Solve :
` 8 ^(x+1) = 16 ^(y+2) and ( (1)/(2)) ^(3+x) = ((1)/(4))^(3y)`

Answer

To solve the equations \( 8^{(x+1)} = 16^{(y+2)} \) and \( \left(\frac{1}{2}\right)^{(3+x)} = \left(\frac{1}{4}\right)^{(3y)} \), we will follow these steps:

### Step 1: Rewrite the bases in terms of powers of 2
We know that:
- \( 8 = 2^3 \)
- \( 16 = 2^4 \)

Thus, we can rewrite the first equation:
\[ 8^{(x+1)} = 16^{(y+2)} \implies (2^3)^{(x+1)} = (2^4)^{(y+2)} \]
This simplifies to:
\[ 2^{3(x+1)} = 2^{4(y+2)} \]

### Step 2: Set the exponents equal to each other
Since the bases are the same, we can set the exponents equal to each other:
\[ 3(x+1) = 4(y+2) \]
Expanding both sides gives:
\[ 3x + 3 = 4y + 8 \]
Rearranging this, we get:
\[ 3x - 4y = 5 \quad \text{(Equation 1)} \]

### Step 3: Rewrite the second equation
Now, let's rewrite the second equation:
\[ \left(\frac{1}{2}\right)^{(3+x)} = \left(\frac{1}{4}\right)^{(3y)} \]
We know that:
- \( \frac{1}{4} = \left(\frac{1}{2}\right)^2 \)

Thus, we can rewrite the second equation as:
\[ \left(\frac{1}{2}\right)^{(3+x)} = \left(\left(\frac{1}{2}\right)^2\right)^{(3y)} \implies \left(\frac{1}{2}\right)^{(3+x)} = \left(\frac{1}{2}\right)^{(6y)} \]

### Step 4: Set the exponents equal to each other
Again, since the bases are the same, we can set the exponents equal:
\[ 3 + x = 6y \quad \text{(Equation 2)} \]

### Step 5: Solve the system of equations
Now we have two equations:
1. \( 3x - 4y = 5 \)
2. \( 3 + x = 6y \)

From Equation 2, we can express \( x \) in terms of \( y \):
\[ x = 6y - 3 \]

### Step 6: Substitute \( x \) in Equation 1
Substituting \( x \) in Equation 1:
\[ 3(6y - 3) - 4y = 5 \]
Expanding this gives:
\[ 18y - 9 - 4y = 5 \]
Combining like terms:
\[ 14y - 9 = 5 \]
Adding 9 to both sides:
\[ 14y = 14 \]
Dividing by 14:
\[ y = 1 \]

### Step 7: Find \( x \)
Now substituting \( y = 1 \) back into Equation 2:
\[ 3 + x = 6(1) \implies 3 + x = 6 \]
Subtracting 3 from both sides:
\[ x = 3 \]

### Final Answer
Thus, the values of \( x \) and \( y \) are:
\[ x = 3, \quad y = 1 \]