Помогите решить системы 4 и 5 Срочно!

$\left \{ {{ \frac{x+y}{x-y}+ \frac{x-y}{x+y}= \frac{10}{3}} \atop {x^2+y^2=5}}\right.$ ; $\left \{ {{ \frac{(x+y)(x+y)+(x-y)(x-y)}{(x-y)(x+y)}= \frac{10}{3}} \atop {x^2+y^2=5}}\right.$ ; $\left \{ {{ \frac{x^2+2xy+y^2+x^2-2xyx+y^2}{x^2-y^2}= \frac{10}{3}} \atop {x^2+y^2=5}}\right.$ ; $\left \{ {{ \frac{3(2x^2+2y^2)-10(x^2-y^2)}{3(x^2-y^2)}=0} \atop {x^2+y^2=5}}\right.$ ; $\left \{ {{ \frac{-4x^2+16y^2}{3(x^2-y^2)}=0} \atop {x^2+y^2=5}}\right.$ ; $\left \{ {{ \frac{-4(5-y^2)+16y^2}{3(5-y^2-y^2)}=0} \atop {x^2=5-y^2}}\right.$ ; $\left \{ {{ \frac{-20+4y^2+16y^2}{15-6y^2}=0} \atop {x^2=5-y^2}}\right.$ ; $\left \{ {{ \frac{-20+20y^2}{15-6y^2}=0} \atop {x^2=5-y^2}}\right.$ . Решим верхнее уравнение
$\frac{-20+20y^2}{15-6y^2}=0$
$y= \sqrt{ \frac{2}{5} }$ и $y=- \sqrt{ \frac{2}{5} }$ не могут быть корнями, т.к. на 0 делить нельзя
$20y^2-20=0$
$y^2=1$
$y_{1}=1$ ; $y_{2}=-1$
$\left \{ {{ y_{1}=1} \atop {x^2=5-1=4}}\right.$ ; $\left \{ {{ y_{1}=1} \atop { x_{1}=2 }}\right.$ ; $\left \{ {{ y_{1}=1} \atop { x_{2}=-2 }}\right.$ ; $\left \{ {{ y_{2}=-1} \atop { x_{3}=2 }}\right.$ ; $\left \{ {{ y_{2}=-1} \atop { x_{4}=-2 }}\right.$