We are given the equation $\frac{x}{1+i} - \frac{y}{2-i} = \frac{1-5i}{3-2i}$ and we need to solve for $x$ and $y$ such that the real parts and imaginary parts are equal on both sides of the equation.

1. Problem Description

We are given the equation

$\frac{x}{1+i} - \frac{y}{2-i} = \frac{1-5i}{3-2i}$

and we need to solve for $x$ and $y$ such that the real parts and imaginary parts are equal on both sides of the equation.

2. Solution Steps

First, we simplify the equation by multiplying both sides by $(1+i)(2-i)$

$x(2-i) - y(1+i) = \frac{1-5i}{3-2i}(1+i)(2-i)$

Next, we simplify the complex fraction:

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

So now we have

$x(2-i) - y(1+i) = (1-i)(1+i)(2-i) = (1 - i^2)(2-i) = (1+1)(2-i) = 2(2-i) = 4-2i$

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

$(2x - y) - (x+y)i = 4 - 2i$.

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

$2x - y = 4$

$x+y = 2$

Adding the two equations we get:

$3x = 6$

$x = 2$

Substituting $x=2$ into $x+y = 2$ we have

$2+y = 2$

$y = 0$

3. Final Answer

$x=2$ and $y=0$.