Given that x, `y in R`. Solve: `x/(1+2i)+y/(3+2i)=(5+6i)/(8i-1)`

Explanation

`(x)/(1+2i)+(y)/(3+2i)=(5+6i)/(8i-1)`
or `(x (1-2i))/(1-4i^(2))+(y(3-2i))/(9-4i^(2)) = ((5+6i)(8i+1))/((8i)^(2)-1^(2))`
`or (x-2x i)/(5) +(15y -10yi)/(13)+(40 i + 5- 48+6i)/(-64-1)`
`or (13x -26x i+15y - 10yi)/(65) = (-43 + 46i)/(-65)`
or (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