# How can I find $x$, $y$ values for $\frac{(1+i)x-2i}{3+i}+\frac{(2-3i)y+i}{3-i}=i$

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

I believe the format I need in order to solve this problem should be such that the real parts and imaginary parts are separated, $\{\mathrm{Real}\}+\{\mathrm{Imaginary}\}i=i$

Then I can equate the real and imaginary parts of the equation and solve for $x$ and $y$. I tried to multiply by the conjugate so that I would get a real number at the denominator.

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

Multiplied by conjugate: $$\frac{x(3-i)+i(x-2)(3-i)}{10}+\frac{2y(3+i)+i(-3y+1)(3+i)}{10}=i$$

And at this step I started to feel as if I made a mistake because I weren't sure how to proceed. Would someone let me know if my approach was correct and show me how to do this?

• @Osama, Just express as $$X+iY=A+iB$$ where $X,Y,A,B$ are real Commented Aug 24, 2014 at 10:58
• Another approach might be to pre-multiply everything with i, and then with (3+i)(3-i) to avoid fractions. Commented Aug 24, 2014 at 14:23

$$\frac{x(3-i)+i(x-2)(3-i)}{10}+\frac{2y(3+i)+i(-3y+1)(3+i)}{10}=i$$ $$\frac{3x-ix+3ix+x-6i-2+6y+2iy-9iy+3y+3i-1}{10}=i$$ $$\frac{4x+9y-3}{10}+\frac{i(2x-7y-3)}{10}=0+i$$
$$\frac{4x+9y-3}{10}=0 \ \mathrm{and}\ \frac{(2x-7y-3)}{10}=1$$ solve simultaneously to obtain values for x and y