**Q.3.** 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) |

**Solution:**

(i) Putting x = 0 and y = 2 in the equation x - 2y = 4, we get,

0 - 2×2 = 4 ⇒ -4 = 4

Since LHS and RHS are not equal, (0, 2) is not a solution of the given equation.

(ii) Putting x = 2 and y = 0 in the equation x - 2y = 4, we get,

2 - 2×0 = 4 ⇒ 2 = 4

Since LHS and RHS are not equal, (2, 0) is not a solution of the given equation.

(iii) Putting x = 4 and y = 0 in the equation x - 2y = 4, we get,

4 - 2×0 = 4 ⇒ 4 = 4

Since LHS and RHS are equal, (4, 0) is a solution of the given equation.

(iv) Putting x = $\sqrt{2}$ and y = 4$\sqrt{2}$ in the equation x - 2y = 4, we get,

$\sqrt{2}$ - 2×4$\sqrt{2}$ = 4

⇒ $\sqrt{2}$ - 8$\sqrt{2}$ = 4

⇒ $\sqrt{2}$(1 - 8) = 4

⇒ -7$\sqrt{2}$ = 4

Since LHS and RHS are not equal, $\sqrt{2}$, 4$\sqrt{2}$) is not a solution of the given equation.

(v) Putting x = 1 and y = 1 in the equation x - 2y = 4, we get,

1 - 2×1 = 4 ⇒ -1 = 4

Since LHS and RHS are not equal, (1, 1) is not a solution of the given equation.