как решить уровнение система 3(х-1)-2(1+х)<1 и 3х-4>0

$\left \{ {{3(x-1)-2(1+x)\ \textless \ 1 } \atop {3x-4\ \textgreater \ 0}} \right.$

Раскрываем скобки, переносим слагаемые, содержащие х влево, числа вправо

$\left \{ {{3x-3-2-2x\ \textless \ 1 } \atop {3x-4\ \textgreater \ 0}} \right. \\ \\ \left \{ {{3x-2x\ \textless \ 1+3+2 } \atop {3x\ \textgreater \ 4}} \right. \\ \\ \left \{ {{x\ \textless \ 6} \atop {x\ \textgreater \ \frac{4}{3} }} \right.$
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
----------------(4/3)------------------------------(6)-------------------------
/\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
Ответ. (4/3 ; 6)
или (1 целая 1/3; 6)