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

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

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

$\Leftrightarrow\begin{cases}-4y-2z+246+y+3z=19\\\\x=-2y-z+123\\\\-2y-z+123-y+z=7\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}-3y+z=19-246\\\\x=-2y-z+123\\\\-3y=7-123\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}-3y+z=-227\\\\x=-2y-z+123\\\\-3y=-116\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}-3y+z=-227\\\\x=-2y-z+123\\\\y=\dfrac{-116}{-3}\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}-3\times\dfrac{116}{3}+z=-227\\\\x=-2\times\dfrac{116}{3}-z+123\\\\y=\dfrac{116}{3}\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}-116+z=-227\\\\x=-\dfrac{232}{3}-z+123\\\\y=\dfrac{116}{3}\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}z=-227+116\\\\x=-\dfrac{232}{3}-z+123\\\\y=\dfrac{116}{3}\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}z=-111\\\\x=-\dfrac{232}{3}-(-111)+123\\\\y=\dfrac{116}{3}\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}z=-111\\\\x=-\dfrac{232}{3}+111+123\\\\y=\dfrac{116}{3}\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}z=-111\\\\x=-\dfrac{232}{3}+234\\\\y=\dfrac{116}{3}\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}z=-111\\\\x=-\dfrac{232}{3}+\dfrac{234\times3}{3}\\\\y=\dfrac{116}{3}\end{cases}\Leftrightarrow$

$\Leftrightarrow\begin{cases}z=-111\\\\x=-\dfrac{232}{3}+\dfrac{702}{3}\\\\y=\dfrac{116}{3}\end{cases}\Leftrightarrow$

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

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

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