• # question_answer The values of $x$ and $y$ satisfying the equation $\frac{(1+i)x-2i}{3+i}$ $+\frac{(2-3i)\,y+i}{3-i}=i$ are [IIT 1980; MNR 1987] A) $x=-1,\,y=3$ B) $x=3,\,y=-1$ C) $x=0,\,y=1$ D) $x=1,y=0$

$\frac{(1+i)x-2i}{3+i}+\frac{(2-3i)y+i}{3-i}=i$ Þ $(4+2i)x+(9-7i)y-3i-3=10i$ Equating real and imaginary parts, we get $2x-7y=13$ and$4x+9y=3$. Hence $x=3$and$y=-1$. Trick : After finding the equations, no need to solve them, put the values of $x$ and $y$ given in the options and get the appropriate option.