Solve the equation (1+2i)(x+yi) = 4+3i for x and y.

Given the equation $(1+2i)(x+yi) = 4+3i$.
Multiply the terms on the left side:

$$(1+2i)(x+yi) = 1(x) + 1(yi) + 2i(x) + 2i(yi)$$ $$= x + yi + 2xi + 2y i^2$$

Since $i^2 = -1$:

$$= x + yi + 2xi - 2y$$

We have $(x - 2y) + (y + 2x)i = 4+3i$.
Equating the real parts:

$$x - 2y = 4 \quad (1)$$

Equating the imaginary parts:

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

From equation (2), express $y$ in terms of $x$:

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

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

$$x - 2(3 - 2x) = 4$$ $$x - 6 + 4x = 4$$ $$5x - 6 = 4$$ $$5x = 4 + 6$$ $$5x = 10$$ $$x = \frac{10}{5}$$ $$x = 2$$

Now substitute the value of $x=2$ back into equation (3) to find $y$:

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

The solution is $x=2$ and $y=-1$.
This matches option A.

The final answer is $\text{A. } x=2, y=-1$.