Решите систему неравенств скобки 2х-6>0, x^2-8x+7>0

$\left \{ {{2x-6\ \textgreater \ 0} \atop {x^2-8x+7\ \textgreater \ 0}} \right. \left \{ {{x\ \textgreater \ 3} \atop {(x-1)(x-7)\ \textgreater \ 0}} \right. \\\\$
(x-1)(x-7)>0
$x_1=1 \\x_2=7$
+ - +
-------------(1)---------------(7)-----------------
$\left \{ {{x\ \textgreater \ 3} \atop {x\ \textless \ 1 ,x\ \textgreater \ 7}} \right.$

ответ: x>7