Solve for $x$ and $y$ in the equation $\frac{x}{1+i} - \frac{y}{2-i} = \frac{1-5i}{3-2i}$.

1. Problem Description

Solve for $x$ and $y$ in the equation $\frac{x}{1+i} - \frac{y}{2-i} = \frac{1-5i}{3-2i}$.

2. Solution Steps

First, let's simplify the equation by rationalizing the denominators:

$\frac{x}{1+i} = \frac{x(1-i)}{(1+i)(1-i)} = \frac{x(1-i)}{1 - i^2} = \frac{x(1-i)}{1 - (-1)} = \frac{x(1-i)}{2}$

$\frac{y}{2-i} = \frac{y(2+i)}{(2-i)(2+i)} = \frac{y(2+i)}{4 - i^2} = \frac{y(2+i)}{4 - (-1)} = \frac{y(2+i)}{5}$

$\frac{1-5i}{3-2i} = \frac{(1-5i)(3+2i)}{(3-2i)(3+2i)} = \frac{3 + 2i - 15i - 10i^2}{9 - 4i^2} = \frac{3 - 13i + 10}{9 + 4} = \frac{13 - 13i}{13} = 1 - i$

Now we have:

$\frac{x(1-i)}{2} - \frac{y(2+i)}{5} = 1 - i$

Multiply both sides by 10:

$5x(1-i) - 2y(2+i) = 10(1-i)$

$5x - 5xi - 4y - 2yi = 10 - 10i$

Now, we can group the real and imaginary terms:

$(5x - 4y) + (-5x - 2y)i = 10 - 10i$

Equating the real and imaginary parts, we have the following system of equations:

$5x - 4y = 10$

$-5x - 2y = -10$

Adding the two equations, we get:

$-6y = 0$

$y = 0$

Substituting $y = 0$ into the first equation:

$5x - 4(0) = 10$

$5x = 10$

$x = 2$

3. Final Answer

$x = 2$ and $y = 0$