Question

$\frac{(x+iy)}{2+3i}+\frac{2+i}{2-3i}=\frac{9}{13}(1+i)$

Answer 1

$\frac{x+iy}{2+3i}+\frac{2+i}{2-3i}=\frac{9}{13}(1+i)$

$\frac{(x+iy)(2-3i)+(2+i)(2+3i)}{9+4}=\frac{9}{13}(1+i)$

$2x-3xi+2iy+3y+4+6i+2i-3=9+9i$

$2x+3y-3xi+2iy+8i+1=9+9i$

$2x+3y+1=9-3x+2y+8=9$

$2x+3y=8-3x+2y=1$

$6x+9y=24-3x+2y=1$

$-6x+4y=23x=3$

$13y=26x=1$

$y=2$

$\therefore x=1;y=2$