Question

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

• A. $(0,0)$
• B. $(3,1)$
• C. $(3,-1)$
• D. $(-3,1)$

Answer 1

The correct option is A $(0,0)$

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

$\frac{\left[\right(1+i)x-2i](3-i)+\left[\right(2-3i)y+i](3+i)}{3^2-i^2}=i$

$\frac{[4x+2+9y-1]+i[2x-6-7y+3]}{10}=i$

$\frac{1}{10}(4x+9y+1)+\frac{i}{10}(2x-7y-3)=i$

Equating real & Imaginary parts

$\frac{1}{10}(4x+9y+1)=0\to 4x+9y+1=0---------\left(1\right)$

$\frac{i}{10}(2x-7y-3)=i$

$2x-7y=7---------\left(2\right)$

Solving $\left(1\right)$ & $\left(2\right)\times 2$

$4x+9y+1=04x-14y-14=0-++-------------23y+15=0$

$y=\frac{-15}{23}$

$2x-7y=7$

$2x-7\left(\frac{-15}{23}\right)=7$

$2x=7-\frac{105}{23}$

$x=\frac{161-105}{2\times 25}$

$x=\frac{28}{23}$