The problem requires solving for $x$ and $y$ such that:

$x(3 - 2i) + y(10 - 5i) - (1 - 4i) = 0.$

Step-by-Step Solution:

1. Rewrite the equation:

   Expand and group the real and imaginary terms:

   $x(3 - 2i) + y(10 - 5i) - (1 - 4i) = 0.$

   This becomes:

   $(3x + 10y - 1) + (-2x - 5y + 4)i = 0.$

2. Separate real and imaginary parts:

   The equation will hold true if the real and imaginary parts are both zero:

   $3x + 10y - 1 = 0 \quad \text{(real part)}.$ $-2x - 5y + 4 = 0 \quad \text{(imaginary part)}.$

3. Solve the system of linear equations:

   The two equations are:

   $3x + 10y = 1 \quad \text{(Equation 1)}.$ $-2x - 5y = -4 \quad \text{(Equation 2)}.$

   Multiply Equation 2 by 2 to eliminate $y$: $-4x - 10y = -8.$

   Add it to Equation 1: $(3x - 4x) + (10y - 10y) = 1 - 8.$

   Simplify: $-x = -7 \quad \Rightarrow \quad x = 7.$

4. Substitute $x = 7$ into Equation 1:

   $3(7) + 10y = 1.$ $21 + 10y = 1 \quad \Rightarrow \quad 10y = -20 \quad \Rightarrow \quad y = -2.$

Final Answer:

$x = 7, \quad y = -2.$

Would you like additional clarification or further details?

Expanding Questions:

1. How would the solution differ if the coefficients were real numbers only?
2. How can you verify the solution by substituting the values back into the equation?
3. What happens if one of the equations is a multiple of the other?
4. Can this problem be solved geometrically using the complex plane?
5. How would the system behave if the determinant of the coefficient matrix were zero?

Tip:

When solving for complex coefficients, always separate real and imaginary parts carefully to avoid missing terms.