Question

Ayúdenme por favor , si no saben no contesten , reporto

Answer 1

Respuesta:

Muy complicado... pero para mi no >;3

Explicación paso a paso:

$\begin{bmatrix}5x+3y+\left(m-1\right)z=2\\ 3x+5y+\left(m-1\right)z=2\\ \left(m-1\right)x+5y+3z=2\end{bmatrix}\\\\\\\\\mathrm{Sustituir\:}x=\frac{2-3y-mz+z}{5}\\\\\\\\\begin{bmatrix}3\cdot \frac{2-3y-mz+z}{5}+5y+\left(m-1\right)z=2\\ \left(m-1\right)\frac{2-3y-mz+z}{5}+5y+3z=2\end{bmatrix}\\\\\\\\\begin{bmatrix}\frac{2mz-2z+16y+6}{5}=2\\ \frac{-m^2z+2mz+14z+2m-3my+28y-2}{5}=2\end{bmatrix}\\\\\\\\\mathrm{Sustituir\:}z=\frac{2\left(-4y+1\right)}{-1+m}$

$\begin{bmatrix}\frac{-m^2\frac{2\left(-4y+1\right)}{-1+m}+2m\frac{2\left(-4y+1\right)}{-1+m}+14\cdot \frac{2\left(-4y+1\right)}{-1+m}+2m-3my+28y-2}{5}=2\end{bmatrix}\\\\\\\\\begin{bmatrix}\frac{m^2y+3my-28y+6}{-1+m}=2\end{bmatrix}\\\\\\\\\mathrm{Para\:}z=\frac{2\left(-4y+1\right)}{-1+m}\\\\\\\\\mathrm{Sustituir\:}y=\frac{2}{m+7}\\\\\\\\z=\frac{2\left(-4\cdot \frac{2}{m+7}+1\right)}{-1+m}\\\\\\\\z=\frac{2}{m+7}\\\\\\\\\mathrm{Para\:}x=\frac{2-3y-mz+z}{5}$

$\mathrm{Sustituir\:}z=\frac{2}{m+7},\:y=\frac{2}{m+7}\\\\\\\\x=\frac{2-3\cdot \frac{2}{m+7}-m\frac{2}{m+7}+\frac{2}{m+7}}{5}\\\\\\\\x=\frac{2}{m+7}\\\\\\\\Las\:soluciones\:para\:el\:sistema\:de\:ecuaciones\:son:\\\\\\\\x=\frac{2}{m+7},\:z=\frac{2}{m+7},\:y=\frac{2}{m+7},\:\quad \:m\ne \:-7,\:m\ne \:4,\:m\ne \:1$

$CORONA\:\:\:\:\:\: >:c$

Answer 2

Respuesta:

haslo por sustitucion