Помогите ПОЖАЛУЙСТА!!!!

$\left \{ {{lg(y-x)=lg4} \atop {log_2 \frac{x}{4}=3-log_2 y }} \right.$

ОДЗ:
$\left \{ {{y-x\ \textgreater \ 0} \atop {x\ \textgreater \ 0},y\ \textgreater \ 0} \right.$
$\left \{ {{y\ \textgreater \ x} \atop {x\ \textgreater \ 0},y\ \textgreater \ 0} \right.$

$\left \{ {{y-x=4} \atop {log_2 \frac{x}{4}=log_2 8-log_2 y }} \right.$
$\left \{ {{y-x=4} \atop {log_2 \frac{x}{4}=log_2 \frac{8}{y}} \right.$
$\left \{ {{y-x=4} \atop {\frac{x}{4}=\frac{8}{y}} \right.$
$\left \{ {{y=4+x} \atop {xy=32} \right.$
$\left \{ {{y=4+x} \atop {x(4+x)=32} \right.$
$\left \{ {{y=4+x} \atop {4x+x^2=32} \right.$
$\left \{ {{y=4+x} \atop {x^2+4x-32=0} \right.$
$\left \{ {{y=4+x} \atop {(x+8)(x-4)=0} \right.$

$\left \{ {{y=4+x} \atop {x_1=-8} \right.$
$\left \{ {{y_1=-4} \atop {x_1=-8} \right.$
y оказался отрицательным и не входит в ОДЗ

$\left \{ {{y=4+x} \atop {x_1=4} \right.$
$\left \{ {{y_1=8} \atop {x_1=4} \right.$
$x\ \textgreater \ 0, y\ \textgreater \ 0, y-x=8-4=4\ \textgreater \ 0$
Корни удовлетворяют ОДЗ

$\left \{ {{x=4} \atop {y=8}} \right.$
$\frac{x}{y} = \frac{4}{8} =0.5$