Question

6. Determine a solução dos sistemas de equações a seguir.
a){x•y=40
x+y=13
b){x+y=-9
x•y=14​

Answer 1

a)

    $\left\{\begin{array}{II}x\times y=40\\\\x+y=13\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{II}x\times y=40\\\\x=13-y\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{II}(13-y)\times y=40\\\\x=13-y\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{II}13y-y^2=40\\\\x=13-y\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{II}y=5\\\\x=13-y\end{array}\right\;\;\vee\;\;\left\{\begin{array}{II}y=8\\\\x=13-y\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{II}y=5\\\\x=13-5\end{array}\right\;\;\vee\;\;\left\{\begin{array}{II}y=8\\\\x=13-8\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{II}y=5\\\\x=8\end{array}\right\;\;\vee\;\;\left\{\begin{array}{II}y=8\\\\x=5\end{array}\right$

$Resposta:\;\;(x\;;\;y)=\{ (5\;;\;8)\;;\;(8\;;\;5)\}$

    Cálculos Auxiliares    

    $-y^2+13y-40=0$

    $y=\frac{-13\pm\sqrt{13^2-4\times(-1)\times(-40)}}{2\times(-1)}\Leftrightarrow$

$\Leftrightarrow y=\frac{-13\pm\sqrt{169+4\times(-40)}}{-2}\Leftrightarrow$

$\Leftrightarrow y=\frac{-13\pm\sqrt{169-160}}{-2}\Leftrightarrow$

$\Leftrightarrow y=\frac{-13\pm\sqrt{9}}{-2}\Leftrightarrow$

$\Leftrightarrow y=\frac{-13\pm3}{-2}\Leftrightarrow$

$\Leftrightarrow y=\frac{-13-3}{-2}\;\;\vee\;\;y=\frac{-13+3}{-2}\Leftrightarrow$

$\Leftrightarrow y=\frac{-16}{-2}\;\;\vee\;\;y=\frac{-10}{-2}\Leftrightarrow$

$\Leftrightarrow y=8\;\;\vee\;\;y=5$

b)

    $\left\{\begin{array}{II}x+y=-9\\\\x\times y=14\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{II}y=-9-x\\\\x\times y=14\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{II}y=-9-x\\\\x\times (-9-x)=14\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{II}y=-9-x\\\\-9x-x^2=14\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{II}y=-9-x\\\\x^2+9x+14=0\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{II}y=-9-x\\\\x=-7\end{array}\right\;\;\vee\;\;\left\{\begin{array}{II}y=-9-x\\\\x=-2\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{II}y=-9-(-7)\\\\x=-7\end{array}\right\;\;\vee\;\;\left\{\begin{array}{II}y=-9-(-2)\\\\x=-2\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{II}y=-9+7\\\\x=-7\end{array}\right\;\;\vee\;\;\left\{\begin{array}{II}y=-9+2\\\\x=-2\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{II}y=-2\\\\x=-7\end{array}\right\;\;\vee\;\;\left\{\begin{array}{II}y=-7\\\\x=-2\end{array}\right$

$Resposta:\;\;(x\;;\;y)=\{ (-7\;;\;-2)\;;\;(-2\;;\;-7)\}$

    Cálculos Auxiliares    

    $x^2+9x+14=0$

    $y=\frac{-9\pm\sqrt{9^2-4\times1\times14}}{2\times1}\Leftrightarrow$

$\Leftrightarrow y=\frac{-9\pm\sqrt{81-4\times14}}{2}\Leftrightarrow$

$\Leftrightarrow y=\frac{-9\pm\sqrt{81-56}}{2}\Leftrightarrow$

$\Leftrightarrow y=\frac{-9\pm\sqrt{25}}{2}\Leftrightarrow$

$\Leftrightarrow y=\frac{-9\pm5}{2}\Leftrightarrow$

$\Leftrightarrow y=\frac{-9-5}{2}\;\;\vee\;\;y=\frac{-9+5}{2}\Leftrightarrow$

$\Leftrightarrow y=\frac{-14}{2}\;\;\vee\;\;y=\frac{-4}{2}\Leftrightarrow$

$\Leftrightarrow y=-7\;\;\vee\;\;y=-2$

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