A) \[x=-1,y=3\]
B) \[x=3,y=-1\]
C) \[x=0,y=1\]
D) \[x=1,y=0\]
Correct Answer: B
Solution :
\[\frac{(1+i)x-2i}{3+i}+\frac{(2-3i)y+i}{3-i}=i\] \[\Rightarrow \] \[\frac{\left[ \begin{align} & (3-i)(1+i)x-2i(3-i) \\ & +(3+i)(2-3i)y+(3+i)i \\ \end{align} \right]}{9+1}\] \[\Rightarrow \]\[(4+2i)x+(9-7i)y-6i-2+3i-1=10i\] \[\Rightarrow \]\[(4+2i)x+(9-7i)y-3i-3=10i\] On equating the real and imaginary parts on both sides, we get \[4x+9y-3=0\] ?(i) and \[2x-7y-3=10\] or \[2x-7y-13=0\] ?(ii) On solving Eqs. (i) and (ii), we get \[x=3,y=-1\]
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