Решить систему 2х-3(2у+1)=15 3(х+у)+3у=2у-2

$\left \{ {{2x-3(2y+1)=15} \atop {3(x+y)+3y=2y-2}} \right.$
$\left \{ {{2x-6y-3=15} \atop {3x+3y+3y=2y-2}} \right.$
$\left \{ {{2x-6y=15+3} \atop {3x+6y-2y=-2}} \right.$
$\left \{ {{2x-6y=18} \atop {3x+4y=-2}} \right.$
$\left \{ {{x=(18+6y):2} \atop {3x+4y=-2}} \right.$
$\left \{ {{x=9+3y} \atop {3x+4y=-2}} \right.$
3 (9 + 3y) + 4y = - 2
27 + 9y + 4y = - 2
13y = - 2 - 27
13y = - 29
y = - 2 3/13
$\left \{ {{x=9+3 * (- 2 \frac{3}{13} )} \atop {y=- 2 \frac{3}{13} } \right.$
$\left \{ {{x=2 \frac{4}{13}} \atop {y=- 2 \frac{3}{13} } \right.$

Ответ: (2 4/13; - 2 3/13).