Find the real values of $x$ and $y$ for which the following equations are satisfied.
$\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i$ .

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Solution

$\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i$ .
$\frac{(1 + i) x (3 + i) - 2 i (3 - i) + (2 - 3 i) (3 + i) y i (3 + i)}{(3 + i) (3 - i)} = i$
$[3 + i + 3 i - 1] x - 6 i + 2 i + (6 + 2 i - 9 i + 3) y + 3 i - 1 = i (9 + 1)$
$(2 + 4 i) x - i + 1 (9 - 7 i) y = 10 i$
$2 x + 1 + 9 y + i (4 x - 8) = 10 i$
$2 x + 9 y + 1 = 0$ $4 x - 8 = 10$
$2 \times 2 + 9 y + 1 = 0$ $x = 2$
$9 y = - 5$
$y = \frac{- 5}{2}$

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