Помогите пожалуйста . буду очень благодарна

1.
$\left \{ {{ \frac{x}{y} - \frac{y}{x} = \frac{5}{6} } \atop {x^2-y^2=5}} \right. \\ \left \{ {{ \frac{x^2-y^2}{xy} = \frac{5}{6} } \atop {x^2-y^2=5}} \right. \\ \left \{ {{ xy=6} \atop {x^2-y^2=5}} \right. \\ \left \{ {{x= \frac{6}{y} } \atop {\frac{36}{y^2} -y^2=5}} \right.$

$\frac{36}{y^2} -y^2=5 \\ y^4 + 5y^2-36=0 \\ D = 25+144=169 \\ y^2= \frac{-5+13}{2} =4 \\ y_2^2= \frac{-5-13}{2} \ \textless \ 0 \\ y_1 = -2 \\ y_2 = 2\\x_1 = \frac{6}{-2} =-3\\x_2=3$
Ответ: (-3; -2), (3; 2)

2.
$\left \{ {{x+y+ \frac{x}{y} =9} \atop { \frac{(x+y)x}{y} =20}} \right. \\ \left \{ {{ x+y =9- \frac{x}{y} } \atop {(9- \frac{x}{y} ) \frac{x}{y} =20}} \right.$
$9 \frac{x}{y} -(\frac{x}{y} )^2=20 \\ ( \frac{x}{y} )^2-9 \frac{x}{y} +20=0\\D=81-80=1\\ (\frac{x}{y})_1 = \frac{9-1}{2} =4\\(\frac{x}{y})_2 = \frac{9+1}{2} =5$
$1) x=5y \\ 5y+y+ \frac{5y}{y} =9\\6y+5=9\\6y=4\\y= \frac{2}{3} \\x= \frac{10}{3} =3 \frac{1}{3}$
$2) x=4y \\ 4y+y+ \frac{4y}{y} =9\\5y+4=9\\5y=5\\y=1\\x=4$
Ответ: $(3 \frac{1}{3}; \frac{2}{3} )$ , (4;1)