Question

{x+y=6
{y+xy=4
Срочно!!!!!!!!!!!!!!

Answer 1

Объяснение:

$\left \{ {{x+y=6} \atop {y+xy=4}} \right. \ \ \ \ \left \{ {{x=6-y} \atop {y+(6-y)*y=4}} \right. \ \ \ \ \left \{ {{x=6-y} \atop {y+6y-y^2=4}} \right. \ \ \ \ \ \left \{ {{x-6-y} \atop {y^2-7x+4=0}} \right.\ \ \ \ \left \{ {{x=6-y} \atop {D=33\ \ \sqrt{D}=\sqrt{33} }} \right.$

$\left \{ {{x_1=\frac{5+\sqrt{33} }{2} \ \ \ \ x_2=\frac{5-\sqrt{33} }{2} } \atop {y_1=\frac{7-\sqrt{33}}{2} \ \ \ \ y_2=\frac{7+\sqrt{33} }{2} }} \right. .$

Answer 2

$\displaystyle\left \{ {{x+y=6} \atop {y+xy=4}} \right. \Leftrightarrow \left \{ {{x=6-y} \atop {y+xy=4}} \right. \Leftrightarrow y+(6-y)y=4~;~y+6y-y^2=4~;\\ 7y-y^2=4~;~7y-y^2-4=0~;~-y^2+7y-4=0~;~y^2-7y+4=0~;~\\ D=(-7)^2-4\cdot1\cdot4=49-16=33\Rightarrow y_{1;2} =\frac{7\pm\sqrt{33} }{2} \Leftrightarrow x_1=6-\frac{7+\sqrt{33} }{2} =\\ =\frac{12-(7+\sqrt{33}) }{2} =\frac{12-7-\sqrt{33} }{2} =\frac{5-\sqrt{33} }{2} \Rightarrow x_2=6-\frac{7-\sqrt{33} }{2} =\frac{5+\sqrt{33} }{2}$

ОТВЕТ: $\displaystyle\bigg(x_1~;~y_1\bigg)=\bigg(\frac{5-\sqrt{33} }{2} ~;~\frac{7+\sqrt{33} }{2} \bigg)~~,~~\bigg(x_2~;~y_2\bigg)=\bigg(\frac{5+\sqrt{33} }{2} ~;~\frac{7-\sqrt{33} }{2} \bigg).$