Find the real values of $x$ and $y$ for which the equation $\;$ $\dfrac{\left(1 + i\right)x - 2i}{3 + i} + \dfrac{\left(2 - 3i\right)y + i}{3 - i} = i$ $\;$ is satisfied.
The given equation is $\;\;$ $\dfrac{\left(1 + i\right)x - 2i}{3 + i} + \dfrac{\left(2 - 3i\right)y + i}{3 - i} = i$
i.e. $\;$ $\left[\left(1 + i\right)x - 2i\right] \left(3 - i\right) + \left[\left(2 - 3i\right)y + i\right] \left(3 + i\right) = i\left(3 + i\right) \left(3 - i\right)$
i.e. $\;$ $x \left(3 + 2 i - i^2\right) -6i + 2i^2 + y \left(6 - 7i -3i^2\right) + 3i + i^2 = 9i - i^3$
i.e. $\;$ $\left(4 + 2i\right)x + \left(9 - 7i\right)y -3i -3 = 10i$ $\;\;\;$ [$\because \; i^3 = -i$]
i.e. $\;$ $\left(4x + 9y - 3\right) + i \left(2x - 7y -13\right) = 0$
$\implies$ $4x + 9y - 3 = 0$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $2x - 7y -13 = 0$ $\;\;\; \cdots \; (2)$
Solving equations $(1)$ and $(2)$ simultaneously, we have
$23 y + 23 = 0$ $\implies$ $y = -1$
Substituting the value of $y$ in equation $(1)$, we get
$4x - 12 = 0$ $\implies$ $x = 3$
$\therefore$ $\;$ The real values of $x$ and $y$ for which the given equation is satisfied are $x = 3$ and $y = -1$