }\end{array}\\ & =\sqrt{2}-8\sqrt{2}=-7\sqrt{2}\ne 4\end{array}$

L.H.S $\ne$ 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, $x$

$-2y1-2(1)=1-2=-1\ne 4=$

L.H.S $\ne$ R.H.S

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

Solution

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

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

Therefore, the value of $k$ is 7 .