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If $\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i,$ then the real value of x and y are given by

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$\begin{aligned} x = 3, y = - 1 \end{aligned}$

$x = 3, y = 1$

$x = 1, y = - 3$

$\begin{aligned} x = - 3, y = - 1 \end{aligned}$

answer is B.

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Detailed Solution

$\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i$

or $(1 + i) (3 - i) x - 2 i (3 - i) + (2 - 3 i) (3 + i) y + i (3 + i)$

$= i (3 + i) (3 - i)$

or $(4 + 2 i) x - 6 i - 2 + (9 - 7 i) y + 3 i - 1 = 10 i$

Equating real and imaginary parts,

$4 x + 9 y - 3 = 0 and 2 x - 7 y - 13 = 0$

Solving, we get $x = 3, y = - 1$

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