Check which of the following are solutions of the equation x – 2y = 4 and which are not:

(i) (0, 2)

(ii) (2, 0)

(iii) (4, 0)

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

(v) (1, 1)

(i) Substituting (0, 2), in L.H.S. of the equation x - 2y = 4, we get :

⇒ x - 2y = 0 - 2(2) = 0 - 4 = -4.

Since, L.H.S. ≠ R.H.S.

Hence, (0, 2) is not a solution for equation x – 2y = 4.

(ii) Substituting (2, 0), in L.H.S. of the equation x - 2y = 4, we get :

⇒ x - 2y = 2 - 2(0) = 2 - 0 = 2

Since, L.H.S. ≠ R.H.S.

Hence, (2, 0) is not a solution for equation x – 2y = 4.

(iii) Substituting (4, 0), in L.H.S. of the equation x - 2y = 4, we get :

⇒ x - 2y = 4 - 2(0) = 4 - 0 = 4

Since, L.H.S. = R.H.S.

Hence, (4, 0) is a solution for equation x – 2y = 4.

(iv) Substituting ($\sqrt{2}, 4\sqrt{2}$), in L.H.S. of the equation x - 2y = 4, we get :

⇒ x - 2y = $\sqrt{2} - 2(4\sqrt{2}) = \sqrt{2} - 8\sqrt{2} = -7\sqrt{2}$.

Since, L.H.S. ≠ R.H.S.

Hence, $(\sqrt{2}, 4\sqrt{2})$ is not a solution for equation x – 2y = 4.

(v) Substituting, (1, 1), in L.H.S. of the equation x - 2y = 4, we get :

⇒ x - 2y = 1 - 2(1) = 1 - 2 = -1

Since, L.H.S. ≠ R.H.S.

Hence, (1, 1) is not a solution for equation x – 2y = 4.