Value of Linear Combination of Real Numbers in Complex Equation

To find the value of $10(x-3y)$, we start with the given equation: $50\left(\frac{2x}{1+3i} - \frac{y}{1-2i}\right) = 31 + 17i$

First, we rationalize the denominators of the complex fractions inside the parenthesis: $\frac{2x}{1+3i} = \frac{2x(1-3i)}{(1+3i)(1-3i)} = \frac{2x(1-3i)}{1 - 9i^2} = \frac{2x(1-3i)}{10} = \frac{x(1-3i)}{5}$

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

Substituting these back into the original equation, we get: $50\left(\frac{x(1-3i)}{5} - \frac{y(1+2i)}{5}\right) = 31 + 17i$

Multiply through by $50$: $10\left[ x(1-3i) - y(1+2i) \right] = 31 + 17i$

Expanding and grouping the real and imaginary parts: $10\left[ (x - y) + i(-3x - 2y) \right] = 31 + 17i$ $10(x - y) + 10(-3x - 2y)i = 31 + 17i$

Since $x$ and $y$ are real numbers, we equate the real and imaginary parts separately:

1. Real Part: $10(x - y) = 31 \implies 10x - 10y = 31 \quad \text{--- (1)}$

2. Imaginary Part: $10(-3x - 2y) = 17 \implies -30x - 20y = 17 \quad \text{--- (2)}$

Now, we can solve these linear equations for $x$ and $y$: From equation (2), we have: $30x + 20y = -17$

Multiplying equation (1) by $3$: $30x - 30y = 93$

Subtracting $30x + 20y = -17$ from $30x - 30y = 93$: $(30x - 30y) - (30x + 20y) = 93 - (-17)$ $-50y = 110 \implies y = -\frac{11}{5}$

Substituting $y = -\frac{11}{5}$ into equation (1) to solve for $10x$: $10x - 10\left(-\frac{11}{5}\right) = 31$ $10x + 22 = 31 \implies 10x = 9$

Finally, we calculate the required value $10(x - 3y)$: $10(x - 3y) = 10x - 30y$ $10(x - 3y) = 9 - 30\left(-\frac{11}{5}\right) = 9 + 66 = 75$

Hence, the value of $10(x-3y)$ is $75$.

Correct Option: D