Предмет: Алгебра
Автор: Аноним
Вопрос задан 1 год назад

СРОЧНО, ПОМОГИТЕ ПОЖАЛУЙСТА С АЛГЕБРОЙ) решение системы уравнений методом подстановки (ДАЮ 100 БАЛЛОВ) ОЧЕНЬ СРОЧНО!!!!!

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Ответы

Ответ дал: Universalka

$\displaystyle\bf\\1)\\\\\left \{ {{y=2+x} \atop {x^{2} -y=10}} \right. \\\\\\\left \{ {{y=2+x} \atop {x^{2} -(2+x)=10}} \right.\\\\\\\left \{ {{y=2+x} \atop {x^{2} -x-12=0}} \right.\\\\\\\left \{ {{y=2+x} \atop {\left[\begin{array}{ccc}x_{1} =4\\x_{2}=-3 \end{array}\right }} \right.\\\\\\x_{1} =4 \ \ \ \Rightarrow \ \ \ y_{1} =2+4=6\\\\\\x_{2} =-3 \ \ \ \Rightarrow \ \ \ y_{2} =2-3=-1\\\\\\Otvet \ : \ (4 \ ; \ 6) \ \ , \ \ (-3 \ ; \ -1)$

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

Ответ : решений нет

$\displaystyle\bf\\3)\\\\\left \{ {{x+1=2y} \atop {5xy+y^{2}=16 }} \right. \\\\\\\left \{ {{x=2y-1} \atop {5\cdot(2y-1)\cdot y+y^{2}=16 }} \right. \\\\\\\left \{ {{x=2y-1} \atop {10y^{2} -5y+y^{2}=16 }} \right. \\\\\\\left \{ {{x=2y-1} \atop {11y^{2}-5y-16=0 }} \right. \\\\\\11y^{2} -5y-16=0\\\\D=(-5)^{2} -4\cdot 11\cdot(-16)=25+704=729=27^{2} \\\\\\y_{1} =\frac{5-27}{22} =-1\\\\\\y_{2} =\frac{5+27}{22} =\frac{32}{22} =\frac{16}{11} =1\frac{5}{11}$

$\displaystyle\bf\\x_{1} =2\cdot(-1)-1=-3\\\\\\x_{2}=2\cdot\frac{16}{11} -1=\frac{32}{11} -1=2\frac{10}{11} -1=1\frac{10}{11} \\\\\\Otvet \ : \ \Big(-3 \ ; \ -1\Big) \ \ , \ \ \Big(1\frac{10}{11} \ ; \ 1\frac{5}{11} \Big)$