Помогите решить, плз Относительно x

Решите задачу:

$\frac{x+y}{x-y} + \frac{x-y}{x+y} = \frac{10}{3} \\$$\left(x - y\right) \left(\frac{x - y}{x + y} + \frac{x + y}{x - y}\right) = \frac{10 x}{3} - \frac{10 y}{3}$$\\ $$x + y + \frac{\left(x - y\right)^{2}}{x + y} = \frac{10 x}{3} - \frac{10 y}{3}$$\\ $$\left(x + y\right) \left(x + y + \frac{\left(x - y\right)^{2}}{x + y}\right) = \left(x + y\right) \left(\frac{10 x}{3} - \frac{10 y}{3}\right)$$\\ $$2 x^{2} + 2 y^{2} = \frac{10 x^{2}}{3} - \frac{10 y^{2}}{3}$$\\$
$-\frac{4 x^{2} }{3} + \frac{16 y^{2} }{3} =0\\ $$x_{1} = \frac{\sqrt{D} - b}{2 a}$$\\ $$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$\\ $$a = - \frac{4}{3}$$\\ $$b = 0$$\\ $$c = \frac{16 y^{2}}{3}$$\\ D = b^2 - 4 * a * c = (0)^2 - 4 * ( -\frac{4}{3} ) * \frac{16 y^{2} }{3} = \frac{256 y^{2} }{9}\\ $$x_{1} = - 2 \sqrt{y^{2}}$$\\ $$x_{2} = 2 \sqrt{y^{2}}$$\\ $$x_{1} = - 2 \Re{y} - 2 i \Im{y}$$\\ $$x_{2} = 2 \Re{y} + 2 i \Im{y}$$\\$