Предмет: Алгебра, автор: kivikivi95

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Автор ответа: Universalka

$\displaystyle\bf\\\left \{ {{y+1=2x} \atop {y^{2}-1=3x+xy }} \right. \\\\\\\left \{ {{y=2x-1} \atop {(2x-1)^{2} -1-3x-x\cdot(2x-1)=0}} \right. \\\\\\\left \{ {{y=2x-1} \atop {4x^{2} -4x+1-1-3x-2x^{2} +x=0}} \right. \\\\\\\left \{ {{y=2x-1} \atop {2x^{2} -6x=0}} \right.\\\\\\\left \{ {{y=2x-1} \atop {x^{2} -3x=0}} \right. \\\\\\\left \{ {{y=2x-1} \atop {x(x-3)=0}} \right.$

$\displaystyle\bf\\\left \{ {{y=2x-1} \atop {\left[\begin{array}{ccc}x=0\\x-3=0\end{array}\right}} \right. \\\\\\\left[\begin{array}{ccc}\left \{ {{x_{1=0} } \atop {y_{1} =2\cdot 0-1}} \right. \\\left \{ {{x_{2}=3 } \atop {y_{2}=2\cdot 3-1 }} \right. \end{array}\right\\\\\\\left[\begin{array}{ccc}\left \{ {{x_{1}=0 } \atop {y_{1} =-1}} \right. \\\left \{ {{x_{2}=3 } \atop {y_{2} =5}} \right. \end{array}\right\\\\\\Otvet:(0 \ ; -1) \ , \ (3 \ ; \ 5)$