Respuesta:
[tex]y=-2,\:z=\frac{1}{2},\:x=\frac{3}{2}[/tex]
Explicación paso a paso:
[tex]\begin{bmatrix}2y+3x+z=1\\ 5x+3y+3z=3\\ x+y+z=0\end{bmatrix}[/tex]
[tex]\mathrm{Sustituir\:}y=\frac{1-3x-z}{2}[/tex]
[tex]\begin{bmatrix}5x+3\cdot \frac{1-3x-z}{2}+3z=3\\ x+\frac{1-3x-z}{2}+z=0\end{bmatrix}[/tex]
[tex]\begin{bmatrix}\frac{x+3z+3}{2}=3\\ \frac{-x+z+1}{2}=0\end{bmatrix}[/tex]
[tex]\mathrm{Sustituir\:}x=-3z+3[/tex]
[tex]\begin{bmatrix}\frac{-\left(-3z+3\right)+z+1}{2}=0\end{bmatrix}[/tex]
[tex]\begin{bmatrix}2z-1=0\end{bmatrix}[/tex]
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Respuesta:
[tex]y=-2,\:z=\frac{1}{2},\:x=\frac{3}{2}[/tex]
Explicación paso a paso:
[tex]\begin{bmatrix}2y+3x+z=1\\ 5x+3y+3z=3\\ x+y+z=0\end{bmatrix}[/tex]
[tex]\mathrm{Sustituir\:}y=\frac{1-3x-z}{2}[/tex]
[tex]\begin{bmatrix}5x+3\cdot \frac{1-3x-z}{2}+3z=3\\ x+\frac{1-3x-z}{2}+z=0\end{bmatrix}[/tex]
[tex]\begin{bmatrix}\frac{x+3z+3}{2}=3\\ \frac{-x+z+1}{2}=0\end{bmatrix}[/tex]
[tex]\mathrm{Sustituir\:}x=-3z+3[/tex]
[tex]\begin{bmatrix}\frac{-\left(-3z+3\right)+z+1}{2}=0\end{bmatrix}[/tex]
[tex]\begin{bmatrix}2z-1=0\end{bmatrix}[/tex]