Solve for x and y in the following simultaneous equatic 3^x * 3^y = 27, 2^x * 2^y = 32

The first equation is $3^{2x} \times 3^y = 27$.
Using the exponent rule $a^m \times a^n = a^{m+n}$, we combine the terms on the left side.

$$3^{2x+y} = 27$$

We know that $27 = 3^3$.

$$3^{2x+y} = 3^3$$

Equating the exponents, we get our first linear equation:

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

The second equation is $2^{x-y} \times 2^x = 32$.
Using the exponent rule $a^m \times a^n = a^{m+n}$, we combine the terms on the left side.

$$2^{(x-y)+x} = 32$$ $$2^{2x-y} = 32$$

We know that $32 = 2^5$.

$$2^{2x-y} = 2^5$$

Equating the exponents, we get our second linear equation:

$$2x-y = 5 \quad (2)$$

We have the system of equations:

1. $2x+y = 3$
2. $2x-y = 5$

Add equation (1) and equation (2) to eliminate $y$:

$$(2x+y) + (2x-y) = 3 + 5$$ $$4x = 8$$ $$x = \frac{8}{4}$$ $$x = 2$$

Using equation (1):

$$2x+y = 3$$

Substitute $x=2$:

$$2(2)+y = 3$$ $$4+y = 3$$ $$y = 3-4$$ $$y = -1$$

The solution is $x=2$ and $y=-1$.

The final answer is $x=2, y=-1$.