Решите пожалуйста 1) 0<(2x+3)/5<1 2)-1<(3x-1)/4<2

$0\ \textless \ \frac{(2x+3)}{5} \ \textless \ 1 |*5\\ 0\ \textless \ 2x+3\ \textless \ 5 \\ \left \{ {{2x+3\ \textgreater \ 0} \atop {2x+3\ \textless \ 5}} \right. \left \{ {{2x\ \textgreater \ -3} \atop {2x\ \textless \ 5-3}} \right. \left \{ {{x\ \textgreater \ -1,5} \atop {2x\ \textless \ 2}} \right. \left \{ {{x\ \textgreater \ -1,5} \atop {x\ \textless \ 1}} \right.$
x∈(-1,5;1)

$-1\ \textless \ \frac{3x-1}{4} \ \textless \ 2 |*4 \\ -4\ \textless \ 3x-1\ \textless \ 8 \\ \left \{ {{3x-1\ \textgreater \ -4} \atop {3x-1\ \textless \ 8}} \right. \left \{ {{3x\ \textgreater \ -4+1} \atop {3x\ \textless \ 8+1}} \right. \left \{ {3x\ \textgreater \ -3} \atop {3x\ \textless \ 9}} \right. \left \{ {{x\ \textgreater \ -1} \atop {x\ \textless \ 3}} \right.$
x∈(-1;3)