Question

Решите , очень прошу

Answer 1

$\left \{ {{4x-y=2} \atop {log_{12} x+log_{12}3}=log_{12}(y+1)} \right.$
ОДЗ:
$\left \{ {{x\ \textgreater \ 0} \atop {y+1\ \textgreater \ 0}} \right.$
$\left \{ {{x\ \textgreater \ 0} \atop {y\ \textgreater \ -1}} \right.$

$\left \{ {{4x-y=2} \atop {log_{12}(3 x)=log_{12}(y+1)} \right.$
$\left \{ {{4x-y=2} \atop {3 x=y+1} \right.$
$\left \{ {{4x-y=2} \atop {3 x-y=1} \right.$
$\left \{ {{-4x+y=-2} \atop {3 x-y=1} \right.$
$\left \{ {{-x=-1} \atop {y=3x-1} \right.$
$\left \{ {{x=1} \atop {y=3*1-1} \right.$
$\left \{ {{x=1} \atop {y=2} \right.$

Ответ: (1;2)