ПОМОГИТЕ!!!!!!!Решите систему уравнений (x+2)(y-1)=0 x^2-xy-12=0

[latex] \left \{ {{(x+2)(y-1)=0} \atop {x^{2}-xy-12=0}} \right. [/latex] [latex] \left \{ {{x+2=0} \atop {x^{2}-xy-12=0}} \right. [/latex] [latex] \left \{ {{y-1=0} \atop {x^{2}-xy-12=0}} \right. [/latex] [latex]\left \{ {{x=-2} \atop {x^{2}-xy-12=0}} \right. [/latex] [latex]\left \{ {{y=1} \atop {x^{2}-xy-12=0}} \right.[/latex] [latex]\left \{ {{x=-2} \atop {4+2y=12}} \right.[/latex] [latex]\left \{ {{y=1} \atop {x^{2}-x-12=0}} \right.[/latex] [latex]\left \{ {{x=-2} \atop {y = 4}} \right.[/latex] Решаем теперь квадратное уравнение во второй системе по обратной теореме Виета: x₁ + x₂ = 1 x₁*x₂ = -12 x₁ = -3 x₂ = 4 [latex]\left \{ {{y=1} \atop {x = -3}} \right.[/latex] [latex]\left \{ {{y=1} \atop {x = 4}} \right.[/latex] Ответ: [latex](-2; 4), (-3; 1), (4; 1).[/latex]