I am new to this forum.
Please help me with this problem-
Find the real values of x and y
if
((1+i)x-2)/(3+i)+((2-3i)y+i)/(3-i)=i where i = imaginary number
What I did is-
$\displaystyle \frac{(1+i)x-2}{3+i}+\frac{(2-3i)y+i}{3-i}= i $
$\displaystyle \frac{4x +2xi-3i-3+9y-7yi}{10}=i$
$\displaystyle 4x +2xi-3i-3+9y-7yi=10i$
$\displaystyle 4x +9y+i(2x-13-7y)=0$
What to do to find real values of x and y