Verified answer

Ответ:

[tex](\frac{1}{2};5),\ (2;2)[/tex]

Объяснение:

[tex]\begin{cases}2x+y=6\\x-xy=-2 \end{cases}\\\\\begin{cases}y=6-2x\\x-xy=-2 \end{cases} \\\\\begin{cases}y=6-2x\\x-x(6-2x)=-2 \end{cases}[/tex]

[tex]x-6x+2x^2=-2\\\\2x^2+x-6x+2=0\\\\2x^2-5x+2=0\\\\a=2 ,\ \ b=-5 ,\ \ c=2\\\\D = b^2 - 4ac = (- 5)^2 - 4\cdot2\cdot2 = 25 - 16 = 9\\\\\sqrt D =\sqrt{9} = 3\\\\\ x_1=\frac{-b-\sqrt{D}}{2a}=\frac{5-3}{2\cdot2}=\frac{2 }{4 }= \frac{ 1 }{ 2 } \\\\ x_2=\frac{-b+\sqrt{D}}{2a}=\frac{5+3}{2\cdot2}=\frac{8}{4}=2[/tex]

[tex]y=6-2x\\\\y_1=6-2\cdot \frac{1}{2}=6-1=5\\\\y_2=6-2\cdot 2=6-4=2[/tex]

[tex](\frac{1}{2};5),\ (2;2)[/tex]