Linear Equations

Option (iii) infinitely many solutions is true.

Reason: A linear equation in two variables ($y = 3x + 5$) represents a straight line. A line consists of infinite points, and every point on the line is a solution to the equation. For every value of $x$, there is a corresponding value of $y$.

(i) $2x + y = 7$ $\Rightarrow y = 7 – 2x$

$\bullet$ If $x=0$, $y = 7 – 0 = 7$ $\rightarrow$ (0, 7)
$\bullet$ If $x=1$, $y = 7 – 2 = 5$ $\rightarrow$ (1, 5)
$\bullet$ If $x=2$, $y = 7 – 4 = 3$ $\rightarrow$ (2, 3)
$\bullet$ If $x=3$, $y = 7 – 6 = 1$ $\rightarrow$ (3, 1)

(ii) $\pi x + y = 9$ $\Rightarrow y = 9 – \pi x$

$\bullet$ If $x=0$, $y = 9$ $\rightarrow$ (0, 9)
$\bullet$ If $x=1$, $y = 9 – \pi$ $\rightarrow$ (1, 9-\pi)
$\bullet$ If $x=2$, $y = 9 – 2\pi$ $\rightarrow$ (2, 9-2\pi)
$\bullet$ If $x=-1$, $y = 9 + \pi$ $\rightarrow$ (-1, 9+\pi)

(iii) $x = 4y$

$\bullet$ If $y=0$, $x = 0$ $\rightarrow$ (0, 0)
$\bullet$ If $y=1$, $x = 4$ $\rightarrow$ (4, 1)
$\bullet$ If $y=2$, $x = 8$ $\rightarrow$ (8, 2)
$\bullet$ If $y=-1$, $x = -4$ $\rightarrow$ (-4, -1)

(i) $(0, 2)$
LHS $= 0 – 2(2) = -4 \neq 4$ (RHS) Not a Solution

(ii) $(2, 0)$
LHS $= 2 – 2(0) = 2 \neq 4$ (RHS) Not a Solution

(iii) $(4, 0)$
LHS $= 4 – 2(0) = 4 = 4$ (RHS) Solution

(iv) $(\sqrt{2}, 4\sqrt{2})$
LHS $= \sqrt{2} – 2(4\sqrt{2}) = \sqrt{2} – 8\sqrt{2} = -7\sqrt{2} \neq 4$ Not a Solution

(v) $(1, 1)$
LHS $= 1 – 2(1) = -1 \neq 4$ (RHS) Not a Solution

Since $x=2$ and $y=1$ is a solution, substitute these values into the equation:

$2(2) + 3(1) = k$
$4 + 3 = k$
$7 = k$

Answer: The value of $k$ is 7.