8^(1−x) = 2^(1-y) and 9^(-y) = 3^(2x+y)

Express 8 as a power of 2: $8 = 2^3$.

$$(2^3)^{4-x} = 2^{2-y}$$ $$2^{3(4-x)} = 2^{2-y}$$ $$2^{12-3x} = 2^{2-y}$$

Since the bases are the same, equate the exponents:

$$12-3x = 2-y$$

Rearrange the terms to form a linear equation:

$$3x - y = 12 - 2$$ $$3x - y = 10 \quad (\text{Equation A})$$

Express 9 as a power of 3: $9 = 3^2$.

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

Since the bases are the same, equate the exponents:

$$-2y = 2x+y$$

Rearrange the terms to form a linear equation:

$$2x + y + 2y = 0$$ $$2x + 3y = 0 \quad (\text{Equation B})$$

We have:
Equation A: $3x - y = 10$
Equation B: $2x + 3y = 0$

From Equation A, express $y$ in terms of $x$:

$$y = 3x - 10$$

Substitute this expression for $y$ into Equation B:

$$2x + 3(3x - 10) = 0$$ $$2x + 9x - 30 = 0$$ $$11x - 30 = 0$$ $$11x = 30$$ $$x = \frac{30}{11}$$

$$y = 3\left(\frac{30}{11}\right) - 10$$ $$y = \frac{90}{11} - \frac{110}{11}$$ $$y = \frac{90 - 110}{11}$$ $$y = \frac{-20}{11}$$

The solutions are:

$$x = \boxed{\frac{30}{11}}$$ $$y = \boxed{-\frac{20}{11}}$$