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(D) Given the equation: $50 \left(\frac{2x}{1 + 3i} - \frac{y}{1 - 2i}\right) = 31 + 17i$.Multiply the numerators and denominators by their conjugates

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(D) Given the equation: $50 \left(\frac{2x}{1 + 3i} - \frac{y}{1 - 2i}\right) = 31 + 17i$.Multiply the numerators and denominators by their conjugates: $\frac{2x(1-3i)}{1^2+3^2} = \frac{2x-6xi}{10}$ and $\frac{y(1+2i)}{1^2+2^2} = \frac{y+2yi}{5}$.Substituting these into the equation: $50 \left(\frac{2x-6xi}{10} - \frac{y+2yi}{5}\right) = 31 + 17i$.$50 \left(\frac{2x-6xi - 2(y+2yi)}{10}\right) = 31 + 17i$.$5(2x - 6xi - 2y - 4yi) = 31 + 17i$.$10x - 30xi - 10y - 20yi = 31 + 17i$.Grouping real and imaginary parts: $(10x - 10y) + i(-30x - 20y) = 31 + 17i$.Comparing real and imaginary parts:$10x - 10y = 31$ (Equation $1$)$-30x - 20y = 17$ (Equation $2$)Multiply Equation $1$ by $2$: $20x - 20y = 62$ (Equation $3$).Add Equation $2$ and Equation $3$: $(-30x - 20y) + (20x - 20y) = 17 + 62 \Rightarrow -10x - 40y = 79$.Alternatively,from Equation $1$,$x - y = 3.1 \Rightarrow x = y + 3.1$.Substitute into Equation $2$: $-30(y + 3.1) - 20y = 17 \Rightarrow -30y - 93 - 20y = 17 \Rightarrow -50y = 110 \Rightarrow y = -2.2$.Then $x = -2.2 + 3.1 = 0.9$.Calculate $10(x - 3y) = 10(0.9 - 3(-2.2)) = 10(0.9 + 6.6) = 10(7.5) = 75$.

Let $x$ and $y$ be real numbers such that $50 \left(\frac{2x}{1 + 3i} - \frac{y}{1 - 2i}\right) = 31 + 17i$,where $i = \sqrt{-1}$. Then the value of $10(x - 3y)$ is:

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