Question

Очень срочно пожалуйста даю 40 балов за 8 завданя

Answer 1

$\displaystyle\bf\\\left \{ {{2x+y=6} \atop {x^{2} +xy=8}} \right. \\\\\\\left \{ {{y=6-2x} \atop {x^{2} +x\cdot(6-2x)=8}} \right. \\\\\\\left \{ {{y=6-2x} \atop {x^{2} +6x-2x^{2} -8=0}} \right. \\\\\\\left \{ {{y=6-2x} \atop {x^{2} -6x+8=0}} \right. \\\\\\\left \{ {{y=6-2x} \atop {(x-2)\cdot(x-4)=0}} \right. \\\\\\\left \{ {{y=6-2x} \atop {\left[\begin{array}{ccc}x-2=0\\x-4=0\end{array}\right }} \right.$

$\displaystyle\bf\\\left \{ {{y=6-2x} \atop {\left[\begin{array}{ccc}x_{1} =2\\x_{2} =4\end{array}\right }} \right.\\\\\\\left[\begin{array}{ccc}\left \{ {{x_{1} =2} \atop {y_{1}=6-2\cdot 2 }} \right. \\\left \{ {{x_{2} =4} \atop {y_{2} =6-2\cdot 4}} \right. \end{array}\right \\\\\\\left[\begin{array}{ccc}\left \{ {{x_{1} =2} \atop {y_{1}=2 }} \right. \\\left \{ {{x_{2} =4} \atop {y_{2} =-2}} \right. \end{array}\right\\\\\\Otvet \ : \ (2 \ ; \ 2) \ , \ (4 \ ; -2)$