$$\frac { x } { 2 + i } = \frac { 1 - 5 i } { 3 - 2 i } + \frac { y } { 2 - i }$$
$x=\left(\frac{3}{5}+\frac{4}{5}i\right)y+\left(3-i\right)$
$$\frac{x}{2+i}=\frac{\left(1-5i\right)\left(3+2i\right)}{\left(3-2i\right)\left(3+2i\right)}+\frac{y}{2-i}$$
$$\frac{x}{2+i}=\frac{13-13i}{13}+\frac{y}{2-i}$$
$$\frac{x}{2+i}=1-i+\frac{y}{2-i}$$
$$\left(\frac{2}{5}-\frac{1}{5}i\right)x=\left(\frac{2}{5}+\frac{1}{5}i\right)y+\left(1-i\right)$$
$$\frac{\left(\frac{2}{5}-\frac{1}{5}i\right)x}{\frac{2}{5}-\frac{1}{5}i}=\frac{\left(\frac{2}{5}+\frac{1}{5}i\right)y+\left(1-i\right)}{\frac{2}{5}-\frac{1}{5}i}$$
$$x=\frac{\left(\frac{2}{5}+\frac{1}{5}i\right)y+\left(1-i\right)}{\frac{2}{5}-\frac{1}{5}i}$$
$$x=\left(\frac{3}{5}+\frac{4}{5}i\right)y+\left(3-i\right)$$
Show Solution
Hide Solution
$y=\left(\frac{3}{5}-\frac{4}{5}i\right)x+\left(-1+3i\right)$
$$1-i+\frac{y}{2-i}=\frac{x}{2+i}$$
$$\frac{y}{2-i}=\frac{x}{2+i}-\left(1-i\right)$$
$$\frac{y}{2-i}=\frac{x}{2+i}+\left(-1+i\right)$$
$$\left(\frac{2}{5}+\frac{1}{5}i\right)y=\left(\frac{2}{5}-\frac{1}{5}i\right)x+\left(-1+i\right)$$
$$\frac{\left(\frac{2}{5}+\frac{1}{5}i\right)y}{\frac{2}{5}+\frac{1}{5}i}=\frac{\left(\frac{2}{5}-\frac{1}{5}i\right)x+\left(-1+i\right)}{\frac{2}{5}+\frac{1}{5}i}$$
$$y=\frac{\left(\frac{2}{5}-\frac{1}{5}i\right)x+\left(-1+i\right)}{\frac{2}{5}+\frac{1}{5}i}$$
$$y=\left(\frac{3}{5}-\frac{4}{5}i\right)x+\left(-1+3i\right)$$
