We are given the equation $\frac{x}{1+i} - \frac{y}{2+i} = \frac{1-5i}{3-2i}$ We are asked to solve this equation. It is assumed that we are looking for real numbers $x$ and $y$ that satisfy the given equation.

1. Problem Description

We are given the equation

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

We are asked to solve this equation. It is assumed that we are looking for real numbers $x$ and $y$ that satisfy the given equation.

2. Solution Steps

First, we multiply both sides of the equation by $(1+i)(2+i)(3-2i)$ to get rid of the denominators:

$x(2+i)(3-2i) - y(1+i)(3-2i) = (1-5i)(1+i)(2+i)$

Let's expand the terms:

$(2+i)(3-2i) = 6 - 4i + 3i - 2i^2 = 6 - i + 2 = 8-i$

$(1+i)(3-2i) = 3 - 2i + 3i - 2i^2 = 3 + i + 2 = 5+i$

$(1-5i)(1+i) = 1 + i - 5i - 5i^2 = 1 - 4i + 5 = 6 - 4i$

$(6-4i)(2+i) = 12 + 6i - 8i - 4i^2 = 12 - 2i + 4 = 16-2i$

So our equation becomes:

$x(8-i) - y(5+i) = 16-2i$

$8x - ix - 5y - iy = 16 - 2i$

$(8x - 5y) - i(x+y) = 16 - 2i$

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

$8x - 5y = 16$

$x+y = 2$

From the second equation, $y = 2-x$. Substituting into the first equation:

$8x - 5(2-x) = 16$

$8x - 10 + 5x = 16$

$13x = 26$

$x=2$

Then $y = 2-x = 2-2 = 0$

3. Final Answer

$x=2$, $y=0$