Решить Систему xy=10y+20 x(y+2)=12(y+5)

Решите задачу:

$\begin{cases}xy=10y+20\\x(y+2)=12(y+5)\end{cases}\ \textless \ =\ \textgreater \ \begin{cases}xy=10y+20\\x(y+2)=12(y+5)\end{cases}\\x=\frac{12(y+5)}{y+2}\\\\\frac{12y(y+5)}{y+2}=10y+20\\12y^2+60y=(10y+20)(y+2)\\12y^2+60y=10y^2+20y+20y+40\\2y^2+20y-40=0\\y^2+10y-20=0\\y_1=-5+3\sqrt{5}\ \ \ \ y_2=-5-3\sqrt{5}$
$x_1=\frac{12(-5+3\sqrt5+5)}{(-5+3\sqrt5+2)}=\frac{36\sqrt5(3\sqrt5+3)}{(-3+3\sqrt5)(3\sqrt5+3)}=\frac{540+108\sqrt5}{36}=15+3\sqrt5\\x_2=\frac{12(-5-3\sqrt5+5)}{(-5-3\sqrt5+2)}=\frac{-36\sqrt5(3\sqrt5-3)}{-(3+3\sqrt5)(3\sqrt5-3)}=-\frac{-540+108\sqrt5}{36}=15-3\sqrt5\\OTBET:(15+3\sqrt5;-5+3\sqrt5);(15-3\sqrt5;-5-3\sqrt5)$