Question

Resolva os sistemas abaixo 2x+y+z=3 -2x+2y-z=0 3x+y+z=1

Answer 1

    $\left\{\begin{array}{llI}2x+y+z=3\\-2x+2y-z=0\\3x+y+z=1\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{llI}z=3-2x-y\\-2x+2y-z=0\\3x+y+z=1\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{llI}z=3-2x-y\\-2x+2y-(3-2x-y)=0\\3x+y+(3-2x-y)=1\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{llI}z=3-2x-y\\-2x+2y-3+2x+y=0\\3x+y+3-2x-y=1\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{llI}z=3-2x-y\\3y-3=0\\x+3=1\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{llI}z=3-2x-y\\3y=3\\x=1-3\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{llI}z=3-2x-y\\y=1\\x=-2\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{llI}z=3-2\times(-2)-1\\y=1\\x=-2\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{llI}z=3-(-4)-1\\y=1\\x=-2\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{llI}z=3+4-1\\y=1\\x=-2\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{llI}z=7-1\\y=1\\x=-2\end{array}\right\Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{llI}z=6\\y=1\\x=-2\end{array}\right$

Resposta:  $(x\;;\;y\;;\;z)=(-2\;;\;1\;;\;6)$