Let x and y be real numbers such that 50((2x)/(1+3i) - (y)/(1-2i))...

Given the equation:

$50\left(\dfrac{2x}{1+3i} - \dfrac{y}{1-2i}\right) = 31 + 17i$

Rationalizing the denominators inside the parentheses:

$\dfrac{2x}{1+3i} = \dfrac{2x(1-3i)}{(1+3i)(1-3i)} = \dfrac{2x(1-3i)}{1^2 + 3^2} = \dfrac{2x(1-3i)}{10} = \dfrac{x(1-3i)}{5}$

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

Substituting these back into the given equation:

$50\left(\dfrac{x(1-3i)}{5} - \dfrac{y(1+2i)}{5}\right) = 31 + 17i$

$10(x(1-3i) - y(1+2i)) = 31 + 17i$

$10(x - y) - 10i(3x + 2y) = 31 + 17i$

Equating the real and imaginary parts on both sides:

$10(x - y) = 31 \Rightarrow 10x - 10y = 31$

$-10(3x + 2y) = 17 \Rightarrow -30x - 20y = 17$

Multiplying the first equation by $3$ gives:

$30x - 30y = 93$

Adding this to the second equation:

$(-30x - 20y) + (30x - 30y) = 17 + 93$

$-50y = 110 \Rightarrow y = -\dfrac{11}{5}$

Substituting $y = -\dfrac{11}{5}$ into $10x - 10y = 31$:

$10x - 10\left(-\dfrac{11}{5}\right) = 31$

$10x + 22 = 31 \Rightarrow 10x = 9 \Rightarrow x = \dfrac{9}{10}$

Now, finding the value of $10(x - 3y)$:

$10(x - 3y) = 10x - 30y = 10\left(\dfrac{9}{10}\right) - 30\left(-\dfrac{11}{5}\right)$

$10(x - 3y) = 9 + 66 = 75$

Answer: $75$