Question

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

• A. $y=\frac{43}{63},x=-\frac{16}{7}$
• B. $y=-\frac{43}{63},x=\frac{16}{7}$
• C. $x=3,y=-1$
• D. $x=1,y=0$

Answer 1

The correct option is C $x=3,y=-1$

$\frac{(1+i)x-2i}{3+i}+\frac{(2-3i)y+i}{3-i}=i\Rightarrow \frac{[x+(x-2\left)i\right](3-i)}{(3^2-i^2)}+\frac{[2y+(1-3y\left)i\right](3+i)}{3^2-i^2}=i\Rightarrow \frac{4x-2+(2x-6)i+9y-1+(3-7y)i}{10}=i\Rightarrow 4x+9y-3+(2x-7y-3)i=10i$

Compare both the sides
$4x+9y=3$ and $2x-7y=13$
By solving above equations, we get
$y=-1$ and $x=3$