Question

O valor de x no sistema linear a seguir é: 2x + y + 3z = 19 x + 2y + z = 123 x − y + z = 7

Answer 1

    $\begin{cases}2x+y+3z=19\\\\x+2y+z=123\\\\x-y+z=7\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}y=19-2x-3z\\\\z=123-x-2y\\\\x=7+y-z\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}y=19-2\times(7+y-z)-3z\\\\z=123-(7+y-z)-2y\\\\x=7+y-z\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}y=19-14-2y+2z-3z\\\\z=123-7-y+z-2y\\\\x=7+y-z\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}y+2y=5-z\\\\z-z=116-3y\\\\x=7+y-z\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}3y=5-z\\\\0=116-3y\\\\x=7+y-z\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}z=5-3y\\\\3y=116\\\\x=7+y-z\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}z=5-3\times\dfrac{116}{3}\\\\y=\dfrac{116}{3}\\\\x=7+\dfrac{116}{3}-z\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}z=5-116\\\\y=\dfrac{116}{3}\\\\x=\dfrac{7\times3}{3}+\dfrac{116}{3}-z\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}z=-111\\\\y=\dfrac{116}{3}\\\\x=\dfrac{21}{3}+\dfrac{116}{3}-(-111)\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}z=-111\\\\y=\dfrac{116}{3}\\\\x=\dfrac{137}{3}+111\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}z=-111\\\\y=\dfrac{116}{3}\\\\x=\dfrac{137}{3}+\dfrac{111\times3}{3}\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}z=-111\\\\y=\dfrac{116}{3}\\\\x=\dfrac{137}{3}+\dfrac{333}{3}\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}z=-111\\\\y=\dfrac{116}{3}\\\\x=\dfrac{470}{3}\end{cases}$

Resposta: O valor de x é  $\frac{470}{3}$.

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