Question

Розвязать систему уравнений.

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Answer 1

Пошаговое объяснение:

$\[\begin{gathered}\left\{\begin{gathered}2xy+x=-9\hfill\\2xy+y=-14\hfill\\\end{gathered}\right.\Leftrightarrow\left\{\begin{gathered}2xy=-9-x\hfill\\2xy=-14-y\hfill\\\end{gathered}\right. \hfill \\-9-x=-14-y \hfill \\-x=-14-y+9 \hfill \\-x=-5-y\;\;\;\;\;\;|*(-1) \hfill \\x=5+y\hfill \\2xy=-14-y\hfill \\2(5+y)y=-14-y\hfill \\\end{gathered}\]$

$\[\begin{gathered}10y+2{y^2} =-14-y\hfill \\2{y^2}+10y+y+14=0 \hfill \\2{y^2}+11y+14=0\hfill \\D={b^2}-4ac\hfill \\D={11^2}-4*2*14=121- 112=9 \hfill \\ \end{gathered} \]$

$\[\begin{gathered}{y_{1,2}}=\frac{{-b\pm\sqrt D}}{{2a}}=\frac{{-11 \pm \sqrt 9}}{{2*2}}=\frac{{-11\pm 3}}{4}\hfill \\{y_1}=\frac{{-11+3}}{4}=\frac{{-8}}{4}=-2 \hfill \\{y_2}=\frac{{-11-3}}{4}=\frac{{-14}}{4}=-\frac{7}{2}=-3.5 \hfill \\ \end{gathered} \]$

$\[\begin{gathered}{x_1} = 5 + {y_1} = 5 + (-2)=5-2=3 \hfill \\{x_2} = 5 + {y_2} =5+ (-3.5)=5-3.5=1.5 \hfill \\\boxed{{x_1} = 3,{y_1} =-2} \hfill \\\boxed{{x_2} = 1.5,{y_2}=-3.5}\hfill \\ \end{gathered}\]$