Cunning Complex 6

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

$= \displaystyle\frac{{\left[ {(1 + i)x - 2i} \right](3 - i)}}{{10}} + \displaystyle\frac{{\left[ {(2 - 3i)y + i} \right](3 - i)}}{{10}}$

$= \displaystyle\frac{2}{5}x + \displaystyle\frac{{ix}}{5} - \displaystyle\frac{1}{5} - \displaystyle\frac{{3i}}{5} + \displaystyle\frac{{3i}}{{10}} - \displaystyle\frac{{7yi}}{{10}} + \displaystyle\frac{{9y}}{{10}} - \displaystyle\frac{1}{{10}}$

$= i\left( {\displaystyle\frac{x}{5} - \displaystyle\frac{{7y}}{{10}} - \displaystyle\frac{3}{{10}}} \right) + \left( {\displaystyle\frac{{2x}}{5} + \displaystyle\frac{{9y}}{{10}} - \displaystyle\frac{3}{{10}}} \right) = 1i+0$

As two complex numbers are equal if and only if their real parts are equal and their imaginary parts are also equal, we have:

$\left\{ \begin{array}{l} \displaystyle\frac{x}{5} - \displaystyle\frac{{7y}}{{10}} - \displaystyle\frac{3}{{10}} = 1 \\ \displaystyle\frac{{2x}}{5} + \displaystyle\frac{{9y}}{{10}} - \displaystyle\frac{3}{{10}} = 0 \\ \end{array} \right.$

$\Leftrightarrow \left\{ \begin{array}{l} x = 3 \\ y = - 1 \\ \end{array} \right.$

Hence, the sum of real values of $x,y$ is $\boxed{2}$

Is there a better solution other than putting x=a+ib and so for y...