Найдите решение системы уравнений способом подстановки: 1) 2*(х+у)-х=-6 3х-(х-у)=0 2)...

1. $\left \{ {{2(x+y)-x=-6} \atop {3x-(x-y)=0}} \right.\\\left \{ {{2x+2y-x=-6} \atop {3x-x+y=0}} \right.\\\left \{ {{x+2y=-6} \atop {2x+y=0}} \right.\\\left \{ {{x=-2y-6} \atop {2(-2y-6)+y=0}} \right.\\\left \{ {{x=-2y-6} \atop {-4y-12+y=0}} \right.\\\left \{ {{x=-2y-6} \atop {-3y=12}} \right.\\\left \{ {{x=-2y-6} \atop {y=-4}} \right.\\\left \{ {{x=2} \atop {y=-4}} \right.\\$

2. $\left \{ {{3(x+2y)-y=27} \atop {4(x+y)-3x=23}} \right.\\\left \{ {{3x+6y-y=27} \atop {4x+4y-3x=23}} \right.\\\left \{ {{3x+5y=27} \atop {x+4y=23}} \right.\\\left \{ {{3x+5y=27} \atop {x=23-4y}} \right.\\\left \{ {{3(23-4y)+5y=27} \atop {x=23-4y}} \right.\\\left \{ {{69-12y+5y=27} \atop {x=23-4y}} \right.\\\left \{ {{7y=42} \atop {x=23-4y}} \right.\\\left \{ {{y=6} \atop {x=-1}} \right.$