Solve for x and y , if :
` (sqrt( 27) )^(x) ÷ 3^(y+ 4) = 1 and 8^(4-(x)/(3)) - 16^y = 0 `.

Answer

To solve for \( x \) and \( y \) in the equations given, we will follow these steps:

### Step 1: Rewrite the first equation
The first equation is:
\[ \frac{(\sqrt{27})^x}{3^{y+4}} = 1 \]
We can express \( \sqrt{27} \) as \( 27^{1/2} \). Since \( 27 = 3^3 \), we have:
\[ \sqrt{27} = (3^3)^{1/2} = 3^{3/2} \]
Thus, the equation becomes:
\[ \frac{(3^{3/2})^x}{3^{y+4}} = 1 \]

### Step 2: Simplify the equation
Using the properties of exponents, we can rewrite the left side:
\[ \frac{3^{(3/2)x}}{3^{y+4}} = 3^{(3/2)x - (y+4)} = 1 \]
Since \( 1 \) can be expressed as \( 3^0 \), we equate the exponents:
\[ \frac{3}{2}x - (y + 4) = 0 \]
This simplifies to:
\[ \frac{3}{2}x - y - 4 = 0 \]
Rearranging gives us:
\[ \frac{3}{2}x - y + 4 = 0 \quad \text{(Equation 1)} \]

### Step 3: Rewrite the second equation
The second equation is:
\[ 8^{4 - \frac{x}{3}} - 16^y = 0 \]
We can express \( 8 \) and \( 16 \) in terms of base \( 2 \):
\[ 8 = 2^3 \quad \text{and} \quad 16 = 2^4 \]
Thus, we rewrite the equation:
\[ (2^3)^{4 - \frac{x}{3}} - (2^4)^y = 0 \]
This simplifies to:
\[ 2^{3(4 - \frac{x}{3})} - 2^{4y} = 0 \]
Which can be rewritten as:
\[ 2^{12 - x} - 2^{4y} = 0 \]
Equating the exponents gives:
\[ 12 - x = 4y \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations
Now we have two equations:
1. \( \frac{3}{2}x - y + 4 = 0 \)
2. \( 12 - x = 4y \)

From Equation 1, we can express \( y \) in terms of \( x \):
\[ y = \frac{3}{2}x + 4 \]

Substituting this expression for \( y \) into Equation 2:
\[ 12 - x = 4\left(\frac{3}{2}x + 4\right) \]
Expanding the right side:
\[ 12 - x = 6x + 16 \]
Rearranging gives:
\[ 12 - 16 = 6x + x \]
\[ -4 = 7x \]
Thus, solving for \( x \):
\[ x = -\frac{4}{7} \]

### Step 5: Find \( y \)
Now substitute \( x \) back into the equation for \( y \):
\[ y = \frac{3}{2}\left(-\frac{4}{7}\right) + 4 \]
Calculating:
\[ y = -\frac{12}{7} + 4 = -\frac{12}{7} + \frac{28}{7} = \frac{16}{7} \]

### Final Solution
Thus, the solutions are:
\[ x = -\frac{4}{7}, \quad y = \frac{16}{7} \]