8^(4-x) / 3^(x+7) = 2^(7-y) / 9^y

Since $8 = 2^3$, we can rewrite the first equation as:

$$(2^3)^{4-x} = 2^{7-y}$$ $$2^{3(4-x)} = 2^{7-y}$$ $$2^{12-3x} = 2^{7-y}$$

Equating the exponents, we get our first linear equation:

$$12 - 3x = 7 - y \quad (A)$$

Since $9 = 3^2$, we can rewrite the second equation as:

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

Equating the exponents, we get our second linear equation:

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

From equation (B), we can express $x$ in terms of $y$:

$$x = -2y - 1$$

Substitute this expression for $x$ into equation (A):

$$12 - 3(-2y - 1) = 7 - y$$ $$12 + 6y + 3 = 7 - y$$ $$15 + 6y = 7 - y$$

Now, gather terms with $y$ on one side and constants on the other:

$$6y + y = 7 - 15$$ $$7y = -8$$ $$y = -\frac{8}{7}$$

$$x = -2\left(-\frac{8}{7}\right) - 1$$ $$x = \frac{16}{7} - 1$$

To subtract, find a common denominator:

$$x = \frac{16}{7} - \frac{7}{7}$$ $$x = \frac{9}{7}$$

The solutions for $x$ and $y$ are:

$$\boxed{x = \frac{9}{7}}$$ $$\boxed{y = -\frac{8}{7}}$$

3 de faites, encore 2 aujourd'hui. Tu progresses.