Решите систему уравнений.Во вложении.......... Created by Klayzi. algebra-ru

Решите систему уравнений.Во вложении............

sangers1959

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Объяснение:

$> 1) Суммируем эти уравнения:<img src=$

2) Вычитаем из первого уравнения второе:

$2xy=6\ |:2\\xy=3.$

3) Составляем систему уравнений:

$\left \{ {{x+y=4} \atop {xy=3}} \right. \ \ \ \ \left \{ {{y=4-x} \atop {x*(4-x)=3}} \right. \ \ \ \ \left \{ {{y=4-x} \atop {4x-x^2=3}} \right.\ \ \ \ \left \{ {{y=4-x} \atop {x^2-4x+3=0}} \right.\ \ \ \ \left \{ {{y=4-x} \atop {D=4\ \ \sqrt{D}=2 }} \right.\ \ \ \ \left \{ {{y_1=3\ \ y_2=1} \atop {x_1=1\ \ x_2=3}} \right. .$

Ответ: (1;3), (3;1).

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Universalka

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$1)\\+\displaystyle\left \{ {{3x+xy+3y=15} \atop {3x-xy+3y=9}} \right. \\----------\\6x+6y=24\\\\\boxed{x+y=4}\\\\2)\\-\displaystyle\left \{ {{3x+xy+3y=15} \atop {3x-xy+3y=9}} \right.\\-----------\\xy+xy=6\\\\2xy=6\\\\\boxed{xy=3}\\\\Teorema \ Vieta:\\\\x_{1}=1 \ ; \ y_{1}=3\\\\x_{2}=3 \ ; \ y_{2}=1\\\\Otvet:\boxed{(1 \ ; \ 3) \ , \ (3 \ ; \ 1)}$

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