# Check which of the following are solutions of the equation x − 2y = 4 and which are not: <br/><br/>(i) $(0,2$ <br/><br/>(ii) $(2,0)$<br/><br/> (iii) $(4,0)$<br/><br/>(iv) $(\sqrt{2}, 4 \sqrt{2})$<br/><br/> (v) $(1,1)$

Solution:

(i) $(0,2)$

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

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

L.H.S ≠ 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,

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

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,

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

$=4=\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{S}$

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

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

Putting $x=\sqrt{2}$ and $y=4 \sqrt{2}$ in the L.H.S of the given equation,

$x-2 y=\sqrt{2}-2(4 \sqrt{2})$

$=\sqrt{2}-8 \sqrt{2}=-7 \sqrt{2} \neq 4$

L.H.S ≠ R.H.S

Therefore, $(\sqrt{2}, 4 \sqrt{2})$ is not a solution of this equation.

(v) $(1,1)$

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 ≠ R.H.S

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