Question

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

• A. $x=-3,y=1$
• B. $x=3,y=-1$
• C. $x=-6,y=2$
• D. $x=6,y=-2$

Answer 1

Answer: (B)

$x=3,y=-1$

Given, $\frac{(1+i)x-2i}{3+i}+\frac{(2-3i)y+i}{3-i}=i.$
or $(1+i)(3-i)x-2i(3-i)+(2-3i)(3+i)y+i(3+i)=i(3+i)=i(3+i)(3-i)$
or $(4+2i)x-6i-2+(9-7i)y+3i-1=10i$
Equating real and imaginary parts,
$4x+9y-3=0$ ...(1)
and $2x-7y-3=10$
or $2x-7y-13=0$ ...(2)
Solving (1) and (2), we get $x=3,y=-1$
Ans: B