Question

x+iy/2+3i + 2+i/2-3i = 9/13(1+i) find x nd y

Answer 1

Answer:

$\bf{x=\frac{89}{26}\:and\:y=\frac{-29}{26}}$

Explanation:

$\frac{x+iy}{2+3i}+\frac{2+i}{2-3i}=\frac{9}{13(1+i)}$

$\implies\:\frac{x+iy}{2+3i}=\frac{9}{13(1+i)}-\frac{2+i}{2-3i}$

$\implies\:\frac{x+iy}{2+3i}=\frac{9(2-3i)-(13+13i)(2+i)}{13(1+i)(2-3i)}$

$\implies\:\frac{x+iy}{2+3i}=\frac{18-27i-(26+13i+26i+13i^2)}{13(2-3i+2i-3i^2)}$

$\implies\:\frac{x+iy}{2+3i}=\frac{18-27i-(26+39i-13)}{13(2-i+3)}$

$\implies\:\frac{x+iy}{2+3i}=\frac{18-27i-(13+39i)}{13(2-i+3)}$

$\implies\:\frac{x+iy}{2+3i}=\frac{18-27i-13-39i)}{13(5-i)}$

$\implies\:\frac{x+iy}{2+3i}=\frac{5-66i}{13(5-i)}$

$\implies\:x+iy=\frac{(5-66i)(2+3i)}{13(5-i)}$

$\implies\:x+iy=\frac{10+15i-132i-198i^2}{13(5-i)}$

$\implies\:x+iy=\frac{10+15i-132i+198}{13(5-i)}$

$\implies\:x+iy=\frac{208-117i}{13(5-i)}$

$\implies\:x+iy=\frac{13(16-9i)}{13(5-i)}$

$\implies\:x+iy=\frac{16-9i}{5-i}*\frac{5+i}{5+i}$

$\implies\:x+iy=\frac{80+16i-45i-9i^2}{5^2-i^2}$

$\implies\:x+iy=\frac{80+16i-45i+9}{25+1}$

$\implies\:x+iy=\frac{89-29i}{26}$

$\implies\:x+iy=\frac{89}{26}-\frac{29i}{26}$

Equating real and imaginary parts we get

$x=\frac{89}{26}\:and\:y=\frac{-29}{26}$

Answer 2

Answer:

$\bf{x=\frac{89}{26}\:and\:y=\frac{-29}{26}}$

Explanation:

$\frac{x+iy}{2+3i}+\frac{2+i}{2-3i}=\frac{9}{13(1+i)}$

$\implies\:\frac{x+iy}{2+3i}=\frac{9}{13(1+i)}-\frac{2+i}{2-3i}$

$\implies\:\frac{x+iy}{2+3i}=\frac{9(2-3i)-(13+13i)(2+i)}{13(1+i)(2-3i)}$

$\implies\:\frac{x+iy}{2+3i}=\frac{18-27i-(26+13i+26i+13i^2)}{13(2-3i+2i-3i^2)}$

$\implies\:\frac{x+iy}{2+3i}=\frac{18-27i-(26+39i-13)}{13(2-i+3)}$

$\implies\:\frac{x+iy}{2+3i}=\frac{18-27i-(13+39i)}{13(2-i+3)}$

$\implies\:\frac{x+iy}{2+3i}=\frac{18-27i-13-39i)}{13(5-i)}$

$\implies\:\frac{x+iy}{2+3i}=\frac{5-66i}{13(5-i)}$

$\implies\:x+iy=\frac{(5-66i)(2+3i)}{13(5-i)}$

$\implies\:x+iy=\frac{10+15i-132i-198i^2}{13(5-i)}$

$\implies\:x+iy=\frac{10+15i-132i+198}{13(5-i)}$

$\implies\:x+iy=\frac{208-117i}{13(5-i)}$

$\implies\:x+iy=\frac{13(16-9i)}{13(5-i)}$

$\implies\:x+iy=\frac{16-9i}{5-i}*\frac{5+i}{5+i}$

$\implies\:x+iy=\frac{80+16i-45i-9i^2}{5^2-i^2}$

$\implies\:x+iy=\frac{80+16i-45i+9}{25+1}$

$\implies\:x+iy=\frac{89-29i}{26}$

$\implies\:x+iy=\frac{89}{26}-\frac{29i}{26}$

Equating real and imaginary parts we get

$x=\frac{89}{26}\:and\:y=\frac{-29}{26}$