Question

Given that $x,y\in R$. Solve:
$\frac{x}{1+2i}+\frac{y}{3+2i}=\frac{5+6i}{8i-1}$

• A. $x=2$ and $y=1$
• B. $x=1$ and $y=2$
• C. $x=-1$ and $y=2$
• D. $x=2$ and $y=-1$

Answer 1

Answer: (B)

$x=1$ and $y=2$

Given, $\frac{x}{1+2i}+\frac{y}{3+2i}=\frac{5+6i}{8i-1}$
$\frac{x(1-2i)}{1-4i^2}+\frac{y(3-2i)}{9-4i^2}=\frac{(5+6i)(8i+1)}{(8i{)}^2-1^2}$
$\frac{x-2xi}{5}+\frac{3y-2yi}{13}=\frac{40i+5-48+6i}{-64-1}$
$\frac{13x-26xi+15y-10yi}{65}=\frac{-43+46i}{-65}$
$(13x+15y)-i(26x+10y)=43-46i$
Equating real and imaginary parts, we get
$13x+15y=43$ (1)
$13x+5y=23$ (2)
Solving for $x$ and $y$, we get $x=1$ and $y=2$
Ans: B