Check which of the following are solutions of the equation $ x-2 y=4 $ and which are not:
(i) $ (0,2) $
(ii) $ (2,0) $
(iii) $ (4,0) $
(iv) $ (\sqrt{2}, 4 \sqrt{2}) $
(v) $ (1,1) $.

To do:

We have to check which of the given solutions are solutions of the equation $x-2y=4$.

Solution:

(i) Given, (0, 2)

That is,

$(x, y) = (0, 2)$

Now, by substituting $(x, y) = (0, 2)$ in equation $x-2y=4$

We get,

$0-2(2)=4$

$0-4=4$

$-4 ≠4$

Therefore,

$(0, 2)$ is not a solution of the given equation $x-2y=4$. 

(ii) Given, (2, 0)

That is,

$(x, y) = (2, 0)$

Now, by substituting $(x, y) = (2, 0)$ in equation $x-2y=4$

We get,

$2-2(0)=4$

$2-0=4$

$2 ≠4$

Therefore,

$(2, 0)$ is not a solution of the given equation $x-2y=4$.

(iii) Given, (4, 0)

That is,

$(x, y) = (4, 0)$

Now, by substituting $(x, y) = (4, 0)$ in equation $x-2y=4$

We get,

$4-2(0)=4$

$4-0=4$

$4 =4$

Therefore,

$(4, 0)$ is a solution of the given equation $x-2y=4$.

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

That is,

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

Now, by substituting $(x, y) = (\sqrt {2}, 4\sqrt{2})$ in equation $\sqrt {2}, 4\sqrt{2})=4$

We get,

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

$-7\sqrt {2}≠4$

Therefore,

$(\sqrt {2}, 4\sqrt {2})$ is not a solution of the equation $x-2y=4$. 

(v) Given, (1, 1)

That is,

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

Now, by substituting $(x, y) = (1, 1)$ in equation $x-2y=4$

We get,

$1-2(1)=4$

$1-2=4$

$-1 ≠4$

Therefore,

$(1, 1)$ is not a solution of the equation $x-2y=4$.