Question

Find the real number $x$ and $y$ such that $\frac{x}{1+2i}+\frac{y}{3+2i}=\frac{5+6i}{-1+8i}$

Answer 1

$\frac{x}{1+2i}+\frac{y}{3+2i}=\frac{5+6i}{-1+8i}$

LHS

$\frac{x}{1+2i}+\frac{y}{3+2i}$

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

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

$=\frac{x(1-2i)}{5}+\frac{y(3-2i)}{13}$

$=\frac{13\times (1-2i)+5y(3-2i)}{65}$

$=\frac{(13x+15y)+i(-26x-10y)}{65}$

RHS : $\frac{(5+6i)(-1-8i)}{(-1+8i)(-1-8i)}=\frac{-5-40i-6i+48}{1+64}$

$=\frac{43-46i}{65}$

Comparing LHS & RHS we can write

$13x+15y=43⟶\left(1\right)$

$26x+10y=46⟶\left(2\right)$

Equ $\left(1\right)\times 2⟶26x+30y=86$

Equ $\left(2\right)\times 1⟶\underset{–}{\underset{(-)}{}26x+10y=46}$

$20y=40$

$y=2$

$\therefore$ $x=1$

$\therefore$ $x=1$ & $y=2$