X+1/4-y-3/3=-1 и x+y=12 решите системное уравнение.СРОЧНО при том что /-это дробная черта...

[latex] \left \{ {{ \frac{x+1}{4}- \frac{y-3}{3}=-1 } \atop {x+y=12}} \right. [/latex] [latex] \left \{ {{12* \frac{x+1}{4}-12* \frac{y-3}{3}=-12*x } \atop {x+y=12}} \right. [/latex] [latex] \left \{ {{3(x+1)-4(y-3)=-12} \atop {x+y=12}} \right. [/latex] [latex] \left \{ {{3x+3-4y+12=-12} \atop {x+y=12}} \right. [/latex] [latex] \left \{ {{3x+15-4y=-12} \atop {x+y=12}} \right. [/latex] [latex] \left \{ {{3x-4y=-12-15} \atop {x+y=12}} \right. [/latex] [latex] \left \{ {{3x-4y=-27} \atop {x+y=12}} \right. [/latex] [latex] \left \{ {{3x-4y=-27} \atop {-3x-3y=-36}} \right. [/latex] [latex]-7y=-63[/latex] [latex]y=9[/latex] [latex]x+9=12[/latex] [latex]x=3[/latex] Ответ: [latex](3;9)[/latex]