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Resposta:

"Regra de Cramer"

[tex]x+y+z=7\\ 2x+y+z=6\\ x+2y+z=8\\ \\ D=\\ \\ \left[\begin{array}{ccccc}1&1&1&1&1\\2&1&1&2&1\\1&2&1&1&2\end{array}\right] \\ \\ (1+1+4)-(1+2+2)=6-5=1\\\\ \\ D=1\\ \\ \\ x=\frac{Dx}{D} \\ \\ Dx=\\ \\ \left[\begin{array}{ccccc}7&1&1&7&1\\6&1&1&6&1\\8&2&1&8&2\end{array}\right] \\ \\ (7+8+12)-(8+14+6)=27-28=-1\\ \\ \\ Dx=-1[/tex]

[tex]Dy=\\ \\ \left[\begin{array}{ccccc}1&7&1&1&7\\2&6&1&2&6\\1&8&1&1&8\end{array}\right] \\ \\ (6+7+16)-(6+8+14)=29-28=1\\ \\ \\ Dy=1\\ \\ \\ Dz=\\ \\ \left[\begin{array}{ccccc}1&1&7&1&1\\2&1&6&2&1\\1&2&8&1&2\end{array}\right] \\ \\ (8+6+28)-(7+12+16)=42-35=7\\ \\ \\ Dz=7[/tex]

[tex]x=\frac{Dx}{D} =\frac{-1}{1} =-1[/tex]

[tex]y=\frac{Dy}{D} =\frac{1}{1} =1[/tex]

[tex]z=\frac{Dz}{D} =\frac{7}{1} =7[/tex]

[tex]S~~\left \{ {{x=-1,~~y=1,~~} {z=7}} \}[/tex]