2^(4−x) = 2^(7-y) and 3^(x+1) = 9^(-y)

$$3^{x+1} = 9^{-y}$$

Since $9 = 3^2$, we can rewrite the equation with a common base:

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

Equating the exponents:

$$x+1 = -2y \quad (*)$$

From $(*)$, we can write $x = -2y - 1$.
Substitute this into the first equation $3^{4-x} = 2^{7-y}$:

$$3^{4 - (-2y - 1)} = 2^{7-y}$$ $$3^{4 + 2y + 1} = 2^{7-y}$$ $$3^{5+2y} = 2^{7-y}$$

Take the natural logarithm (ln) of both sides:

$$\ln(3^{5+2y}) = \ln(2^{7-y})$$

Using the logarithm property $\ln(a^b) = b\ln a$:

$$(5+2y)\ln 3 = (7-y)\ln 2$$

Distribute the logarithms:

$$5\ln 3 + 2y\ln 3 = 7\ln 2 - y\ln 2$$

Group terms containing $y$ on one side and constant terms on the other:

$$2y\ln 3 + y\ln 2 = 7\ln 2 - 5\ln 3$$

Factor out $y$:

$$y(2\ln 3 + \ln 2) = 7\ln 2 - 5\ln 3$$

Use logarithm properties $n\ln a = \ln a^n$ and $\ln a + \ln b = \ln(ab)$:

$$y(\ln 3^2 + \ln 2) = \ln 2^7 - \ln 3^5$$ $$y(\ln 9 + \ln 2) = \ln 128 - \ln 243$$ $$y(\ln(9 \times 2)) = \ln\left(\frac{128}{243}\right)$$ $$y\ln 18 = \ln\left(\frac{128}{243}\right)$$

Solve for $y$:

$$y = \frac{\ln\left(\frac{128}{243}\right)}{\ln 18}$$

This can also be written using the change of base formula for logarithms:

$$y = \log_{18}\left(\frac{128}{243}\right)$$

We can express $128 = 2^7$ and $243 = 3^5$:

$$y = \log_{18}\left(\frac{2^7}{3^5}\right)$$

$$x = -2y - 1$$ $$x = -2 \left( \log_{18}\left(\frac{2^7}{3^5}\right) \right) - 1$$

Using logarithm properties:

$$x = \log_{18}\left(\left(\frac{2^7}{3^5}\right)^{-2}\right) - 1$$ $$x = \log_{18}\left(\left(\frac{3^5}{2^7}\right)^{2}\right) - 1$$ $$x = \log_{18}\left(\frac{3^{10}}{2^{14}}\right) - \log_{18} 18$$ $$x = \log_{18}\left(\frac{3^{10}}{2^{14} \times 18}\right)$$

Since $18 = 2 \times 3^2$:

$$x = \log_{18}\left(\frac{3^{10}}{2^{14} \times 2 \times 3^2}\right)$$ $$x = \log_{18}\left(\frac{3^{10}}{2^{15} \times 3^2}\right)$$ $$x = \log_{18}\left(\frac{3^{10-2}}{2^{15}}\right)$$ $$x = \log_{18}\left(\frac{3^8}{2^{15}}\right)$$

The final answers are:

$$\boxed{x = \log_{18}\left(\frac{3^8}{2^{15}}\right)}$$ $$\boxed{y = \log_{18}\left(\frac{2^7}{3^5}\right)}$$