Solve the system of exponential equations: 8^x = 4^(y+1) and 3^(2x−y) = (1)/(27).

The given equation is $8^x = 4^{(y+1)}$.
We can express both bases as powers of 2: $8 = 2^3$ and $4 = 2^2$.

$$(2^3)^x = (2^2)^{(y+1)}$$

Apply the power rule $(a^m)^n = a^{mn}$:

$$2^{3x} = 2^{2(y+1)}$$

Since the bases are the same, the exponents must be equal:

$$3x = 2(y+1)$$ $$3x = 2y + 2$$

Rearrange to form a linear equation (Equation 1):

$$3x - 2y = 2 \quad (1)$$

The given equation is $3^{(2x-y)} = \frac{1}{27}$.
We can express $\frac{1}{27}$ as a power of 3: $\frac{1}{27} = \frac{1}{3^3} = 3^{-3}$.

$$3^{(2x-y)} = 3^{-3}$$

Since the bases are the same, the exponents must be equal:

$$2x - y = -3 \quad (2)$$

We have the system:

1. $3x - 2y = 2$
2. $2x - y = -3$

From Equation (2), we can express $y$ in terms of $x$:

$$y = 2x + 3$$

Substitute this expression for $y$ into Equation (1):

$$3x - 2(2x + 3) = 2$$ $$3x - 4x - 6 = 2$$ $$-x - 6 = 2$$ $$-x = 2 + 6$$ $$-x = 8$$ $$x = -8$$

Now substitute the value of $x$ back into the expression for $y$:

$$y = 2(-8) + 3$$ $$y = -16 + 3$$ $$y = -13$$

The solution is $x = -8$ and $y = -13$.

The final answer is $x = -8, y = -13$.