Решите систему уравнений и найти ОДЗ log2(x-y)=3 ; 4^log2(корень x+y)=10

$\left \{ {{log_2(x-y)=3} \atop {4^{log_2\sqrt{x+y}}=10}} \right.$
ОДЗ: $x-y\ \textgreater \ 0;x\ \textgreater \ y$
$\sqrt{x+y}\ \textgreater \ 0;x+y\ \textgreater \ 0;x\ \textgreater \ -y$

$\left \{ {{log_2(x-y)=log_2 8} \atop {2^{2log_2\sqrt{x+y}}=10}} \right.$
$\left \{ {{(x-y)=8} \atop {2^{log_2(x+y)}=10}} \right.$
$\left \{ {{x-y=8} \atop {x+y=10}} \right.$
$\left \{ {{2x=18} \atop {x+y=10}} \right.$
$\left \{ {{x=9} \atop {9+y=10}} \right.$
$\left \{ {{x=9} \atop {y=1}} \right.$