Solve for x and y, if :

$\Big(\sqrt{27}\Big)^x \div 3^{y+4} = 1$ and

$8^{{4}-\dfrac{x}{3}}- 16^y = 0$

$\Big(\sqrt{27}\Big)^x \div 3^{y+4} = 1\\[1em] ⇒ 27^{\dfrac{x}{2}} \div 3^{y+4} = 1\\[1em] ⇒ (3^3)^{\dfrac{x}{2}} \div 3^{y+4} = 3^0\\[1em] ⇒ 3^{\dfrac{3x}{2}} \div 3^{y+4} = 3^0\\[1em] ⇒ 3^{\dfrac{3x}{2} - {(y+4)}} = 3^0\\[1em] ⇒ \dfrac{3x}{2} - y - 4 = 0\\[1em] ⇒ \dfrac{3x}{2} - 4 = y………….(1)$

$8^{{4}-\dfrac{x}{3}}- 16^y = 0\\[1em] ⇒ 8^{{4}-\dfrac{x}{3}} = 16^y\\[1em] ⇒ (2^3)^{{4}-\dfrac{x}{3}} = (2^4)^y\\[1em] ⇒ (2)^{12-\dfrac{3x}{3}} = (2)^{4y}\\[1em] ⇒ (2)^{12 - x} = (2)^{4y}\\[1em] ⇒ 12 - x = 4y\\[1em] ⇒ x + 4y = 12 ………………(2)$

Multiply equation (1) by 2 and equation (2) by 3, then subtract equation (2) from equation (1).

$(\dfrac{3x}{2} - 4 = y) \times 2$ = 3x - 2y = 8

(x + 4y = 12) x 3 = 3x + 12y = 36

$\begin{matrix} & 3x & - & 2y & = & 8 \ & 3x & + & 12y & = & 36 \ & - & - & & & - \ \hline & & & -14y & = & 8 - 36 \ \Rightarrow & & - & 14y & = & -28 \end{matrix}$

⇒ y = $\dfrac{-28}{-14}$

⇒ y = 2

Substituting the value of y in equation (1), we get:

⇒ 3x - 2 $\times$ 2 = 8

⇒ 3x - 4 = 8

⇒ 3x = 8 + 4

⇒ 3x = 12

⇒ x = $\dfrac{12}{3}$

⇒ x = 4

Hence, the value of x = 4 and y = 2.