Solve the equation $\frac{x}{1+i} - \frac{4}{2-i} = \frac{1-5i}{3-2i}$ for $x$.

1. Problem Description

Solve the equation $\frac{x}{1+i} - \frac{4}{2-i} = \frac{1-5i}{3-2i}$ for $x$.

2. Solution Steps

First, we isolate the term with $x$:

$$\frac{x}{1+i} = \frac{1-5i}{3-2i} + \frac{4}{2-i}$$

Next, we simplify the terms on the right side. To divide complex numbers, we multiply the numerator and denominator by the complex conjugate of the denominator.

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

$$\frac{4}{2-i} = \frac{4(2+i)}{(2-i)(2+i)} = \frac{8+4i}{4-i^2} = \frac{8+4i}{4+1} = \frac{8+4i}{5} = \frac{8}{5} + \frac{4}{5}i$$

Substituting these back into the equation:

$$\frac{x}{1+i} = (1-i) + (\frac{8}{5} + \frac{4}{5}i) = (1+\frac{8}{5}) + (-1+\frac{4}{5})i = \frac{13}{5} - \frac{1}{5}i$$

Multiply both sides by $1+i$ to solve for $x$:

$$x = (1+i)(\frac{13}{5} - \frac{1}{5}i) = \frac{13}{5} - \frac{1}{5}i + \frac{13}{5}i - \frac{1}{5}i^2 = \frac{13}{5} + \frac{12}{5}i + \frac{1}{5} = \frac{14}{5} + \frac{12}{5}i$$

3. Final Answer

$x = \frac{14}{5} + \frac{12}{5}i$