Решите системное уравнение способом подстановки выразив одну переменную через другую:...

Решите задачу:

$\left \{ {{x+y=4} \atop {xy=-12}} \right. \; \left \{ {{x=4-y} \atop {y(4-y)=-12}} \right. \; \left \{ {{x=4-y} \atop {-y^2+4y=-12}} \right. \; \left \{ {{x=4-y} \atop {y^2-4y-12=0}} \right. \\\\y^2-4y-12=0\\\\D=16+48=64,\; \; y_1=\frac{4-8}{2}=-2\; ,\; \; y_2=6\\\\x_1=4-(-2)=6\; ,\; \; x_2=4-6=-2\\\\Otvet:\; \; (6,-2)\; ,\; \; (-2,6).\\$

$2) \left \{ {{y=-\frac{12}{x}} \atop {x+y=4}} \right. \left \{ {{y=-\frac{12}{x}} \atop {x-\frac{12}{x}-4=0}} \right. \\\\x-\frac{12}{x}-4=0\\\\\frac{x^2-4x-12}{x}=0\to \\\\\left \{ {{x^2-4x-12=0} \atop {x\ne 0}} \right. \\\\ \left \{ {{x_1=-2,x_2=6} \atop {y_1=6,y_2=-2}} \right.$