Решите систему управлений 2х-у=1, 3х+2у=12

$\left \{ {{2x-y=1} \atop {3x+2y=12}} \right.$
$\left \{ {{y=2x-1} \atop {3x+2y=12}} \right.$
$\left \{ {{y=2x-1} \atop {3x+2(2x-1)=12}} \right.$
$\left \{ {{y=2x-1} \atop {3x+4x-2=12}} \right.$
$\left \{ {{y=2x-1} \atop {7x=14}} \right.$
$\left \{ {{y=2x-1} \atop {x=2}} \right.$
$\left \{ {{y=2*2-1} \atop {x=2}} \right.$
$\left \{ {{y=3} \atop {x=2}} \right.$
Ответ: x=2, y=3