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Q.
If $4x+i(3x-y)=3+i(-6)$, where $x$ and $y$ are real numbers, then find the values of $x$ and $y$ respectively.
Complex Numbers and Quadratic Equations
Solution:
We have, $4x+i\left(3x-y\right)=3+i\left(-6\right)$
Equating the real and imaginary parts, we get
$4x = 3$, $3 x - y =-6$
which, on solving simultaneously, gives
$x=\frac{3}{4}$ and $y=\frac{33}{4}\cdot$