Q. 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
A) $x = - 1,\ y = 3$
B) $x = 3,\ y = - 1$
C) $x = 0,\ y = 1$
D) $x = 1,\ y = 0$
Explanation
(b)
Given, $\frac{(1 + i)x - 2i}{3 + i} + \frac{(2 - 3i)y + i}{3 - i} = i$
$\Rightarrow \ \ \ \ \ \ (4 + 2i)x + (9 - 7i)y - 3i - 3 = 10i$
Equating real and imaginary parts, we get
$2x - 7y = 13$ and $4x + 9y = 3$, hence $x = 3$ and $y = - 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.